6.3 Complex Numbers

There is no real number $x$ such that $x^2 = -1$. However, it turns out to be useful^[1] to invent such a number, called the imaginary unit and denoted by the letter $i$. Thus, $i^2 = -1$, and hence $i = \sqrt{-1}$. If $a$ and $b$ are real numbers, then a number of the form $a + bi$ is called a complex number, and if $b \ne 0$ then it is called an imaginary number (and pure imaginary if $a=0$ and $b \ne 0$). The real number $a$ is called the real part of the complex number $a+bi$, and $bi$ is called its imaginary part.

What does it mean to add $a$ to $bi$ in the definition $a+bi$ of a complex number, i.e. adding a real number and an imaginary number? You can think of it as a way of extending the set of real numbers. If $b=0$ then $a+bi = a+0i = a$ (since $0i$ is defined as $0$), so that every real number is a complex number. The imaginary part $bi$ in $a+bi$ can be thought of as a way of taking the one-dimensional set of all real numbers and extending it to a two-dimensional set: there is a natural correspondence between a complex number $a+bi$ and a point $(a,b)$ in the (two-dimensional) $xy$-coordinate plane.

Before exploring that correspondence further, we will first state some fundamental properties of and operations on complex numbers:

Let $a+bi$ and $c+di$ be complex numbers. Then:

$a+bi ~=~ c+di$ if and only if $a=c$ and $b=d~$ (i.e. the real parts are equal and the imaginary parts are equal)
$(a+bi) \;+\; (c+di) ~=~ (a+c) \;+\; (b+d)i~$ (i.e. add the real parts together and add the imaginary parts together)
$(a+bi) \;-\; (c+di) ~=~ (a-c) \;+\; (b-d)i$
$(a+bi)\,(c+di) ~=~ (ac-bd) \;+\; (ad+bc)i$
$(a+bi)\,(a-bi) ~=~ a^2 \;+\; b^2$
$\dfrac{a+bi}{c+di} ~=~ \dfrac{(ac+bd) \;+\; (bc-ad)i}{c^2 + d^2}$

The first three items above are just definitions of equality, addition, and subtraction of complex numbers. The last three items can be derived by treating the multiplication and division of complex numbers as you would normally treat factors of real numbers:

\begin{align*} (a+bi)\,(c+di) ~&=~ a\,(c+di) \;+\; bi\,(c+di)\\ &=~ ac \;+\; adi \;+\; bci \;+\; bdi^2 ~=~ ac \;+\; adi \;+\; bci \;+\; bd(-1)\\ &=~ (ac - bd) \;+\; (ad+bc)i \end{align*}

The fifth item is a special case of the multiplication formula:

\begin{align*} (a+bi)\,(a-bi) ~&=~ ((a)(a) - (b)(-b)) \;+\; ((a)(-b) + (b)(a))i\\ &=~ ( a^2 + b^2 ) \;+\; (-ab + ba)i ~=~ ( a^2 + b^2 ) \;+\; 0i\\ &=~ a^2 \;+\; b^2 \end{align*}

The sixth item comes from using the previous items:

\begin{align*} \dfrac{a+bi}{c+di} ~&=~ \dfrac{a+bi}{c+di} \,\cdot\, \dfrac{c-di}{c-di}\\ &=~ \dfrac{(ac - b(-d)) \;+\; (a(-d) + bc)i}{c^2 + d^2}\\ &=~ \dfrac{(ac+bd) \;+\; (bc-ad)i}{c^2 + d^2} \end{align*}

The conjugate $\overline{a+bi}$ of a complex number $a+bi$ is defined as $\overline{a+bi} = a-bi$. Notice that $(a+bi) \;+\; \overline{(a+bi)} ~=~ 2a$ is a real number, $(a+bi) \;-\; \overline{(a+bi)} ~=~ 2bi$ is an imaginary number if $b \ne 0$, and $(a+bi) \overline{(a+bi)} ~=~ a^2 + b^2$ is a real number. So for a complex number $z=a+bi$, $z\,\overline{z} = a^2 + b^2 \,$ and thus we can define the modulus of $z$ to be $\sqrt{z\,\overline{z}} = \sqrt{a^2 + b^2}$, which we denote by $\abs{z}$.

Example 6.9

Let $z_1 = -2+3i$ and $z_2 = 3+4i$. Find $z_1 + z_2$, $z_1 - z_2$, $z_1 \, z_2$, $z_1 / z_2$, $\abs{z_1}$, and $\abs{z_2}$.

Solution: Using our rules and definitions, we have:

\begin{align*} z_1 \;+\; z_2 ~&=~ (-2+3i) \;+\; (3+4i)\\ &=~ 1 + 7i\\ z_1 \;-\; z_2 ~&=~ (-2+3i) \;-\; (3+4i)\\ &=~ -5 - i\\ z_1 \, z_2 ~&=~ (-2+3i)\, (3+4i)\\ &=~ ((-2)(3) - (3)(4)) \;+\; ((-2)(4) + (3)(3))i\\ &=~ -18 + i\\ \dfrac{z_1}{z_2} ~&=~ \dfrac{-2+3i}{3+4i}\\ &=~ \dfrac{(-2)(3) + (3)(4) \;+\; ((3)(3) - (-2)(4))i}{3^2 + 4^2}\\ &=~ \dfrac{6}{25} \;+\; \dfrac{17}{25}\,i\\ \abs{z_1} ~&=~ \sqrt{(-2)^2 + 3^2}\\ &=~ \sqrt{13}\\ \abs{z_2} ~&=~ \sqrt{3^2 + 4^2}\\ &=~ 5 \end{align*}

We know that any point $(x,y)$ in the $xy$-coordinate plane that is a distance $r >0$ from the origin has coordinates $x=r\,\cos\;\theta$ and $y=r\,\sin\;\theta$, where $\theta$ is the angle in standard position as in Figure 6.3.1(a).

tikzpicture[every node/.style=font=] [line width=0.5pt,-latex] (0.8,0) arc (0:55:0.8); [black!60,line width=0.3pt,-latex] (-0.5,0) -- (3,0) node [right] x; [black!60,line width=0.3pt,-latex] (0,-0.5) -- (0,2.6) node [above] y; [black!60,below left] at (0,0) 0; [right] at (30:0.4) ; [dashed] (0,0) -- (55:2.5) node[above left,midway] r; (55:2.5) circle (2pt) node[above right] (x,y)=(r\, \; , r\, \; ); tikzpicture — Figure 6.3.1

Let $z=x+yi$ be a complex number. We can represent $z$ as a point in the complex plane, where the horizontal $x$-axis represents the real part of $z$, and the vertical $y$-axis represents the pure imaginary part of $z$, as in Figure 6.3.1(b). The distance $r$ from $z$ to the origin is, by the Pythagorean Theorem, $r = \sqrt{x^2 + y^2}$, which is just the modulus of $z$. And we see from Figure 6.3.1(b) that $x=r\,\cos\;\theta$ and $y=r\,\sin\;\theta$, where $\theta$ is the angle formed by the positive $x$-axis and the line segment from the origin to $z$. We call this angle $\theta$ the argument of $z$. Thus, we get the trigonometric form (sometimes called the polar form) of the complex number $z$:

For any complex number $z=x+yi$, we can write

\begin{align} z ~&=~ r\,(\cos\;\theta \;+\; i\,\sin\;\theta)~~,~\text{where}\tag{6.3}\\ r ~&=~ \abs{z} ~=~ \sqrt{x^2 + y^2}~~\text{and}\\ \theta ~&=~ \text{the argument of $z$}~.\\ \intertext{The representation $z=r\,(\cos\;\theta \;+\; i\,\sin\;\theta)$ is often abbreviated as:} z ~&=~ r\,\text{cis}\;\theta \tag{6.4}\end{align}

In the special case $z=0 = 0+0i$, the argument $\theta$ is undefined since $r=\abs{z}=0$. Also, note that the argument $\theta$ can be replaced by $\theta \;+\; 360\Degrees k$ or $\theta \;+\; \pi k$, depending on whether you are using degrees or radians, respectively, for $k=0$, $\pm\,1$, $\pm\,2$, $...$ . Note also that for $z=x+yi$ with $r=\abs{z}$, $\theta$ must satisfy

\begin{displaymath} \tan\;\theta ~=~ \tfrac{y}{x}~~,~ \cos\;\theta ~=~ \tfrac{x}{r}~~,~ \sin\;\theta ~=~ \tfrac{y}{r}~. \end{displaymath}

Example 6.10

tikzpicture[every node/.style=font=] [black!60,solid,line width=0.3pt,-latex] (-2.1,0) -- (0.7,0) node[right] x; [black!60,solid,line width=0.3pt,-latex] (0,-1.1) -- (0,0.7) node[above] y; [black,below right] at (0,0) 0; (-2,-1) circle (2pt); [dashed] (-2,-1) -- (0,0) node [midway,below] r; [dashed] (-2,-1) -- (-2,0) node [midway,left] 1; [above] at (-1.2,0) 2; [below] at (-1.4,-1) z=-2-i; [-latex] (0:0.45) arc (0:206.56:0.45); [left] at (120:0.5) ; tikzpicture — Figure 6.3.2

Represent the complex number $-2 - i$ in trigonometric form.

Solution: Let $z=-2-i=x+yi$, so that $x=-2$ and $y=-1$. Then $\theta$ is in QIII, as we see in Figure 6.3.2. So since $\tan\;\theta = \tfrac{y}{x} = \tfrac{-1}{-2} = \tfrac{1}{2}$, we have $\theta = 206.6\Degrees$. Also,

\begin{displaymath} r ~=~ \sqrt{x^2 + y^2} ~=~ \sqrt{(-2)^2 + (-1)^2} ~=~ \sqrt{5} ~. \end{displaymath}

Thus, $\boxed{-2 - i = \sqrt{5}\;(\cos\;206.6\Degrees \;+\; i\,\sin\;206.6\Degrees)}\;$, or $\sqrt{5}\;\text{cis}\;206.6\Degrees$.

For complex numbers in trigonometric form, we have the following formulas for multiplication and division:

Let $z_1 = r_1 \,(\cos\;\theta_1 \;+\; i\,\sin\;\theta_1 )$ and $z_2 = r_2 \,(\cos\;\theta_2 \;+\; i\,\sin\;\theta_2 )$ be complex numbers. Then

\begin{align} z_1 \, z_2 ~&=~ r_1 \, r_2 \,(\cos\;(\theta_1 + \theta_2 ) \;+\; i\,\sin\;(\theta_1 + \theta_2 ))~\text{, and}\tag{6.5}\\ \frac{z_1}{z_2} ~&=~ \frac{r_1}{r_2} \,(\cos\;(\theta_1 - \theta_2 ) \;+\; i\,\sin\;(\theta_1 - \theta_2 ))\quad\text{if $z_2 \ne 0$.} \tag{6.6}\end{align}

The proofs of these formulas are straightforward:

\begin{align*} z_1 \, z_2 ~&=~ r_1 \,(\cos\;\theta_1 \;+\; i\,\sin\;\theta_1 ) \;\cdot\; r_2 \,(\cos\;\theta_2 \;+\; i\,\sin\;\theta_2 )\\ &=~ r_1 \, r_2 \,\left[ (\cos\;\theta_1 ~ \cos\;\theta_2 \;-\; \sin\;\theta_1 ~ \sin\;\theta_2 ) \;+\; i\,(\sin\;\theta_1 ~ \cos\;\theta_2 \;+\; \cos\;\theta_1 ~ \sin\;\theta_2 ) \right]\\ &=~ r_1 \, r_2 \,(\cos\;(\theta_1 + \theta_2 ) \;+\; i\,\sin\;(\theta_1 + \theta_2 ))\\ \intertext{by the addition formulas for sine and cosine. And} \frac{z_1}{z_2} ~&=~ \frac{r_1 \,(\cos\;\theta_1 \;+\; i\,\sin\;\theta_1 )}{ r_2 \,(\cos\;\theta_2 \;+\; i\,\sin\;\theta_2 )}\\ &=~ \frac{r_1}{r_2} \;\cdot\; \frac{\cos\;\theta_1 \;+\; i\,\sin\;\theta_1}{ \cos\;\theta_2 \;+\; i\,\sin\;\theta_2} \;\cdot\; \frac{\cos\;\theta_2 \;-\; i\,\sin\;\theta_2}{ \cos\;\theta_2 \;-\; i\,\sin\;\theta_2}\\ &=~ \frac{r_1}{r_2} \;\cdot\; \frac{(\cos\;\theta_1 ~ \cos\;\theta_2 \;+\; \sin\;\theta_1 ~ \sin\;\theta_2 ) \;+\; i\,(\sin\;\theta_1 ~ \cos\;\theta_2 \;-\; \cos\;\theta_1 ~ \sin\;\theta_2 )}{\cos^2 \,\theta_2 \;+\; \sin^2 \,\theta_2}\\ &=~ \frac{r_1}{r_2} \,(\cos\;(\theta_1 - \theta_2 ) \;+\; i\,\sin\;(\theta_1 - \theta_2 )) \end{align*}

by the subtraction formulas for sine and cosine, and since $\cos^2 \,\theta_2 \;+\; \sin^2 \,\theta_2 = 1$. $\text{qed}$

Note that formulas (6.5) and (6.6) say that when multiplying complex numbers the moduli are multiplied and the arguments are added, while when dividing complex numbers the moduli are divided and the arguments are subtracted. This makes working with complex numbers in trigonometric form fairly simple.

Example 6.11

Let $z_1 = 6\,(\cos\;70\Degrees \;+\; i\,\sin\;70\Degrees )$ and $z_1 = 2\,(\cos\;31\Degrees \;+\; i\,\sin\;31\Degrees )$. Find $z_1 \, z_2$ and $\frac{z_1}{z_2}$.

Solution: By formulas (6.5) and (6.6) we have

\begin{alignat*}{3} z_1 \, z_2 ~&=~ (6) \, (2) \, (\cos\;(70\Degrees + 31\Degrees ) \;+\; i\,\sin\;(70\Degrees + 31\Degrees )) \quad&&\Rightarrow\quad \boxed{z_1 \, z_2 ~=~ 12 \, (\cos\;101\Degrees \;+\; i\,\sin\;101\Degrees )} ~\text{, and}\\ \frac{z_1}{z_2} ~&=~ \frac{6}{2} \, (\cos\;(70\Degrees - 31\Degrees ) \;+\; i\,\sin\;(70\Degrees - 31\Degrees )) \quad&&\Rightarrow\quad \boxed{\frac{z_1}{z_2} ~=~ 3 \, (\cos\;39\Degrees \;+\; i\,\sin\;39\Degrees )} ~. \end{alignat*}

For the special case when $z_1 = z_2 = z = r\,(\cos\;\theta \;+\; i\,\sin\;\theta)$ in formula (6.5), we have

\begin{align*} \left[ r\,(\cos\;\theta \;+\; i\,\sin\;\theta)\right]^2 ~&=~ r \cdot r \,(\cos\;(\theta + \theta ) \;+\; i\,\sin\;(\theta + \theta))\\ &=~ r^2 \,(\cos\;2\theta \;+\; i\,\sin\;2\theta) ~,\\ \intertext{and so} \left[ r\,(\cos\;\theta \;+\; i\,\sin\;\theta)\right]^3 ~&=~ \left[ r\,(\cos\;\theta \;+\; i\,\sin\;\theta)\right]^2 \;\cdot\; r\,(\cos\;\theta \;+\; i\,\sin\;\theta )\\ &=~ r^2 \,(\cos\;2\theta \;+\; i\,\sin\;2\theta) \;\cdot\; r\,(\cos\;\theta \;+\; i\,\sin\;\theta )\\ &=~ r^3 \,(\cos\;(2\theta + \theta) \;+\; i\,\sin\;(2\theta + \theta) )\\ &=~ r^3 \,(\cos\;3\theta \;+\; i\,\sin\;3\theta) ~, \end{align*}

and continuing like this (i.e. by mathematical induction), we get:

Theorem 6.1

De Moivre’s Theorem:^[2] For any integer $n \ge 1$,

\begin{equation} \left[ r\,(\cos\;\theta \;+\; i\,\sin\;\theta )\right]^n ~=~ r^n \,(\cos\;n\theta \;+\; i\,\sin\;n\theta ) ~. \tag{6.7}\end{equation}

We define $z^0 = 1$ and $z^{-n} = 1/z^n$ for all integers $n \ge 1$. So by De Moivre’s Theorem and formula (6.5), for any $z=r\,(\cos\;\theta \;+\; i\,\sin\;\theta)$ and integer $n \ge 1$ we get

\begin{align*} z^{-n} ~&=~ \frac{1}{z^n}\\ &=~ \frac{1\,(\cos\;0\Degrees \;+\; i\,\sin\;0\Degrees )}{r^n \,(\cos\;n\theta \;+\; i\,\sin\;n\theta )}\\ &=~ \frac{1}{r^n} \,(\cos\;(0\Degrees - n\theta) \;+\; i\,\sin\;(0\Degrees - n\theta))\\ &=~ r^{-n} \, (\cos\;(- n\theta) \;+\; i\,\sin\;(- n\theta)) ~, \end{align*}

and so De Moivre’s Theorem in fact holds for all integers.^[3]

Example 6.12

Find $(1+i)^{10}$.

Solution: Since $1+i = \sqrt{2}\;(\cos\;45\Degrees \;+\; i\,\sin\;45\Degrees )$ (why?), by De Moivre’s Theorem we have

\begin{displaymath} (1+i)^{10} ~=~ (\sqrt{2})^{10} \;(\cos\;450\Degrees \;+\; i\,\sin\;450\Degrees ) ~=~ 2^{10/2} \;(0 \;+\; i\,(1)) ~=~ 2^5 \,\cdot\, i ~=~ \boxed{32i} ~. \end{displaymath}

We can use De Moivre’s Theorem to find the $n^{th}$ roots of a complex number. That is, given any complex number $z$ and positive integer $n$, find all complex numbers $w$ such that $w^n = z$. Let $z=r\,(\cos\;\theta \;+\; i\,\sin\;\theta)$. Since the cosine and sine functions repeat every $360\Degrees$, we know that $z=r\,(\cos\;(\theta + 360\Degrees k)\;+\; i\,\sin\;(\theta + 360\Degrees k))$ for $k=0$, $\pm\,1$, $\pm\,2$, $...$. Now let $w=r_0 \,(\cos\;\theta_0 \;+\; i\,\sin\;\theta_0 )$ be an $n^{th}$ root of $z$. Then

\begin{align*} w^n ~=~ z \quad&\Rightarrow\quad \left[ r_0 \,(\cos\;\theta_0 \;+\; i\,\sin\;\theta_0 )\right]^n ~=~ r\,(\cos\;(\theta + 360\Degrees k)\;+\; i\,\sin\;(\theta + 360\Degrees k))\\ &\Rightarrow\quad r_0^n \,(\cos\;n\theta_0 \;+\; i\,\sin\;n\theta_0 ) ~=~ r\,(\cos\;(\theta + 360\Degrees k)\;+\; i\,\sin\;(\theta + 360\Degrees k))\\ &\Rightarrow\quad r_0^n ~=~ r \quad\text{and}\quad n\theta_0 ~=~ \theta + 360\Degrees k\\ &\Rightarrow\quad r_0 ~=~ r^{1/n} \quad\text{and}\quad \theta_0 ~=~ \frac{\theta + 360\Degrees k}{n} ~. \end{align*}

Since the cosine and sine of $\frac{\theta + 360\Degrees k}{n}$ will repeat for $k \ge n$, we get the following formula for the $n^{th}$ roots of $z$:

For any nonzero complex number $z=r\,(\cos\;\theta \;+\; i\,\sin\;\theta)$ and positive integer $n$, the $n$ distinct $n^{th}$ roots of $z$ are

\begin{equation} r^{1/n} \, \left[ \cos\;\left(\frac{\theta + 360\Degrees k}{n}\right) \;+\; i\,\sin\;\left(\frac{\theta + 360\Degrees k}{n}\right) \right] \tag{6.8}\end{equation}

for $k=0$, $1$, $2$, $...$, $n-1$.

Note: An $n^{th}$ root of $z$ is usually written as $z^{1/n}$ or $\sqrt[n]{z}$. The number $r^{1/n}$ in the above formula is the usual real $n^{th}$ root of the real number $r=\abs{z}$.

Example 6.13

Find the three cube roots of $i$.

Solution: Since $i = 1\,(\cos\;90\Degrees \;+\; i\,\sin\;90\Degrees)$, the three cube roots of $i$ are:

\begin{alignat*}{3} \sqrt[3]{1} \;\left[ \cos\;\left(\frac{90\Degrees + 360\Degrees (0)}{3}\right) \;+\; i\,\sin\;\left(\frac{90\Degrees + 360\Degrees (0)}{3}\right) \right] ~&=~ \cos\;30\Degrees \;+\; i\,\sin\;30\Degrees ~&&=~ \boxed{\frac{\sqrt{3}}{2} \;+\; \frac{1}{2}\,i}~,\\[3pt] \sqrt[3]{1} \;\left[ \cos\;\left(\frac{90\Degrees + 360\Degrees (1)}{3}\right) \;+\; i\,\sin\;\left(\frac{90\Degrees + 360\Degrees (1)}{3}\right) \right] ~&=~ \cos\;150\Degrees \;+\; i\,\sin\;150\Degrees ~&&=~ \boxed{-\frac{\sqrt{3}}{2} \;+\; \frac{1}{2}\,i}~,\\[3pt] \sqrt[3]{1} \;\left[ \cos\;\left(\frac{90\Degrees + 360\Degrees (2)}{3}\right) \;+\; i\,\sin\;\left(\frac{90\Degrees + 360\Degrees (2)}{3}\right) \right] ~&=~ \cos\;270\Degrees \;+\; i\,\sin\;270\Degrees ~&&=~ \boxed{-i} \end{alignat*}

tikzpicture[every node/.style=font=] [black!60,solid,line width=0.3pt,-latex] (-1.4,0) -- (1.6,0) node[right] x; [black!60,solid,line width=0.3pt,-latex] (0,-1.4) -- (0,1.6) node[above] y; [line width=1pt] (0,0) circle (1.2); [linecolor] (270:1.2) circle (2pt); [below left] at (270:1.2) -i; [linecolor] (30:1.2) circle (2pt); [right] at (30:1.2) 32 + i2; [linecolor] (150:1.2) circle (2pt); [left] at (150:1.2) - 32 + i2; [above right] at (80:1.2) z=1; [linecolor,dashed] (150:1.2) -- (0,0) -- (30:1.2); [linecolor,dashed,latex-latex] (30:0.4) arc (30:150:0.4); [linecolor,fill=white] at (90:0.7) 120 ; [linecolor,dashed,latex-latex] (150:0.4) arc (150:270:0.4); [linecolor] at (210:0.8) 120 ; [linecolor,dashed,latex-latex] (270:0.4) arc (270:390:0.4); [linecolor] at (330:0.8) 120 ; tikzpicture — Figure 6.3.3

Notice from Example 6.13 that the three cube roots of $i$ are equally spaced points along the unit circle $\abs{z}=1$ in the complex plane, as shown in Figure 6.3.3. We see that consecutive cube roots are $120\Degrees$ apart. In general, the $n$ $n^{th}$ roots of a complex number $z$ will be equally spaced points along the circle of radius $\abs{z}^{1/n}$ in the complex plane, with consecutive roots separated by $\tfrac{360\Degrees}{n}$.

In higher mathematics the Fundamental Theorem of Algebra states that every polynomial of degree $n$ with complex coefficients has $n$ complex roots (some of which may repeat). In particular, every real number $a$ has $n$ $n^{th}$ roots (being the roots of $z^n - a$). For example, the square roots of $1$ are $\pm\,1$, and the square roots of $-1$ are $\pm\,i$.

Exercises

For Exercises 1-16, calculate the given expression.

$(2+3i) \;+\; (-3-2i)$
$(2+3i) \;-\; (-3-2i)$
$(2+3i) \;\cdot\; (-3-2i)$
$(2+3i)/(-3-2i)$

4
$\overline{(2+3i)} \;+\; \overline{(-3-2i)}$
$\overline{(2+3i)} \;-\; \overline{(-3-2i)}$
$(1+i)/(1-i)$
$\abs{-3+2i}$

8
$i^3$
$i^4$
$i^5$
$i^6$
$i^7$
$i^8$
$i^9$
$i^{2009}$

For Exercises 17-24, prove the given identity for all complex numbers.

$\overline{\left( \overline{z} \right)} \;=\; z$
$\overline{z_1 + z_2} \;=\; \overline{z_1} + \overline{z_2}$
$\overline{z_1 - z_2} \;=\; \overline{z_1} - \overline{z_2}$
$\overline{z_1 \, z_2} \;=\; \overline{z_1} ~ \overline{z_2}$

4
$\overline{\left( \dfrac{z_1}{z_2} \right)} \;=\; \dfrac{\overline{z_1}}{\overline{z_2}}$
$\abs{z} \;=\; \abs{\overline{z}}\phantom{\dfrac{\abs{1_1}}{\abs{1_2}}}$
$\abs{z_1 \, z_2} \;=\; \abs{z_1}\,\abs{z_2}\phantom{\dfrac{\abs{1_1}}{\abs{1_2}}}$
$\left| \dfrac{z_1}{z_2} \right| \;=\; \dfrac{\abs{z_1}}{\abs{z_2}}$

For Exercises 25-30, put the given number in trigonometric form.

$2+3i$
$-3-2i$
$1-i$
$-i$
$1$
$-1$
Verify that De Moivre’s Theorem holds for the power $n=0$.

For Exercises 32-35, calculate the given number.

$3\,(\cos\;14\Degrees \;+\; i\,\sin\;14\Degrees ) \;\cdot\; 2\,(\cos\;121\Degrees \;+\; i\,\sin\;121\Degrees )$ 3
$\lbrack 3\,(\cos\;14\Degrees \;+\; i\,\sin\;14\Degrees )\rbrack^4\phantom{\dfrac{3}{4}}$
$\lbrack 3\,(\cos\;14\Degrees \;+\; i\,\sin\;14\Degrees )\rbrack^{-4}\phantom{\dfrac{3}{4}}$
$\dfrac{3\,(\cos\;14\Degrees \;+\; i\,\sin\;14\Degrees )}{ 2\,(\cos\;121\Degrees \;+\; i\,\sin\;121\Degrees )}$

2
Find the three cube roots of $-i$.
Find the three cube roots of $1+i$.

2
Find the three cube roots of $1$.
Find the three cube roots of $-1$.

2
Find the five fifth roots of $1$.
Find the five fifth roots of $-1$.
Find the two square roots of $-2 + 2\sqrt{3}\,i$.
Prove that if $z$ is an $n^{th}$ root of a real number $a$, then so is $\overline{z}$. (Hint: Use Exercise 20.)

Especially in electrical engineering, physics, and various fields of mathematics. ↩
Named after the French statistician and mathematician Abraham de Moivre (1667-1754). ↩
There is a way of defining $z^n$ when $n$ is a real (or complex) number, so that De Moivre’s Theorem holds for any real number $n$. See pp. 59-60 in R.V. Churchill, Complex Variables and Applications, 2nd ed., New York: McGraw-Hill Book Co., 1960. ↩