Theorem 2.4.3. Euler’s Formula.
\begin{equation*}
e^{i\theta}=\cos(\theta)+i \cdot \sin(\theta)
\end{equation*}
Section 2.4 Complex Numbers and Polar Coordinates
Subsection 2.4.1 Forms of Complex Numbers
Recall that a complex number has the form \(a+bi\) where \(a,b \in \mathbb{R}\) and \(i=\sqrt{-1}\text{.}\) Complex numbers have two parts: a real part and an imaginary part. For the number \(a+bi\text{,}\) the real part is \(a\) and the imaginary part is \(b\text{.}\) Because they have two parts, we can use the two dimensional rectangular coordinate plane to represent complex numbers. We use the horizontal axis to represent the real part and the vertical axis to represent the complex part. Thus, the complex number \(a+bi\) can be represented by the point \(\left( a,b \right) \) on the rectangular coordinate plane; see Figure 2.4.1.
As we’ve studied in this course, the rectangular ordered pair \(\left( a,b\right) \) can be represented in polar coordinates \(\left( r,\theta \right) \) where \(r\) represents the distance the point is from the origin and \(\theta\) represents the angle between the positive \(x\)-axis and the segment connecting the origin and the point; see Figure 2.4.2.
We know that if the rectangular pair \(\left( a,b\right) \) represents the same point as the polar pair \(\left( r,\theta \right) \text{,}\) then \(a=r\cos(\theta)\) and \(b=r\sin(\theta)\text{.}\) Thus,
\begin{align*}
a+bi \amp=r\cos(\theta)+r\sin(\theta) \cdot i\\
\amp=r \mathopen{}\left( \cos \mathopen{}\left( \theta \right)\mathclose{}+i\cdot\sin \mathopen{}\left( \theta \right)\mathclose{} \right)\mathclose{}
\end{align*}
i.e., we can express a complex number using the "polar information" \(r\) and \(\theta\text{.}\)
The expression “\(r \mathopen{}\left( \cos \mathopen{}\left( \theta \right)\mathclose{}+i\cdot\sin \mathopen{}\left( \theta \right)\mathclose{} \right)\mathclose{}\)” is what a textbook might decribe as the “polar form of a complex number.” But a more appropriate expression to label as “the polar form of a complex number” involves Euler’s Formula. Euler’s Formula is an identity that establishes a surprising connection between the exponential function \(e^x\) and complex numbers.
This is Euler’s Formula. For reference purposes, we state this in a theorem.
Notice that if we multiply both sides of Euler’s formula by \(r\text{,}\) we obtain a formula that allows us to write any complex number in polar form:
\begin{align*}
e^{i\theta} \amp=\cos(\theta)+i \cdot \sin(\theta)\\
\implies r \cdot e^{i\theta} \amp= r \cdot \mathopen{}\left( \cos \mathopen{}\left( \theta \right)\mathclose{}+i\cdot\sin \mathopen{}\left( \theta \right)\mathclose{} \right)\mathclose{}\\
\implies r(e^{i\theta}) \amp=r\cos(\theta)+r\sin(\theta) \cdot i
\end{align*}
Definition 2.4.4.
The polar form of the complex number \(z=r\cos(\theta)+r\sin(\theta)\cdot i\) is \(z=re^{i\theta}\text{.}\)
Let’s review what we’ve established: First, we observed that we can write a complex number of the form “\(a+bi\)” in the form “\(r \cdot \mathopen{}\left( \cos \mathopen{}\left( \theta \right)\mathclose{}+i\cdot\sin \mathopen{}\left( \theta \right)\mathclose{} \right)\mathclose{}\)”. Then we noticed that we can write an expression of the form “\(r \cdot \mathopen{}\left( \cos \mathopen{}\left( \theta \right)\mathclose{}+i\cdot\sin \mathopen{}\left( \theta \right)\mathclose{} \right)\mathclose{}\)” in the form “\(re^{i\theta}\)”. Finally, we realized that we can write a complex number “\(a+bi\)” in the form “\(re^{i\theta}\)” so we defined “\(re^{i\theta}\)” as being the polar form of the complex number \(a+bi\text{.}\)
Example 2.4.5.
Express in "rectangular form" (i.e. in the form \(z=a+bi\)) the complex number \(z=6e^{\frac{5\pi}{6}\cdot i}\text{,}\) given in polar form.
Solution.
\begin{align*}
z \amp=6e^{\frac{5\pi}{6}\cdot i} \\
\amp=6\cos\mathopen{}\left( \frac{5\pi}{6} \right)\mathclose{} +6\sin \mathopen{}\left( \frac{5\pi}{6} \right)\mathclose{} \cdot i \\
\amp=6 \mathopen{}\left(-\frac{\sqrt{3}}{2} \right)\mathclose{} +6 \mathopen{}\left( \frac{1}{2} \right)\mathclose{} \cdot i \\
\amp=-3\sqrt{3}+3i
\end{align*}
Thus, the complex number \(z=6e^{\frac{5\pi}{6}\cdot i}\) can be expressed in "rectangular form" as \(z=-3\sqrt{3}+3i\text{.}\)
Example 2.4.6.
Express in polar form (i.e. in the form \(z=re^{i\theta}\)) the complex number \(z=3-3i\text{,}\) given in "rectangular form."
Solution.
We can associate the complex number \(z=3-3i\) with the rectangular ordered pair \((3,-3)\text{,}\) and then translate this ordered pair into polar coordinates \((r,\theta)\text{,}\) and finally use the polar ordered pair to obtain the polar form \(z=re^{i\theta}\text{.}\) First, let’s find \(r\text{:}\)
\begin{align*}
r \amp= \sqrt{(3)^2+(-3)^2} \\
\amp= \sqrt{9+9} \\
\amp=3\sqrt{2} \text{.}
\end{align*}
Now, let’s find \(\theta\text{:}\)
\begin{align*}
\tan(\theta) \amp=\frac{-3}{3} \\
\implies \theta \amp=\tan^{-1}(-1) \\
\implies \theta \amp=-\frac{\pi}{4}
\end{align*}
Thus, the complex number \(z=3-3i\) can be expressed in polar form as \(z=3\sqrt{2}e^{-\frac{\pi}{4} \cdot i}\text{.}\)
Subsection 2.4.2 Using the Polar Form to Find Complex Roots
Example 2.4.7.
Find the two square roots of \(-1+\sqrt{3}i\) using the polar form of \(-1+\sqrt{3}i\text{.}\)
Solution.
Recall that there are two distinct square roots of any positive real number (e.g., the two square roots of \(4\) are \(2\) and \(-2\)). The same is true for any complex number. We can find two different square roots of a complex number by using two difference polar forms of the number.
To find polar forms of \(-1+\sqrt{3}i\text{,}\) we first associate the number with the rectangular ordered pair \(\left( -1,\sqrt{3}\right) \text{,}\) and then translate it into polar coordinates \(\left( r,\theta\right) \text{.}\) First let’s find \(r\text{:}\)
\begin{align*}
r \amp= \sqrt{ \left( -1\right) ^2+ \left( \sqrt{3}\right) ^2} \\
\amp= \sqrt{1+3} \\
\amp=2 \text{.}
\end{align*}
Now, let’s find \(\theta\text{.}\) \(\tan(\theta)=-\sqrt{3}\) with \(\theta\) in Quadrant II.
Both \(\theta=\frac{2\pi}{3}\) and \(\theta=-\frac{4\pi}{3}\) satisfy the condition, so we’ll use these two angles to obtain two polar forms of \(-1+\sqrt{3}i\text{:}\)
\begin{equation*}
-1+\sqrt{3}i=2e^{\frac{2\pi}{3} \cdot i} \text{and} -1+\sqrt{3}i=2e^{-\frac{4\pi}{3} \cdot i}
\end{equation*}
Therefore,
\begin{align*}
\left( -1+\sqrt{3}i \right) ^{1/2} \amp= (2e^{\frac{2\pi}{3}\cdot i})^{1/2} \\
\amp=2^{\frac{1}{2}}e^{\frac{2\pi}{3} \cdot \frac{1}{2}i} \\
\amp=\sqrt{2}e^{\frac{\pi}{3} \cdot i} \\
\amp=\sqrt{2} \mathopen{}\left( \cos \mathopen{}\left( \frac{\pi}{3} \right)\mathclose{} +i\cdot \sin \mathopen{}\left( \frac{\pi}{3} \right)\mathclose{} \right)\mathclose{}\\
\amp=\sqrt{2} \mathopen{}\left( \frac{1}{2}+\frac{\sqrt{3}}{2}i \right)\mathclose{} \\
\amp=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}i
\end{align*}
and
\begin{align*}
\left( -1+\sqrt{3}i \right) ^{1/2} \amp= \left( 2e^{-\frac{4\pi}{3} \cdot i} \right) ^{1/2} \\
\amp=2^{\frac{1}{2}}e^{-\frac{4\pi}{3} \cdot \frac{1}{2}i} \\
\amp=\sqrt{2}e^{-\frac{2\pi}{3} \cdot i} \\
\amp=\sqrt{2}\mathopen{}\left( \cos \mathopen{}\left( -\frac{2\pi}{3} \right)\mathclose{} +i\cdot \sin \mathopen{}\left( -\frac{2\pi}{3} \right)\mathclose{} \right)\mathclose{}\\
\amp=\sqrt{2} \mathopen{}\left( -\frac{1}{2}-\frac{\sqrt{3}}{2}i \right)\mathclose{} \\
\amp=-\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{2}i.
\end{align*}
So both \(\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}i\) and \(-\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{2}i\) are square roots of \(-1+\sqrt{3}i\text{.}\) But just as \(2\text{,}\) not \(-2\text{,}\) is called the principal square root of \(4\text{,}\) only one of thes two square roots that we found is the principal square root of \(-1+\sqrt{3}i\text{.}\) The principal square root (or principal \(n\)th root) of a complex number is the root with the greatest real component. And if there is a tie between two roots for having the greatest real component, the one with positive imaginary component is the principal root. So the first root we found (i.e., the one we found using \(\theta=\frac{2\pi}{3}\)) is the principal square root of \(-1+\sqrt{3}i\text{,}\) because its real part is \(\frac{\sqrt{2}}{2}\) which is greater than the other root’s real part, \(-\frac{\sqrt{2}}{2}\text{.}\) The princial root is the one represented by the radical symbol, so we can write
\begin{equation*}
\sqrt{-1+\sqrt{3}i}=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}i.
\end{equation*}
Example 2.4.8.
Find \(\sqrt[3]{-4\sqrt{2}+4\sqrt{2}i}\) using the polar form of \(-4\sqrt{2}+4\sqrt{2}i\text{.}\)
Solution.
To find polar forms of \(-4\sqrt{2}+4\sqrt{2}i\text{,}\) we first associate the number with the rectangular ordered pair \(\left( -4\sqrt{2},4\sqrt{2} \right) \text{,}\) and then translate it into polar coordinates \(\left( r,\theta \right) \text{.}\) First let’s find \(r\text{:}\)
\begin{align*}
r \amp= \sqrt{\left( -4\sqrt{2} \right) ^2+ \left( 4\sqrt{2} \right) ^2} \\
\amp= \sqrt{4^2 \cdot 2 + 4^2 \cdot 2} \\
\amp=4 \sqrt{2+2} \\
\amp=8
\end{align*}
Now, let’s find \(\theta\text{:}\)
\begin{align*}
\tan(\theta) \amp= \frac{4\sqrt{2}}{-4\sqrt{2}} \\
\implies \theta \amp= \tan^{-1}(-1)+\pi \amp\amp \text{(add } \pi \text{ since} \left( -4\sqrt{2},4\sqrt{2} \right) \text{is in Quadrant II)}\\
\implies \theta \amp= -\frac{\pi}{4}+\pi \\
\implies \theta \amp= \frac{3\pi}{4}
\end{align*}
Note: we add \(\pi\) since \(\left( -4\sqrt{2},4\sqrt{2} \right) \) is in Quadrant II, outside the range of arctangent.
So the polar form of \(-4\sqrt{2}+4\sqrt{2}i\) is \(z=8e^{\frac{3\pi}{4} \cdot i}\text{.}\) Therefore,
\begin{align*}
\sqrt[3]{-4\sqrt{2}+4\sqrt{2}i} \amp= \left( 8e^{\frac{3\pi}{4} \cdot i} \right) ^{1/3} \\
\amp=8^{\frac{1}{3}}e^{\frac{3\pi}{4} \cdot \frac{1}{3}i} \\
\amp=2e^{\frac{\pi}{4} \cdot i} \\
\amp=2 \mathopen{}\left( \cos \mathopen{}\left( \frac{\pi}{4} \right)\mathclose{} +i \cdot \sin \mathopen{}\left( \frac{\pi}{4} \right)\mathclose{} \right)\mathclose{}\\
\amp=2 \mathopen{}\left( \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i \right)\mathclose{} \\
\amp=\sqrt{2}+\sqrt{2}i
\end{align*}
Exercises 2.4.3 Exercises
Polar Form.
In Exercises 1–3, find the polar form \(z=re^{i\theta}\) of the following complex numbers given in rectangular form.
1.
\(z=6+6\sqrt{3} \cdot i\)
2.
\(z=-2\sqrt{3}+2i\)
3.
\(z=5\sqrt{2}-5 \sqrt{2} \cdot i\)
Rectangular Form.
In Exercises 4–6, find the rectangular form \(z=a+bi\) of the following complex numbers given in polar form.
4.
\(z=8e^{\frac{\pi}{6} \cdot i}\)
5.
\(z=4e^{\pi \cdot i}\)
6.
\(z=5e^{\frac{4\pi}{3} \cdot i}\)
Principal Roots.
In Exercises 7–10, find the following principal roots by first converting to the polar form of each complex number.
7.
\(\sqrt{18-18\sqrt{3}\cdot i}\)
8.
\(\sqrt[3]{-16+16i}\)
9.
\(\sqrt{-i}\)
10.
\(\sqrt[5]{-16\sqrt{3}-16i}\)
11.
Find all three cube roots of \(27i\text{.}\)
12.
Find both of the square roots of \(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\text{.}\)
13.
Find all three solutions to the equation of \(z^3+1=0\text{.}\)
