11.7 Polar Form of Complex Numbers

In this section, we return to our study of complex numbers which were first introduced in Section. Recall that a complex number is a number of the form $z = a + b i$ where $a$ and $b$ are real numbers and $i$ is the imaginary unit defined by $i = \sqrt{- 1}$ . The number $a$ is called the real part of $z$ , denoted $Re (z)$ , while the real number $b$ is called the imaginary part of $z$ , denoted $Im (z)$ . From Intermediate Algebra, we know that if $z = a + b i = c + d i$ where $a$ , $b$ , $c$ and $d$ are real numbers, then $a = c$ and $b = d$ , which means $Re (z)$ and $Im (z)$ are well-defined.¹ To start off this section, we associate each complex number $z = a + b i$ with the point $(a, b)$ on the coordinate plane. In this case, the $x$ -axis is relabeled as the real axis, which corresponds to the real number line as usual, and the $y$ -axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit $i$ . The plane determined by these two axes is called the complex plane.

Coordinate-plane figure. — **Figure 11.251**

The Complex Plane

Since the ordered pair $(a, b)$ gives the rectangular coordinates associated with the complex number $z = a + b i$ , the expression $z = a + b i$ is called the rectangular form of $z$ . Of course, we could just as easily associate $z$ with a pair of polar coordinates $(r, θ)$ . Although it is not as straightforward as the definitions of $Re (z)$ and $Im (z)$ , we can still give $r$ and $θ$ special names in relation to $z$ .

Definition

The Modulus and Argument of Complex Numbers: Let $z = a + b i$ be a complex number with $a = Re (z)$ and $b = Im (z)$ . Let $(r, θ)$ be a polar representation of the point with rectangular coordinates $(a, b)$ where $r greater than or equal to 0$ .

The modulus of $z$ , denoted $| z |$ , is defined by $| z | = r$ .
The angle $θ$ is an argument of $z$ . The set of all arguments of $z$ is denoted $arg (z)$ .
If $z not equal to 0$ and $- π < θ \leq π$ , then $θ$ is the principal argument of $z$ , written $θ = Arg (z)$ .

Some remarks about Definition are in order. We know from Section that every point in the plane has infinitely many polar coordinate representations $(r, θ)$ which means it's worth our time to make sure the quantities `modulus', `argument' and `principal argument' are well-defined. Concerning the modulus, if $z equals 0$ then the point associated with $z$ is the origin. In this case, the only $r$ -value which can be used here is $r equals 0$ . Hence for $z equals 0$ , $vertical bar z vertical bar equals 0$ is well-defined. If $z not equal to 0$ , then the point associated with $z$ is not the origin, and there are two possibilities for $r$ : one positive and one negative. However, we stipulated $r greater than or equal to 0$ in our definition so this pins down the value of $| z |$ to one and only one number. Thus the modulus is well-defined in this case, too.² Even with the requirement $r greater than or equal to 0$ , there are infinitely many angles $θ$ which can be used in a polar representation of a point $(r, θ)$ . If $z not equal to 0$ then the point in question is not the origin, so all of these angles $θ$ are coterminal. Since coterminal angles are exactly $2 ⁢ π$ radians apart, we are guaranteed that only one of them lies in the interval $(- π, π]$ , and this angle is what we call the principal argument of $z$ , $Arg (z)$ . In fact, the set $arg (z)$ of all arguments of $z$ can be described using set-builder notation as $arg (z) = {Arg (z) + 2 π k | k is an integer}$ . Note that since $arg (z)$ is a set, we will write ` $θ ∈ arg ⁢ open parenthesis z close parenthesis$ ' to mean ` $θ$ is in³ the set of arguments of $z$ '. If $z equals 0$ then the point in question is the origin, which we know can be represented in polar coordinates as $open parenthesis 0, θ close parenthesis$ for any angle $θ$ . In this case, we have $arg (0) = (- \infty, \infty)$ and since there is no one value of $θ$ which lies $(- π, π]$ , we leave $Arg ⁢ open parenthesis 0 close parenthesis$ undefined.⁴ It is time for an example.

Example 1

For each of the following complex numbers find $Re (z)$ , $Im (z)$ , $| z |$ , $arg (z)$ and $Arg (z)$ . Plot $z$ in the complex plane.

$z = \sqrt{3} - i$
$z = - 2 + 4 i$
$z equals 3 ⁢ i$
$z = - 117$

For $z = \sqrt{3} - i = \sqrt{3} + (- 1) i$ , we have $Re (z) = \sqrt{3}$ and $Im (z) = - 1$ . To find $| z |$ , $arg (z)$ and $Arg (z)$ , we need to find a polar representation $(r, θ)$ with $r greater than or equal to 0$ for the point $P (\sqrt{3}, - 1)$ associated with $z$ . We know $r^{2} = {(\sqrt{3})}^{2} + {(- 1)}^{2} = 4$ , so $r = \pm 2$ . Since we require $r greater than or equal to 0$ , we choose $r equals 2$ , so $vertical bar z vertical bar equals 2$ . Next, we find a corresponding angle $θ$ . Since $r greater than 0$ and $P$ lies in Quadrant IV, $θ$ is a Quadrant IV angle. We know $\tan (θ) = \frac{- 1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$ , so $θ = - \frac{π}{6} + 2 π k$ for integers $k$ . Hence, $arg (z) = {- \frac{π}{6} + 2 π k | k is an integer}$ . Of these values, only $θ = - \frac{π}{6}$ satisfies the requirement that $- π < θ \leq π$ , hence $Arg (z) = - \frac{π}{6}$ .
The complex number $z = - 2 + 4 i$ has $Re (z) = - 2$ , $Im ⁢ open parenthesis z close parenthesis equals 4$ , and is associated with the point $P ⁢ open parenthesis minus 2, 4 close parenthesis$ . Our next task is to find a polar representation $(r, θ)$ for $P$ where $r greater than or equal to 0$ . Running through the usual calculations gives $r = 2 \sqrt{5}$ , so $| z | = 2 \sqrt{5}$ . To find $θ$ , we get $\tan (θ) = - 2$ , and since $r greater than 0$ and $P$ lies in Quadrant II, we know $θ$ is a Quadrant II angle. We find $θ = π + \arctan (- 2) + 2 π k$ , or, more succinctly $θ = π - \arctan (2) + 2 π k$ for integers $k$ . Hence $arg (z) = {π - \arctan (2) + 2 π k | k is an integer}$ . Only $θ = π - \arctan (2)$ satisfies the requirement $- π < θ \leq π$ , so $Arg (z) = π - \arctan (2)$ .
We rewrite $z equals 3 ⁢ i$ as $z = 0 + 3 i$ to find $Re ⁢ open parenthesis z close parenthesis equals 0$ and $Im ⁢ open parenthesis z close parenthesis equals 3$ . The point in the plane which corresponds to $z$ is $open parenthesis 0, 3 close parenthesis$ and while we could go through the usual calculations to find the required polar form of this point, we can almost `see' the answer. The point $open parenthesis 0, 3 close parenthesis$ lies $3$ units away from the origin on the positive $y$ -axis. Hence, $r = | z | = 3$ and $θ = \frac{π}{2} + 2 π k$ for integers $k$ . We get $arg (z) = {\frac{π}{2} + 2 π k | k is an integer}$ and $Arg (z) = \frac{π}{2}$ .
As in the previous problem, we write $z = - 117 = - 117 + 0 i$ so $Re (z) = - 117$ and $Im ⁢ open parenthesis z close parenthesis equals 0$ . The number $z = - 117$ corresponds to the point $open parenthesis minus 117, 0 close parenthesis$ , and this is another instance where we can determine the polar form `by eye'. The point $open parenthesis minus 117, 0 close parenthesis$ is $117$ units away from the origin along the negative $x$ -axis. Hence, $r = | z | = 117$ and $θ = π + 2 π = (2 k + 1) π k$ for integers $k$ . We have $arg (z) = {(2 k + 1) π | k is an integer}$ . Only one of these values, $θ = π$ , just barely lies in the interval $(- π, π]$ which means and $Arg (z) = π$ . We plot $z$ along with the other numbers in this example below.
Figure 11.252

Now that we've had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem.

Theorem

Properties of the Modulus: Let $z$ and $w$ be complex numbers.

$| z |$ is the distance from $z$ to $0$ in the complex plane
$vertical bar z vertical bar greater than or equal to 0$ and $vertical bar z vertical bar equals 0$ if and only if $z equals 0$
$| z | = \sqrt{Re {(z)}^{2} + Im {(z)}^{2}}$
Product Rule: $| z w | = | z | | w |$
Power Rule: $| z^{n} | = {| z |}^{n}$ for all natural numbers, $n$
Quotient Rule: $| \frac{z}{w} | = \frac{| z |}{| w |}$ , provided $w not equal to 0$

To prove the first three properties in Theorem, suppose $z = a + b i$ where $a$ and $b$ are real numbers. To determine $| z |$ , we find a polar representation $(r, θ)$ with $r greater than or equal to 0$ for the point $(a, b)$ . From Section, we know $r^{2} = a^{2} + b^{2}$ so that $r = \pm \sqrt{a^{2} + b^{2}}$ . Since we require $r greater than or equal to 0$ , then it must be that $r = \sqrt{a^{2} + b^{2}}$ , which means $| z | = \sqrt{a^{2} + b^{2}}$ . Using the distance formula, we find the distance from $open parenthesis 0, 0 close parenthesis$ to $(a, b)$ is also $\sqrt{a^{2} + b^{2}}$ , establishing the first property.⁵ For the second property, note that since $| z |$ is a distance, $vertical bar z vertical bar greater than or equal to 0$ . Furthermore, $vertical bar z vertical bar equals 0$ if and only if the distance from $z$ to $0$ is $0$ , and the latter happens if and only if $z equals 0$ , which is what we were asked to show.⁶ For the third property, we note that since $a = Re (z)$ and $b = Im (z)$ , $z = \sqrt{a^{2} + b^{2}} = \sqrt{Re {(z)}^{2} + Im {(z)}^{2}}$ .

To prove the product rule, suppose $z = a + b i$ and $w = c + d i$ for real numbers $a$ , $b$ , $c$ and $d$ . Then $z w = (a + b i) (c + d i)$ . After the usual arithmetic⁷ we get $z w = (a c - b d) + (a d + b c) i$ . Therefore,

$\begin{matrix} | z w | & = & \sqrt{{(a c - b d)}^{2} + {(a d + b c)}^{2}} \\ = & \sqrt{a^{2} c^{2} - 2 a b c d + b^{2} d^{2} + a^{2} d^{2} + 2 a b c d + b^{2} c^{2}} & Expand \\ = & \sqrt{a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2}} & Rearrange terms \\ = & \sqrt{a^{2} (c^{2} + d^{2}) + b^{2} (c^{2} + d^{2})} & Factor \\ = & \sqrt{(a^{2} + b^{2}) (c^{2} + d^{2})} & Factor \\ = & \sqrt{a^{2} + b^{2}} \sqrt{c^{2} + d^{2}} & Product Rule for Radicals \\ = & | z | | w | & Definition of | z | and | w | \end{matrix}$

Hence $| z w | = | z | | w |$ as required.

Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction⁸ to prove the power rule. Let $P (n)$ be the statement $| z^{n} | = {| z |}^{n}$ . Then $P ⁢ open parenthesis 1 close parenthesis$ is true since $| z^{1} | = | z | = {| z |}^{1}$ . Next, assume $P (k)$ is true. That is, assume $| z^{k} | = {| z |}^{k}$ for some $k greater than or equal to 1$ . Our job is to show that $P ⁢ open parenthesis k plus 1 close parenthesis$ is true, namely $| z^{k + 1} | = {| z |}^{k + 1}$ . As is customary with induction proofs, we first try to reduce the problem in such a way as to use the Induction Hypothesis.

$\begin{matrix} | z^{k + 1} | & = & | z^{k} z | & Properties of Exponents \\ = & | z^{k} | | z | & Product Rule \\ = & {| z |}^{k} | z | & Induction Hypothesis \\ = & {| z |}^{k + 1} & Properties of Exponents \end{matrix}$

Hence, $P ⁢ open parenthesis k plus 1 close parenthesis$ is true, which means $| z^{n} | = {| z |}^{n}$ is true for all natural numbers $n$ .

Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume $w not equal to 0$ (so $vertical bar w vertical bar not equal to 0$ ) and we get

$\begin{matrix} | \frac{z}{w} | & = & | (z) (\frac{1}{w}) | \\ = & | z | | \frac{1}{w} | & Product Rule. \end{matrix}$

Hence, the proof really boils down to showing $| \frac{1}{w} | = \frac{1}{| w |}$ . This is left as an exercise.

Next, we characterize the argument of a complex number in terms of its real and imaginary parts.

Theorem

Properties of the Argument: Let $z$ be a complex number.

If $Re ⁢ open parenthesis z close parenthesis not equal to 0$ and $θ ∈ arg ⁢ open parenthesis z close parenthesis$ , then $\tan (θ) = \frac{Im (z)}{Re (z)}$ .
If $Re ⁢ open parenthesis z close parenthesis equals 0$ and $Im ⁢ open parenthesis z close parenthesis greater than 0$ , then $arg (z) = {\frac{π}{2} + 2 π k | k is an integer}$ .
If $Re ⁢ open parenthesis z close parenthesis equals 0$ and $Im ⁢ open parenthesis z close parenthesis less than 0$ , then $arg (z) = {- \frac{π}{2} + 2 π k | k is an integer}$ .
If $Re (z) = Im (z) = 0$ , then $z equals 0$ and $arg (z) = (- \infty, \infty)$ .

To prove Theorem, suppose $z = a + b i$ for real numbers $a$ and $b$ . By definition, $a = Re (z)$ and $b = Im (z)$ , so the point associated with $z$ is $(a, b) = (Re (z), Im (z))$ . From Section, we know that if $(r, θ)$ is a polar representation for $(Re (z), Im (z))$ , then $\tan (θ) = \frac{Im (z)}{Re (z)}$ , provided $Re ⁢ open parenthesis z close parenthesis not equal to 0$ . If $Re ⁢ open parenthesis z close parenthesis equals 0$ and $Im ⁢ open parenthesis z close parenthesis greater than 0$ , then $z$ lies on the positive imaginary axis. Since we take $r greater than 0$ , we have that $θ$ is coterminal with $\frac{π}{2}$ , and the result follows. If $Re ⁢ open parenthesis z close parenthesis equals 0$ and $Im ⁢ open parenthesis z close parenthesis less than 0$ , then $z$ lies on the negative imaginary axis, and a similar argument shows $θ$ is coterminal with $- \frac{π}{2}$ . The last property in the theorem was already discussed in the remarks following Definition.

Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end, consider the figure below.

Polar coordinates, $(r, θ)$ associated with $z = a + b i$ with $r greater than or equal to 0$ .

We know from Theorem that $a = r \cos (θ)$ and $b = r \sin (θ)$ . Making these substitutions for $a$ and $b$ gives $z = a + b i = r \cos (θ) + r \sin (θ) i = r [\cos (θ) + i \sin (θ)]$ . The expression ` $\cos (θ) + i \sin (θ)$ ' is abbreviated $cis (θ)$ so we can write $z = r cis (θ)$ . Since $r = | z |$ and $θ ∈ arg ⁢ open parenthesis z close parenthesis$ , we get

Definition

A Polar Form of a Complex Number: Suppose $z$ is a complex number and $θ ∈ arg ⁢ open parenthesis z close parenthesis$ . The expression:

$| z | cis (θ) = | z | [\cos (θ) + i \sin (θ)]$

is called a polar form for $z$ .

Since there are infinitely many choices for $θ ∈ arg ⁢ open parenthesis z close parenthesis$ , there infinitely many polar forms for $z$ , so we used the indefinite article `a' in Definition. It is time for an example.

Example 2

Find the rectangular form of the following complex numbers. Find $Re (z)$ and $Im (z)$ .
1. $z = 4 cis (\frac{2 π}{3})$
2. $z = 2 cis (- \frac{3 π}{4})$
3. $z equals 3 ⁢ cis ⁢ open parenthesis 0 close parenthesis$
4. $z = cis (\frac{π}{2})$
Use the results from Example Example 1 to find a polar form of the following complex numbers.
1. $z = \sqrt{3} - i$
2. $z = - 2 + 4 i$
3. $z equals 3 ⁢ i$
4. $z = - 117$

The key to this problem is to write out $cis (θ)$ as $\cos (θ) + i \sin (θ)$ .
1. By definition, $z = 4 cis (\frac{2 π}{3}) = 4 [\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3})]$ . After some simplifying, we get $z = - 2 + 2 i \sqrt{3}$ , so that $Re (z) = - 2$ and $Im (z) = 2 \sqrt{3}$ .
2. Expanding, we get $z = 2 cis (- \frac{3 π}{4}) = 2 [\cos (- \frac{3 π}{4}) + i \sin (- \frac{3 π}{4})]$ . From this, we find $z = - \sqrt{2} - i \sqrt{2}$ , so $Re (z) = - \sqrt{2} = Im (z)$ .
3. We get $z = 3 cis (0) = 3 [\cos (0) + i \sin (0)] = 3$ . Writing $3 = 3 + 0 i$ , we get $Re ⁢ open parenthesis z close parenthesis equals 3$ and $Im ⁢ open parenthesis z close parenthesis equals 0$ , which makes sense seeing as $3$ is a real number.
4. Lastly, we have $z = cis (\frac{π}{2}) = \cos (\frac{π}{2}) + i \sin (\frac{π}{2}) = i$ . Since $i = 0 + 1 i$ , we get $Re ⁢ open parenthesis z close parenthesis equals 0$ and $Im ⁢ open parenthesis z close parenthesis equals 1$ . Since $i$ is called the `imaginary unit,' these answers make perfect sense.
To write a polar form of a complex number $z$ , we need two pieces of information: the modulus $| z |$ and an argument (not necessarily the principal argument) of $z$ . We shamelessly mine our solution to Example Example 1 to find what we need.
1. For $z = \sqrt{3} - i$ , $vertical bar z vertical bar equals 2$ and $θ = - \frac{π}{6}$ , so $z = 2 cis (- \frac{π}{6})$ . We can check our answer by converting it back to rectangular form to see that it simplifies to $z = \sqrt{3} - i$ .
2. For $z = - 2 + 4 i$ , $| z | = 2 \sqrt{5}$ and $θ = π - \arctan (2)$ . Hence, $z = 2 \sqrt{5} cis (π - \arctan (2))$ . It is a good exercise to actually show that this polar form reduces to $z = - 2 + 4 i$ .
3. For $z equals 3 ⁢ i$ , $vertical bar z vertical bar equals 3$ and $θ = \frac{π}{2}$ . In this case, $z = 3 cis (\frac{π}{2})$ . This can be checked geometrically. Head out $3$ units from $0$ along the positive real axis. Rotating $\frac{π}{2}$ radians counter-clockwise lands you exactly $3$ units above $0$ on the imaginary axis at $z equals 3 ⁢ i$ .
4. Last but not least, for $z = - 117$ , $vertical bar z vertical bar equals 117$ and $θ = π$ . We get $z equals 117 ⁢ cis ⁢ open parenthesis π close parenthesis$ . As with the previous problem, our answer is easily checked geometrically.

The following theorem summarizes the advantages of working with complex numbers in polar form.

Theorem

Products, Powers and Quotients Complex Numbers in Polar Form: Suppose $z$ and $w$ are complex numbers with polar forms $z = | z | cis (α)$ and $w = | w | cis (β)$ . Then

Product Rule: $z w = | z | | w | cis (α + β)$
Power Rule ( DeMoivre's Theorem ) : $z^{n} = {| z |}^{n} cis (n θ)$ for every natural number $n$
Quotient Rule: $\frac{z}{w} = \frac{| z |}{| w |} cis (α - β)$ , provided $vertical bar w vertical bar not equal to 0$

The proof of Theorem requires a healthy mix of definition, arithmetic and identities. We first start with the product rule.

$\begin{matrix} z w & = & [| z | cis (α)] [| w | cis (β)] \\ = & | z | | w | [\cos (α) + i \sin (α)] [\cos (β) + i \sin (β)] \end{matrix}$

We now focus on the quantity in brackets on the right hand side of the equation.

$\begin{matrix} [\cos (α) + i \sin (α)] [\cos (β) + i \sin (β)] & = & \cos (α) \cos (β) + i \cos (α) \sin (β) \\ + i \sin (α) \cos (β) + i^{2} \sin (α) \sin (β) \\ = & \cos (α) \cos (β) + i^{2} \sin (α) \sin (β) & Rearranging terms \\ + i \sin (α) \cos (β) + i \cos (α) \sin (β) \\ = & (\cos (α) \cos (β) - \sin (α) \sin (β)) & Since i^{2} = - 1 \\ + i (\sin (α) \cos (β) + \cos (α) \sin (β)) & Factor out i \\ = & \cos (α + β) + i \sin (α + β) & Sum identities \\ = & cis (α + β) & Definition of ‘cis’ \end{matrix}$

Putting this together with our earlier work, we get $z w = | z | | w | cis (α + β)$ , as required.

Moving right along, we next take aim at the Power Rule, better known as DeMoivre's Theorem.⁹ We proceed by induction on $n$ . Let $P (n)$ be the sentence $z^{n} = {| z |}^{n} cis (n θ)$ . Then $P ⁢ open parenthesis 1 close parenthesis$ is true, since $z^{1} = z = | z | cis (θ) = {| z |}^{1} cis (1 \cdot θ)$ . We now assume $P (k)$ is true, that is, we assume $z^{k} = {| z |}^{k} cis (k θ)$ for some $k greater than or equal to 1$ . Our goal is to show that $P ⁢ open parenthesis k plus 1 close parenthesis$ is true, or that $z^{k + 1} = {| z |}^{k + 1} cis ((k + 1) θ)$ . We have

$\begin{matrix} z^{k + 1} & = & z^{k} z & Properties of Exponents \\ = & ({| z |}^{k} cis (k θ)) (| z | cis (θ)) & Induction Hypothesis \\ = & ({| z |}^{k} | z |) cis (k θ + θ) & Product Rule \\ = & {| z |}^{k + 1} cis ((k + 1) θ) \end{matrix}$

Hence, assuming $P (k)$ is true, we have that $P ⁢ open parenthesis k plus 1 close parenthesis$ is true, so by the Principle of Mathematical Induction, $z^{n} = {| z |}^{n} cis (n θ)$ for all natural numbers $n$ .

The last property in Theorem to prove is the quotient rule. Assuming $vertical bar w vertical bar not equal to 0$ we have

$\begin{matrix} \frac{z}{w} & = & \frac{| z | cis (α)}{| w | cis (β)} \\ = & (\frac{| z |}{| w |}) \frac{\cos (α) + i \sin (α)}{\cos (β) + i \sin (β)} \end{matrix}$

Next, we multiply both the numerator and denominator of the right hand side by $(\cos (β) - i \sin (β))$ which is the complex conjugate of $(\cos (β) + i \sin (β))$ to get

$\begin{matrix} \frac{z}{w} & = & (\frac{| z |}{| w |}) \frac{\cos (α) + i \sin (α)}{\cos (β) + i \sin (β)} \cdot \frac{\cos (β) - i \sin (β)}{\cos (β) - i \sin (β)} \end{matrix}$

If we let the numerator be $N = [\cos (α) + i \sin (α)] [\cos (β) - i \sin (β)]$ and simplify we get

$\begin{matrix} N & = & [\cos (α) + i \sin (α)] [\cos (β) - i \sin (β)] \\ = & \cos (α) \cos (β) - i \cos (α) \sin (β) + i \sin (α) \cos (β) - i^{2} \sin (α) \sin (β) & Expand \\ = & [\cos (α) \cos (β) + \sin (α) \sin (β)] + i [\sin (α) \cos (β) - \cos (α) \sin (β)] & Rearrange and Factor \\ = & \cos (α - β) + i \sin (α - β) & Difference Identities \\ = & cis (α - β) & Definition of ‘cis’ \end{matrix}$

If we call the denominator $D$ then we get

$\begin{matrix} D & = & [\cos (β) + i \sin (β)] [\cos (β) - i \sin (β)] \\ = & \cos^{2} (β) - i \cos (β) \sin (β) + i \cos (β) \sin (β) - i^{2} \sin^{2} (β) & Expand \\ = & \cos^{2} (β) - i^{2} \sin^{2} (β) & Simplify \\ = & \cos^{2} (β) + \sin^{2} (β) & Again, i^{2} = - 1 \\ = & 1 & Pythagorean Identity \end{matrix}$

Putting it all together, we get

$\begin{matrix} \frac{z}{w} & = & (\frac{| z |}{| w |}) \frac{\cos (α) + i \sin (α)}{\cos (β) + i \sin (β)} \cdot \frac{\cos (β) - i \sin (β)}{\cos (β) - i \sin (β)} \\ = & (\frac{| z |}{| w |}) \frac{cis (α - β)}{1} \\ = & \frac{| z |}{| w |} cis (α - β) \end{matrix}$

and we are done. The next example makes good use of Theorem.

Example 3

Let $z = 2 \sqrt{3} + 2 i$ and $w = - 1 + i \sqrt{3}$ . Use Theorem to find the following.

$z w$
$w to the power 5$
$\frac{z}{w}$

Write your final answers in rectangular form.

In order to use Theorem, we need to write $z$ and $w$ in polar form. For $z = 2 \sqrt{3} + 2 i$ , we find $| z | = \sqrt{{(2 \sqrt{3})}^{2} + {(2)}^{2}} = \sqrt{16} = 4$ . If $θ ∈ arg ⁢ open parenthesis z close parenthesis$ , we know $\tan (θ) = \frac{Im (z)}{Re (z)} = \frac{2}{2 \sqrt{3}} = \frac{\sqrt{3}}{3}$ . Since $z$ lies in Quadrant I, we have $θ = \frac{π}{6} + 2 π k$ for integers $k$ . Hence, $z = 4 cis (\frac{π}{6})$ . For $w = - 1 + i \sqrt{3}$ , we have $| w | = \sqrt{{(- 1)}^{2} + {(\sqrt{3})}^{2}} = 2$ . For an argument $θ$ of $w$ , we have $\tan (θ) = \frac{\sqrt{3}}{- 1} = - \sqrt{3}$ . Since $w$ lies in Quadrant II, $θ = \frac{2 π}{3} + 2 π k$ for integers $k$ and $w = 2 cis (\frac{2 π}{3})$ . We can now proceed.

We get $z w = (4 cis (\frac{π}{6})) (2 cis (\frac{2 π}{3})) = 8 cis (\frac{π}{6} + \frac{2 π}{3}) = 8 cis (\frac{5 π}{6}) = 8 [\cos (\frac{5 π}{6}) + i \sin (\frac{5 π}{6})]$ . After simplifying, we get $z w = - 4 \sqrt{3} + 4 i$ .
We use DeMoivre's Theorem which yields $w^{5} = {[2 cis (\frac{2 π}{3})]}^{5} = 2^{5} cis (5 \cdot \frac{2 π}{3}) = 32 cis (\frac{10 π}{3})$ . Since $\frac{10 π}{3}$ is coterminal with $\frac{4 π}{3}$ , we get $w^{5} = 32 [\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3})] = - 16 - 16 i \sqrt{3}$ .
Last, but not least, we have $\frac{z}{w} = \frac{4 cis (\frac{π}{6})}{2 cis (\frac{2 π}{3})} = \frac{4}{2} cis (\frac{π}{6} - \frac{2 π}{3}) = 2 cis (- \frac{π}{2})$ . Since $- \frac{π}{2}$ is a quadrantal angle, we can `see' the rectangular form by moving out $2$ units along the positive real axis, then rotating $\frac{π}{2}$ radians clockwise to arrive at the point $2$ units below $0$ on the imaginary axis. The long and short of it is that $\frac{z}{w} = - 2 i$ .

Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers – especially if they aren't given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers $z$ and $w$ in Example Example 3 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out $z w = (2 \sqrt{3} + 2 i) (- 1 + i \sqrt{3})$ the old-fashioned way. However, Theorem pays huge dividends when computing powers of complex numbers. Consider how we computed $w to the power 5$ above and compare that to using the Binomial Theorem, Theorem, to accomplish the same feat by expanding ${(- 1 + i \sqrt{3})}^{5}$ . Division is tricky in the best of times, and we saved ourselves a lot of time and effort using Theorem to find and simplify $\frac{z}{w}$ using their polar forms as opposed to starting with $\frac{2 \sqrt{3} + 2 i}{- 1 + i \sqrt{3}}$ , rationalizing the denominator, and so forth.

There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem. Take the product rule, for instance. If $z = | z | cis (α)$ and $w = | w | cis (β)$ , the formula $z w = | z | | w | cis (α + β)$ can be viewed geometrically as a two step process. The multiplication of $| z |$ by $| w |$ can be interpreted as magnifying¹⁰ the distance $| z |$ from $z$ to $0$ , by the factor $| w |$ . Adding the argument of $w$ to the argument of $z$ can be interpreted geometrically as a rotation of $β$ radians counter-clockwise.¹¹ Focusing on $z$ and $w$ from Example Example 3, we can arrive at the product $z w$ by plotting $z$ , doubling its distance from $0$ (since $vertical bar w vertical bar equals 2$ ), and rotating $\frac{2 π}{3}$ radians counter-clockwise. The sequence of diagrams below attempts to describe this process geometrically.

Multiplying $z$ by $vertical bar w vertical bar equals 2$ .

Rotating counter-clockwise by $Arg (w) = \frac{2 π}{3}$ radians.

Visualizing $z w$ for $z = 4 cis (\frac{π}{6})$ and $w = 2 cis (\frac{2 π}{3})$ .

We may also visualize division similarly. Here, the formula $\frac{z}{w} = \frac{| z |}{| w |} cis (α - β)$ may be interpreted as shrinking¹² the distance from $0$ to $z$ by the factor $| w |$ , followed up by a clockwise ¹³ rotation of $β$ radians. In the case of $z$ and $w$ from Example Example 3, we arrive at $\frac{z}{w}$ by first halving the distance from $0$ to $z$ , then rotating clockwise $\frac{2 π}{3}$ radians.

Dividing $z$ by $vertical bar w vertical bar equals 2$ .

Rotating clockwise by $Arg (w) = \frac{2 π}{3}$ radians.

Visualizing $\frac{z}{w}$ for $z = 4 cis (\frac{π}{6})$ and $w = 2 cis (\frac{2 π}{3})$ .

Our last goal of the section is to reverse DeMoivre's Theorem to extract roots of complex numbers.

Unlike Definition in Section, we do not specify one particular prinicpal $n^{th}$ root, hence the use of the indefinite article `an' as in `an $n^{th}$ root of $z$ '. Using this definition, both $4$ and $minus 4$ are square roots of $16$ , while $the square root of 16$ means the principal square root of $16$ as in $\sqrt{16} = 4$ . Suppose we wish to find all complex third (cube) roots of $8$ . Algebraically, we are trying to solve $w cubed equals 8$ . We know that there is only one real solution to this equation, namely $w = \sqrt[3]{8} = 2$ , but if we take the time to rewrite this equation as $w^{3} - 8 = 0$ and factor, we get $(w - 2) (w^{2} + 2 w + 4) = 0$ . The quadratic factor gives two more cube roots $w = - 1 \pm i \sqrt{3}$ , for a total of three cube roots of $8$ . In accordance with Theorem, since the degree of $p (w) = w^{3} - 8$ is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of $8$ . Let us now solve this same problem using the machinery developed in this section. To do so, we express $z equals 8$ in polar form. Since $z equals 8$ lies $8$ units away on the positive real axis, we get $z equals 8 ⁢ cis ⁢ open parenthesis 0 close parenthesis$ . If we let $w = | w | cis (α)$ be a polar form of $w$ , the equation $w cubed equals 8$ becomes

$\begin{matrix} w^{3} & = & 8 \\ {(| w | cis (α))}^{3} & = & 8 cis (0) \\ {| w |}^{3} cis (3 α) & = & 8 cis (0) & DeMoivre’s Theorem \end{matrix}$

The complex number on the left hand side of the equation corresponds to the point with polar coordinates $open parenthesis vertical bar w vertical bar cubed, 3 ⁢ α close parenthesis$ , while the complex number on the right hand side corresponds to the point with polar coordinates $open parenthesis 8, 0 close parenthesis$ . Since $vertical bar w vertical bar greater than or equal to 0$ , so is ${| w |}^{3}$ , which means $open parenthesis vertical bar w vertical bar cubed, 3 ⁢ α close parenthesis$ and $open parenthesis 8, 0 close parenthesis$ are two polar representations corresponding to the same complex number, both with positive $r$ values. From Section, we know $vertical bar w vertical bar cubed equals 8$ and $3 α = 0 + 2 π k$ for integers $k$ . Since $| w |$ is a real number, we solve $vertical bar w vertical bar cubed equals 8$ by extracting the principal cube root to get $| w | = \sqrt[3]{8} = 2$ . As for $α$ , we get $α = \frac{2 π k}{3}$ for integers $k$ . This produces three distinct points with polar coordinates corresponding to $k equals 0$ , $1$ and $2$ : specifically $open parenthesis 2, 0 close parenthesis$ , $(2, \frac{2 π}{3})$ and $(2, \frac{4 π}{3})$ . These correspond to the complex numbers $w sub 0 equals 2 ⁢ cis ⁢ open parenthesis 0 close parenthesis$ , $w_{1} = 2 cis (\frac{2 π}{3})$ and $w_{2} = 2 cis (\frac{4 π}{3})$ , respectively. Writing these out in rectangular form yields $w sub 0 equals 2$ , $w_{1} = - 1 + i \sqrt{3}$ and $w_{2} = - 1 - i \sqrt{3}$ . While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of $32$ . (Try using Chapter techniques on that!) If we start with a generic complex number in polar form $z = | z | cis (θ)$ and solve $w^{n} = z$ in the same manner as above, we arrive at the following theorem.

Theorem

The $n^{th}$ roots of a Complex Number: Let $z not equal to 0$ be a complex number with polar form $z = r cis (θ)$ . For each natural number $n$ , $z$ has $n$ distinct $n^{th}$ roots, which we denote by $w sub 0$ , $w sub 1$ , …, $w sub n minus 1$ , and they are given by the formula

$w_{k} = \sqrt[n]{r} cis (\frac{θ}{n} + \frac{2 π}{n} k)$

The proof of Theorem breaks into to two parts: first, showing that each $w_{k}$ is an $n^{th}$ root, and second, showing that the set ${w_{k} | k = 0, 1, \dots, (n - 1)}$ consists of $n$ different complex numbers. To show $w_{k}$ is an $n^{th}$ root of $z$ , we use DeMoivre's Theorem to show ${(w_{k})}^{n} = z$ .

$\begin{matrix} {(w_{k})}^{n} & = & {(\sqrt[n]{r} cis (\frac{θ}{n} + \frac{2 π}{n} k))}^{n} \\ = & {(\sqrt[n]{r})}^{n} cis (n \cdot [\frac{θ}{n} + \frac{2 π}{n} k]) & DeMoivre’s Theorem \\ = & r cis (θ + 2 π k) \end{matrix}$

Since $k$ is a whole number, $\cos (θ + 2 π k) = \cos (θ)$ and $\sin (θ + 2 π k) = \sin (θ)$ . Hence, it follows that $cis (θ + 2 π k) = cis (θ)$ , so ${(w_{k})}^{n} = r cis (θ) = z$ , as required. To show that the formula in Theorem generates $n$ distinct numbers, we assume $n greater than or equal to 2$ (or else there is nothing to prove) and note that the modulus of each of the $w_{k}$ is the same, namely $\sqrt[n]{r}$ . Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal – that is, if the arguments differ by an integer multiple of $2 ⁢ π$ . Suppose $k$ and $j$ are whole numbers between $0$ and $open parenthesis n minus 1 close parenthesis$ , inclusive, with $k \neq j$ . Since $k$ and $j$ are different, let's assume for the sake of argument that $k > j$ . Then $(\frac{θ}{n} + \frac{2 π}{n} k) - (\frac{θ}{n} + \frac{2 π}{n} j) = 2 π (\frac{k - j}{n})$ . For this to be an integer multiple of $2 ⁢ π$ , $(k - j)$ must be a multiple of $n$ . But because of the restrictions on $k$ and $j$ , $0 < k - j \leq n - 1$ . (Think this through.) Hence, $(k - j)$ is a positive number less than $n$ , so it cannot be a multiple of $n$ . As a result, $w_{k}$ and $w_{j}$ are different complex numbers, and we are done. By Theorem, we know there at most $n$ distinct solutions to $w^{n} = z$ , and we have just found all of them. We illustrate Theorem in the next example.

Example 4

Use Theorem to find the following:

both square roots of $z = - 2 + 2 i \sqrt{3}$
the four fourth roots of $z = - 16$
the three cube roots of $z = \sqrt{2} + i \sqrt{2}$
the five fifth roots of $z equals 1$ .

We start by writing $z = - 2 + 2 i \sqrt{3} = 4 cis (\frac{2 π}{3})$ . To use Theorem, we identify $r equals 4$ , $θ = \frac{2 π}{3}$ and $n equals 2$ . We know that $z$ has two square roots, and in keeping with the notation in Theorem, we'll call them $w sub 0$ and $w sub 1$ . We get $w_{0} = \sqrt{4} cis (\frac{(2 π / 3)}{2} + \frac{2 π}{2} (0)) = 2 cis (\frac{π}{3})$ and $w_{1} = \sqrt{4} cis (\frac{(2 π / 3)}{2} + \frac{2 π}{2} (1)) = 2 cis (\frac{4 π}{3})$ . In rectangular form, the two square roots of $z$ are $w_{0} = 1 + i \sqrt{3}$ and $w_{1} = - 1 - i \sqrt{3}$ . We can check our answers by squaring them and showing that we get $z = - 2 + 2 i \sqrt{3}$ .
Proceeding as above, we get $z = - 16 = 16 cis (π)$ . With $r equals 16$ , $θ = π$ and $n equals 4$ , we get the four fourth roots of $z$ to be $w_{0} = \sqrt[4]{16} cis (\frac{π}{4} + \frac{2 π}{4} (0)) = 2 cis (\frac{π}{4})$ , $w_{1} = \sqrt[4]{16} cis (\frac{π}{4} + \frac{2 π}{4} (1)) = 2 cis (\frac{3 π}{4})$ , $w_{2} = \sqrt[4]{16} cis (\frac{π}{4} + \frac{2 π}{4} (2)) = 2 cis (\frac{5 π}{4})$ and $w_{3} = \sqrt[4]{16} cis (\frac{π}{4} + \frac{2 π}{4} (3)) = 2 cis (\frac{7 π}{4})$ . Converting these to rectangular form gives $w_{0} = \sqrt{2} + i \sqrt{2}$ , $w_{1} = - \sqrt{2} + i \sqrt{2}$ , $w_{2} = - \sqrt{2} - i \sqrt{2}$ and $w_{3} = \sqrt{2} - i \sqrt{2}$ .
For $z = \sqrt{2} + i \sqrt{2}$ , we have $z = 2 cis (\frac{π}{4})$ . With $r equals 2$ , $θ = \frac{π}{4}$ and $n equals 3$ the usual computations yield $w_{0} = \sqrt[3]{2} cis (\frac{π}{12})$ , $w_{1} = \sqrt[3]{2} cis (\frac{9 π}{12}) = \sqrt[3]{2} cis (\frac{3 π}{4})$ and $w_{2} = \sqrt[3]{2} cis (\frac{17 π}{12})$ . If we were to convert these to rectangular form, we would need to use either the Sum and Difference Identities in Theorem or the Half-Angle Identities in Theorem to evaluate $w sub 0$ and $w sub 2$ . Since we are not explicitly told to do so, we leave this as a good, but messy, exercise.
To find the five fifth roots of $1$ , we write $1 equals 1 ⁢ cis ⁢ open parenthesis 0 close parenthesis$ . We have $r equals 1$ , $θ equals 0$ and $n equals 5$ . Since $\sqrt[5]{1} = 1$ , the roots are $w_{0} = cis (0) = 1$ , $w_{1} = cis (\frac{2 π}{5})$ , $w_{2} = cis (\frac{4 π}{5})$ , $w_{3} = cis (\frac{6 π}{5})$ and $w_{4} = cis (\frac{8 π}{5})$ . The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of $\frac{2 π}{5}$ . At this stage, we could approximate our answers using a calculator, and we leave this as an exercise.

Now that we have done some computations using Theorem, we take a step back to look at things geometrically. Essentially, Theorem says that to find the $n^{th}$ roots of a complex number, we first take the $n^{th}$ root of the modulus and divide the argument by $n$ . This gives the first root $w sub 0$ . Each succeessive root is found by adding $\frac{2 π}{n}$ to the argument, which amounts to rotating $w sub 0$ by $\frac{2 π}{n}$ radians. This results in $n$ roots, spaced equally around the complex plane. As an example of this, we plot our answers to number in Example Example 4 below.

The four fourth roots of $z = - 16$ equally spaced $\frac{2 π}{4} = \frac{π}{2}$ around the plane.

We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications.¹⁴ For now, the following exercises will have to suffice.

Exercises

In Exercises -, find a polar representation for the complex number $z$ and then identify $Re (z)$ , $Im (z)$ , $| z |$ , $arg (z)$ and $Arg (z)$ .

$z = 9 + 9 i$
$z = 5 + 5 i \sqrt{3}$
$z equals 6 ⁢ i$
$z = - 3 \sqrt{2} + 3 i \sqrt{2}$
$z = - 6 \sqrt{3} + 6 i$
$z = - 2$
$z = - \frac{\sqrt{3}}{2} - \frac{1}{2} i$
$z = - 3 - 3 i$
$z = - 5 i$
$z = 2 \sqrt{2} - 2 i \sqrt{2}$
$z equals 6$
$z = i \sqrt[3]{7}$
$z = 3 + 4 i$
$z = \sqrt{2} + i$
$z = - 7 + 24 i$
$z = - 2 + 6 i$
$z = - 12 - 5 i$
$z = - 5 - 2 i$
$z = 4 - 2 i$
$z = 1 - 3 i$
$z equals 6 ⁢ cis ⁢ open parenthesis 0 close parenthesis$
$z = 2 cis (\frac{π}{6})$
$z = 7 \sqrt{2} cis (\frac{π}{4})$
$z = 3 cis (\frac{π}{2})$
$z = 4 cis (\frac{2 π}{3})$
$z = \sqrt{6} cis (\frac{3 π}{4})$
$z equals 9 ⁢ cis ⁢ open parenthesis π close parenthesis$
$z = 3 cis (\frac{4 π}{3})$
$z = 7 cis (- \frac{3 π}{4})$
$z = \sqrt{13} cis (\frac{3 π}{2})$
$z = \frac{1}{2} cis (\frac{7 π}{4})$
$z = 12 cis (- \frac{π}{3})$
$z = 8 cis (\frac{π}{12})$
$z = 2 cis (\frac{7 π}{8})$
$z = 5 cis (\arctan (\frac{4}{3}))$
$z = \sqrt{10} cis (\arctan (\frac{1}{3}))$
$z = 15 cis (\arctan (- 2))$
$z = \sqrt{3} (\arctan (- \sqrt{2}))$
$z = 50 cis (π - \arctan (\frac{7}{24}))$
$z = \frac{1}{2} cis (π + \arctan (\frac{5}{12}))$
$z w$
$\frac{z}{w}$
$\frac{w}{z}$
$z to the power 4$
$w^{3}$
$z to the power 5 ⁢ w squared$
$z^{3} w^{2}$
$\frac{z^{2}}{w}$
$\frac{w}{z^{2}}$
$\frac{z^{3}}{w^{2}}$
$\frac{w^{2}}{z^{3}}$
${(\frac{w}{z})}^{6}$
${(- 2 + 2 i \sqrt{3})}^{3}$
${(- \sqrt{3} - i)}^{3}$
${(- 3 + 3 i)}^{4}$
${(\sqrt{3} + i)}^{4}$
${(\frac{5}{2} + \frac{5}{2} i)}^{3}$
${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{6}$
${(\frac{3}{2} - \frac{3}{2} i)}^{3}$
${(\frac{\sqrt{3}}{3} - \frac{1}{3} i)}^{4}$
${(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)}^{4}$
${(2 + 2 i)}^{5}$
${(\sqrt{3} - i)}^{5}$
${(1 - i)}^{8}$
the two square roots of $z equals 4 ⁢ i$
the two square roots of $z = - 25 i$
the two square roots of $z = 1 + i \sqrt{3}$
the two square roots of $\frac{5}{2} - \frac{5 \sqrt{3}}{2} i$
the three cube roots of $z equals 64$
the three cube roots of $z = - 125$
the three cube roots of $z = i$
the three cube roots of $z = - 8 i$
the four fourth roots of $z equals 16$
the four fourth roots of $z = - 81$
the six sixth roots of $z equals 64$
the six sixth roots of $z = - 729$
Use the Sum and Difference Identities in Theorem or the Half Angle Identities in Theorem to express the three cube roots of $z = \sqrt{2} + i \sqrt{2}$ in rectangular form. (See Example Example 4, number.)
Use a calculator to approximate the five fifth roots of $1$ . (See Example Example 4, number.)
According to Theorem in Section, the polynomial $p (x) = x^{4} + 4$ can be factored into the product linear and irreducible quadratic factors. In Exercise in Section, we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of non-linear equations. Now that we can compute the complex fourth roots of $minus 4$ directly, we can simply apply the Complex Factorization Theorem, Theorem, to obtain the linear factorization $p (x) = (x - (1 + i)) (x - (1 - i)) (x - (- 1 + i)) (x - (- 1 - i))$ . By multiplying the first two factors together and then the second two factors together, thus pairing up the complex conjugate pairs of zeros Theorem told us we'd get, we have that $p (x) = (x^{2} - 2 x + 2) (x^{2} + 2 x + 2)$ . Use the 12 complex $12 to the power th$ roots of 4096 to factor $p (x) = x^{12} - 4096$ into a product of linear and irreducible quadratic factors.
Complete the proof of Theorem by showing that if $w not equal to 0$ than $| \frac{1}{w} | = \frac{1}{| w |}$ .
Recall from Section that given a complex number $z = a + b i$ its complex conjugate, denoted $\bar{z}$ , is given by $\bar{z} = a - b i$ .
1. Prove that $| \bar{z} | = | z |$ .
2. Prove that $| z | = \sqrt{z \bar{z}}$
3. Show that $Re (z) = \frac{z + \bar{z}}{2}$ and $Im (z) = \frac{z - \bar{z}}{2 i}$
4. Show that if $θ ∈ arg ⁢ open parenthesis z close parenthesis$ then $minus θ ∈ arg ⁢ open parenthesis z bar close parenthesis$ . Interpret this result geometrically.
5. Is it always true that $Arg (\bar{z}) = - Arg (z)$ ?
Given any natural number $n greater than or equal to 2$ , the $n$ complex $n^{th}$ roots of the number $z equals 1$ are called the $n^{th}$ Roots of Unity. In the following exercises, assume that $n$ is a fixed, but arbitrary, natural number such that $n greater than or equal to 2$ .
1. Show that $w equals 1$ is an $n^{th}$ root of unity.
2. Show that if both $w_{j}$ and $w_{k}$ are $n^{th}$ roots of unity then so is their product $w_{j} w_{k}$ .
3. Show that if $w_{j}$ is an $n^{th}$ root of unity then there exists another $n^{th}$ root of unity $w_{j^{'}}$ such that $w sub j ⁢ w sub j prime equals 1$ . Hint: If $w_{j} = cis (θ)$ let $w_{j^{'}} = cis (2 π - θ)$ . You'll need to verify that $w_{j^{'}} = cis (2 π - θ)$ is indeed an $n^{th}$ root of unity.
Another way to express the polar form of a complex number is to use the exponential function. For real numbers $t$ , Euler 's Formula defines $e^{i t} = \cos (t) + i \sin (t)$ .
1. Use Theorem to show that $e^{i x} e^{i y} = e^{i (x + y)}$ for all real numbers $x$ and $y$ .
2. Use Theorem to show that ${(e^{i x})}^{n} = e^{i (n x)}$ for any real number $x$ and any natural number $n$ .
3. Use Theorem to show that $\frac{e^{i x}}{e^{i y}} = e^{i (x - y)}$ for all real numbers $x$ and $y$ .
4. If $z = r cis (θ)$ is the polar form of $z$ , show that $z = r e^{i t}$ where $θ = t$ radians.
5. Show that $e^{i π} + 1 = 0$ . (This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics.)
6. Show that $\cos (t) = \frac{e^{i t} + e^{- i t}}{2}$ and that $\sin (t) = \frac{e^{i t} - e^{- i t}}{2 i}$ for all real numbers $t$ .

In Exercises -, find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values.

For Exercises -, use $z = - \frac{3 \sqrt{3}}{2} + \frac{3}{2} i$ and $w = 3 \sqrt{2} - 3 i \sqrt{2}$ to compute the quantity. Express your answers in polar form using the principal argument.

In Exercises -, use DeMoivre's Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form.

In Exercises -, find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form.

Answers

$z = 9 + 9 i = 9 \sqrt{2} cis (\frac{π}{4})$ , $Re ⁢ open parenthesis z close parenthesis equals 9$ , $Im ⁢ open parenthesis z close parenthesis equals 9$ , $| z | = 9 \sqrt{2}$
$arg (z) = {\frac{π}{4} + 2 π k | k is an integer}$ and $Arg (z) = \frac{π}{4}$ .
$z = 5 + 5 i \sqrt{3} = 10 cis (\frac{π}{3})$ , $Re ⁢ open parenthesis z close parenthesis equals 5$ , $Im (z) = 5 \sqrt{3}$ , $vertical bar z vertical bar equals 10$
$arg (z) = {\frac{π}{3} + 2 π k | k is an integer}$ and $Arg (z) = \frac{π}{3}$ .
$z = 6 i = 6 cis (\frac{π}{2})$ , $Re ⁢ open parenthesis z close parenthesis equals 0$ , $Im ⁢ open parenthesis z close parenthesis equals 6$ , $vertical bar z vertical bar equals 6$
$arg (z) = {\frac{π}{2} + 2 π k | k is an integer}$ and $Arg (z) = \frac{π}{2}$ .
$z = - 3 \sqrt{2} + 3 i \sqrt{2} = 6 cis (\frac{3 π}{4})$ , $Re (z) = - 3 \sqrt{2}$ , $Im (z) = 3 \sqrt{2}$ , $vertical bar z vertical bar equals 6$
$arg (z) = {\frac{3 π}{4} + 2 π k | k is an integer}$ and $Arg (z) = \frac{3 π}{4}$ .
$z = - 6 \sqrt{3} + 6 i = 12 cis (\frac{5 π}{6})$ , $Re (z) = - 6 \sqrt{3}$ , $Im ⁢ open parenthesis z close parenthesis equals 6$ , $vertical bar z vertical bar equals 12$
$arg (z) = {\frac{5 π}{6} + 2 π k | k is an integer}$ and $Arg (z) = \frac{5 π}{6}$ .
$z = - 2 = 2 cis (π)$ , $Re (z) = - 2$ , $Im ⁢ open parenthesis z close parenthesis equals 0$ , $vertical bar z vertical bar equals 2$
$arg (z) = {(2 k + 1) π | k is an integer}$ and $Arg (z) = π$ .
$z = - \frac{\sqrt{3}}{2} - \frac{1}{2} i = cis (\frac{7 π}{6})$ , $Re (z) = - \frac{\sqrt{3}}{2}$ , $Im (z) = - \frac{1}{2}$ , $vertical bar z vertical bar equals 1$
$arg (z) = {\frac{7 π}{6} + 2 π k | k is an integer}$ and $Arg (z) = - \frac{5 π}{6}$ .
$z = - 3 - 3 i = 3 \sqrt{2} cis (\frac{5 π}{4})$ , $Re (z) = - 3$ , $Im (z) = - 3$ , $| z | = 3 \sqrt{2}$
$arg (z) = {\frac{5 π}{4} + 2 π k | k is an integer}$ and $Arg (z) = - \frac{3 π}{4}$ .
$z = - 5 i = 5 cis (\frac{3 π}{2})$ , $Re ⁢ open parenthesis z close parenthesis equals 0$ , $Im (z) = - 5$ , $vertical bar z vertical bar equals 5$
$arg (z) = {\frac{3 π}{2} + 2 π k | k is an integer}$ and $Arg (z) = - \frac{π}{2}$ .
$z = 2 \sqrt{2} - 2 i \sqrt{2} = 4 cis (\frac{7 π}{4})$ , $Re (z) = 2 \sqrt{2}$ , $Im (z) = - 2 \sqrt{2}$ , $vertical bar z vertical bar equals 4$
$arg (z) = {\frac{7 π}{4} + 2 π k | k is an integer}$ and $Arg (z) = - \frac{π}{4}$ .
$z = 6 = 6 cis (0)$ , $Re ⁢ open parenthesis z close parenthesis equals 6$ , $Im ⁢ open parenthesis z close parenthesis equals 0$ , $vertical bar z vertical bar equals 6$
$arg ⁢ open parenthesis z close parenthesis equals open brace 2 ⁢ π ⁢ k vertical bar k is an integer close brace$ and $Arg ⁢ open parenthesis z close parenthesis equals 0$ .
$z = i \sqrt[3]{7} = \sqrt[3]{7} cis (\frac{π}{2})$ , $Re ⁢ open parenthesis z close parenthesis equals 0$ , $Im (z) = \sqrt[3]{7}$ , $| z | = \sqrt[3]{7}$
$arg (z) = {\frac{π}{2} + 2 π k | k is an integer}$ and $Arg (z) = \frac{π}{2}$ .
$z = 3 + 4 i = 5 cis (\arctan (\frac{4}{3}))$ , $Re ⁢ open parenthesis z close parenthesis equals 3$ , $Im ⁢ open parenthesis z close parenthesis equals 4$ , $vertical bar z vertical bar equals 5$
$arg (z) = {\arctan (\frac{4}{3}) + 2 π k | k is an integer}$ and $Arg (z) = \arctan (\frac{4}{3})$ .
$z = \sqrt{2} + i = \sqrt{3} cis (\arctan (\frac{\sqrt{2}}{2}))$ , $Re (z) = \sqrt{2}$ , $Im ⁢ open parenthesis z close parenthesis equals 1$ , $| z | = \sqrt{3}$
$arg (z) = {\arctan (\frac{\sqrt{2}}{2}) + 2 π k | k is an integer}$ and $Arg (z) = \arctan (\frac{\sqrt{2}}{2})$ .
$z = - 7 + 24 i = 25 cis (π - \arctan (\frac{24}{7}))$ , $Re (z) = - 7$ , $Im ⁢ open parenthesis z close parenthesis equals 24$ , $vertical bar z vertical bar equals 25$
$arg (z) = {π - \arctan (\frac{24}{7}) + 2 π k | k is an integer}$ and $Arg (z) = π - \arctan (\frac{24}{7})$ .
$z = - 2 + 6 i = 2 \sqrt{10} cis (π - \arctan (3))$ , $Re (z) = - 2$ , $Im ⁢ open parenthesis z close parenthesis equals 6$ , $| z | = 2 \sqrt{10}$
$arg (z) = {π - \arctan (3) + 2 π k | k is an integer}$ and $Arg (z) = π - \arctan (3)$ .
$z = - 12 - 5 i = 13 cis (π + \arctan (\frac{5}{12}))$ , $Re (z) = - 12$ , $Im (z) = - 5$ , $vertical bar z vertical bar equals 13$
$arg (z) = {π + \arctan (\frac{5}{12}) + 2 π k | k is an integer}$ and $Arg (z) = \arctan (\frac{5}{12}) - π$ .
$z = - 5 - 2 i = \sqrt{29} cis (π + \arctan (\frac{2}{5}))$ , $Re (z) = - 5$ , $Im (z) = - 2$ , $| z | = \sqrt{29}$
$arg (z) = {π + \arctan (\frac{2}{5}) + 2 π k | k is an integer}$ and $Arg (z) = \arctan (\frac{2}{5}) - π$ .
$z = 4 - 2 i = 2 \sqrt{5} cis (\arctan (- \frac{1}{2}))$ , $Re ⁢ open parenthesis z close parenthesis equals 4$ , $Im (z) = - 2$ , $| z | = 2 \sqrt{5}$
$arg (z) = {\arctan (- \frac{1}{2}) + 2 π k | k is an integer}$ and $Arg (z) = \arctan (- \frac{1}{2}) = - \arctan (\frac{1}{2})$ .
$z = 1 - 3 i = \sqrt{10} cis (\arctan (- 3))$ , $Re ⁢ open parenthesis z close parenthesis equals 1$ , $Im (z) = - 3$ , $| z | = \sqrt{10}$
$arg (z) = {\arctan (- 3) + 2 π k | k is an integer}$ and $Arg (z) = \arctan (- 3) = - \arctan (3)$ .
$z = 6 cis (0) = 6$
$z = 2 cis (\frac{π}{6}) = \sqrt{3} + i$
$z = 7 \sqrt{2} cis (\frac{π}{4}) = 7 + 7 i$
$z = 3 cis (\frac{π}{2}) = 3 i$
$z = 4 cis (\frac{2 π}{3}) = - 2 + 2 i \sqrt{3}$
$z = \sqrt{6} cis (\frac{3 π}{4}) = - \sqrt{3} + i \sqrt{3}$
$z = 9 cis (π) = - 9$
$z = 3 cis (\frac{4 π}{3}) = - \frac{3}{2} - \frac{3 i \sqrt{3}}{2}$
$z = 7 cis (- \frac{3 π}{4}) = - \frac{7 \sqrt{2}}{2} - \frac{7 \sqrt{2}}{2} i$
$z = \sqrt{13} cis (\frac{3 π}{2}) = - i \sqrt{13}$
$z = \frac{1}{2} cis (\frac{7 π}{4}) = \frac{\sqrt{2}}{4} - i \frac{\sqrt{2}}{4}$
$z = 12 cis (- \frac{π}{3}) = 6 - 6 i \sqrt{3}$
$z = 8 cis (\frac{π}{12}) = 4 \sqrt{2 + \sqrt{3}} + 4 i \sqrt{2 - \sqrt{3}}$
$z = 2 cis (\frac{7 π}{8}) = - \sqrt{2 + \sqrt{2}} + i \sqrt{2 - \sqrt{2}}$
$z = 5 cis (\arctan (\frac{4}{3})) = 3 + 4 i$
$z = \sqrt{10} cis (\arctan (\frac{1}{3})) = 3 + i$
$z = 15 cis (\arctan (- 2)) = 3 \sqrt{5} - 6 i \sqrt{5}$
$z = \sqrt{3} cis (\arctan (- \sqrt{2})) = 1 - i \sqrt{2}$
$z = 50 cis (π - \arctan (\frac{7}{24})) = - 48 + 14 i$
$z = \frac{1}{2} cis (π + \arctan (\frac{5}{12})) = - \frac{6}{13} - \frac{5 i}{26}$
$z w = 18 cis (\frac{7 π}{12})$
$\frac{z}{w} = \frac{1}{2} cis (- \frac{11 π}{12})$
$\frac{w}{z} = 2 cis (\frac{11 π}{12})$
$z^{4} = 81 cis (- \frac{2 π}{3})$
$w^{3} = 216 cis (- \frac{3 π}{4})$
$z^{5} w^{2} = 8748 cis (- \frac{π}{3})$
$z cubed ⁢ w squared equals 972 ⁢ cis ⁢ open parenthesis 0 close parenthesis$
$\frac{z^{2}}{w} = \frac{3}{2} cis (- \frac{π}{12})$
$\frac{w}{z^{2}} = \frac{2}{3} cis (\frac{π}{12})$
$\frac{z^{3}}{w^{2}} = \frac{3}{4} cis (π)$
$\frac{w^{2}}{z^{3}} = \frac{4}{3} cis (π)$
${(\frac{w}{z})}^{6} = 64 cis (- \frac{π}{2})$
${(- 2 + 2 i \sqrt{3})}^{3} = 64$
${(- \sqrt{3} - i)}^{3} = - 8 i$
${(- 3 + 3 i)}^{4} = - 324$
${(\sqrt{3} + i)}^{4} = - 8 + 8 i \sqrt{3}$
${(\frac{5}{2} + \frac{5}{2} i)}^{3} = - \frac{125}{4} + \frac{125}{4} i$
${(- \frac{1}{2} - \frac{i \sqrt{3}}{2})}^{6} = 1$
${(\frac{3}{2} - \frac{3}{2} i)}^{3} = - \frac{27}{4} - \frac{27}{4} i$
${(\frac{\sqrt{3}}{3} - \frac{1}{3} i)}^{4} = - \frac{8}{81} - \frac{8 i \sqrt{3}}{81}$
${(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)}^{4} = - 1$
${(2 + 2 i)}^{5} = - 128 - 128 i$
${(\sqrt{3} - i)}^{5} = - 16 \sqrt{3} - 16 i$
${(1 - i)}^{8} = 16$
Since $z = 4 i = 4 cis (\frac{π}{2})$ we have
$w_{0} = 2 cis (\frac{π}{4}) = \sqrt{2} + i \sqrt{2}$
$w_{1} = 2 cis (\frac{5 π}{4}) = - \sqrt{2} - i \sqrt{2}$
Since $z = - 25 i = 25 cis (\frac{3 π}{2})$ we have
$w_{0} = 5 cis (\frac{3 π}{4}) = - \frac{5 \sqrt{2}}{2} + \frac{5 \sqrt{2}}{2} i$
$w_{1} = 5 cis (\frac{7 π}{4}) = \frac{5 \sqrt{2}}{2} - \frac{5 \sqrt{2}}{2} i$
Since $z = 1 + i \sqrt{3} = 2 cis (\frac{π}{3})$ we have
$w_{0} = \sqrt{2} cis (\frac{π}{6}) = \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2} i$
$w_{1} = \sqrt{2} cis (\frac{7 π}{6}) = - \frac{\sqrt{6}}{2} - \frac{\sqrt{2}}{2} i$
Since $z = \frac{5}{2} - \frac{5 \sqrt{3}}{2} i = 5 cis (\frac{5 π}{3})$ we have
$w_{0} = \sqrt{5} cis (\frac{5 π}{6}) = - \frac{\sqrt{15}}{2} + \frac{\sqrt{5}}{2} i$
$w_{1} = \sqrt{5} cis (\frac{11 π}{6}) = \frac{\sqrt{15}}{2} - \frac{\sqrt{5}}{2} i$
Since $z = 64 = 64 cis (0)$ we have
$w_{0} = 4 cis (0) = 4$
$w_{1} = 4 cis (\frac{2 π}{3}) = - 2 + 2 i \sqrt{3}$
$w_{2} = 4 cis (\frac{4 π}{3}) = - 2 - 2 i \sqrt{3}$
Since $z = - 125 = 125 cis (π)$ we have
$w_{0} = 5 cis (\frac{π}{3}) = \frac{5}{2} + \frac{5 \sqrt{3}}{2} i$
$w_{1} = 5 cis (π) = - 5$
$w_{2} = 5 cis (\frac{5 π}{3}) = \frac{5}{2} - \frac{5 \sqrt{3}}{2} i$
Since $z = i = cis (\frac{π}{2})$ we have
$w_{0} = cis (\frac{π}{6}) = \frac{\sqrt{3}}{2} + \frac{1}{2} i$
$w_{1} = cis (\frac{5 π}{6}) = - \frac{\sqrt{3}}{2} + \frac{1}{2} i$
$w_{2} = cis (\frac{3 π}{2}) = - i$
Since $z = - 8 i = 8 cis (\frac{3 π}{2})$ we have
$w_{0} = 2 cis (\frac{π}{2}) = 2 i$
$w_{1} = 2 cis (\frac{7 π}{6}) = - \sqrt{3} - i$
$w_{2} = cis (\frac{11 π}{6}) = \sqrt{3} - i$
Since $z = 16 = 16 cis (0)$ we have
$w_{0} = 2 cis (0) = 2$
$w_{1} = 2 cis (\frac{π}{2}) = 2 i$
$w_{2} = 2 cis (π) = - 2$
$w_{3} = 2 cis (\frac{3 π}{2}) = - 2 i$
Since $z = - 81 = 81 cis (π)$ we have
$w_{0} = 3 cis (\frac{π}{4}) = \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2}}{2} i$
$w_{1} = 3 cis (\frac{3 π}{4}) = - \frac{3 \sqrt{2}}{2} + \frac{3 \sqrt{2}}{2} i$
$w_{2} = 3 cis (\frac{5 π}{4}) = - \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2}}{2} i$
$w_{3} = 3 cis (\frac{7 π}{4}) = \frac{3 \sqrt{2}}{2} - \frac{3 \sqrt{2}}{2} i$
Since $z = 64 = 64 cis (0)$ we have
$w_{0} = 2 cis (0) = 2$
$w_{1} = 2 cis (\frac{π}{3}) = 1 + \sqrt{3} i$
$w_{2} = 2 cis (\frac{2 π}{3}) = - 1 + \sqrt{3} i$
$w_{3} = 2 cis (π) = - 2$
$w_{4} = 2 cis (- \frac{2 π}{3}) = - 1 - \sqrt{3} i$
$w_{5} = 2 cis (- \frac{π}{3}) = 1 - \sqrt{3} i$
Since $z = - 729 = 729 cis (π)$ we have
$w_{0} = 3 cis (\frac{π}{6}) = \frac{3 \sqrt{3}}{2} + \frac{3}{2} i$
$w_{1} = 3 cis (\frac{π}{2}) = 3 i$
$w_{2} = 3 cis (\frac{5 π}{6}) = - \frac{3 \sqrt{3}}{2} + \frac{3}{2} i$
$w_{3} = 3 cis (\frac{7 π}{6}) = - \frac{3 \sqrt{3}}{2} - \frac{3}{2} i$
$w_{4} = 3 cis (- \frac{3 π}{2}) = - 3 i$
$w_{5} = 3 cis (- \frac{11 π}{6}) = \frac{3 \sqrt{3}}{2} - \frac{3}{2} i$
Note: In the answers for $w sub 0$ and $w sub 2$ the first rectangular form comes from applying the appropriate Sum or Difference Identity ( $\frac{π}{12} = \frac{π}{3} - \frac{π}{4}$ and $\frac{17 π}{12} = \frac{2 π}{3} + \frac{3 π}{4}$ , respectively) and the second comes from using the Half-Angle Identities.
$w_{0} = \sqrt[3]{2} cis (\frac{π}{12}) = \sqrt[3]{2} (\frac{\sqrt{6} + \sqrt{2}}{4} + i (\frac{\sqrt{6} - \sqrt{2}}{4})) = \sqrt[3]{2} (\frac{\sqrt{2 + \sqrt{3}}}{2} + i \frac{\sqrt{2 - \sqrt{3}}}{2})$
$w_{1} = \sqrt[3]{2} cis (\frac{3 π}{4}) = \sqrt[3]{2} (- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)$
$w_{2} = \sqrt[3]{2} cis (\frac{17 π}{12}) = \sqrt[3]{2} (\frac{\sqrt{2} - \sqrt{6}}{4} + i (\frac{- \sqrt{2} - \sqrt{6}}{4})) = \sqrt[3]{2} (\frac{\sqrt{2 - \sqrt{3}}}{2} + i \frac{\sqrt{2 + \sqrt{3}}}{2})$
$w_{0} = cis (0) = 1$
$w_{1} = cis (\frac{2 π}{5}) \approx 0.309 + 0.951 i$
$w_{2} = cis (\frac{4 π}{5}) \approx - 0.809 + 0.588 i$
$w_{3} = cis (\frac{6 π}{5}) \approx - 0.809 - 0.588 i$
$w_{4} = cis (\frac{8 π}{5}) \approx 0.309 - 0.951 i$
$p (x) = x^{12} - 4096 = (x - 2) (x + 2) (x^{2} + 4) (x^{2} - 2 x + 4) (x^{2} + 2 x + 4) (x^{2} - 2 \sqrt{3} x + 4) (x^{2} + 2 \sqrt{3} + 4)$

In Exercises -, we have that $z = - \frac{3 \sqrt{3}}{2} + \frac{3}{2} i = 3 cis (\frac{5 π}{6})$ and $w = 3 \sqrt{2} - 3 i \sqrt{2} = 6 cis (- \frac{π}{4})$ so we get the following.

`Well-defined' means that no matter how we express $z$ , the number $Re (z)$ is always the same, and the number $Im (z)$ is always the same. In other words, $Re$ and $Im$ are functions of complex numbers. ↩
In case you're wondering, the use of the absolute value notation $| z |$ for modulus will be explained shortly. ↩
Recall the symbol being used here, ` $∈$ ,' is the mathematical symbol which denotes membership in a set. ↩
If we had Calculus, we would regard $Arg ⁢ open parenthesis 0 close parenthesis$ as an `indeterminate form.' But we don't, so we won't. ↩
Since the absolute value $| x |$ of a real number $x$ can be viewed as the distance from $x$ to $0$ on the number line, this first property justifies the notation $| z |$ for modulus. We leave it to the reader to show that if $z$ is real, then the definition of modulus coincides with absolute value so the notation $| z |$ is unambiguous. ↩
This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is true. We know $vertical bar z vertical bar equals 0$ if and only if $\sqrt{a^{2} + b^{2}} = 0$ if and only if $a^{2} + b^{2} = 0$ , which is true if and only if $a = b = 0$ . The latter happens if and only if $z = a + b i = 0$ . There. ↩
See Example in Section for a review of complex number arithmetic. ↩
See Section for a review of this technique. ↩
Compare this proof with the proof of the Power Rule in Theorem . ↩
Assuming $vertical bar w vertical bar greater than 1$ . ↩
Assuming $β greater than 0$ . ↩
Again, assuming $vertical bar w vertical bar greater than 1$ . ↩
Again, assuming $β greater than 0$ . ↩
For more on this, see the beautifully written epilogue to Section found on page . ↩

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.