9.4: Polar Form of Complex Numbers (2024)

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Learning Objectives

Write complex numbers in polar form.
Convert a complex number from polar to rectangular form.
Find products of complex numbers in polar form.
Find quotients of complex numbers in polar form.
Find powers of complex numbers in polar form.
Find roots of complex numbers in polar form.

"God made the integers; all else is the work of man."

This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science.

Writing Complex Numbers in Polar Form

Given a complex number expressed as \(z = x + y \, i\), we can use the material from the previous section to visual \( z \) as the point \( \left( x,y \right) \) on the Cartesian plane.¹ We then observe that we can rewrite both \( x \) and \( y \) in trigonometric form:\[\begin{array}{rcl} x & = & r \cos\left( \theta \right) \\ y & = & r \sin\left( \theta \right) \\ r & = & \sqrt{x^2+y^2} \end{array} \nonumber \]where \( \theta \) is the angle in standard position with terminal side being the line segment connecting the origin to the point \( \left( x,y \right) \). We review these relationships in Figure \(\PageIndex{1}\).

Figure \( \PageIndex{ 1 } \)

Thus, we can rewrite the complex number, \( z = x + y \, i \), in the form\[ z = r \cos\left( \theta \right) + \left( r \sin\left( \theta \right) \right) i = r \left( \cos\left( \theta \right) + i \, \sin\left( \theta \right) \right). \nonumber \]This form is known as the polar form of the complex number and it is often written as\[ z = r \, \mathrm{cis}\left( \theta \right) \nonumber \](\( \mathrm{cis} \) is read as "cosine plus i sine of theta").

As we did in the previous section, we use the term modulus to represent the "size" of a complex number, or the distance from the origin to the point \(\left(x,y\right)\) on the complex plane. The modulus, therefore, is the same as \(r\), the radius in polar form. We use \(\theta\) to indicate the angle of direction (just as with polar coordinates). The following theorem is provided to summarize these findings.

Theorem: Polar Form of a Complex Number

The complex number, \( z = x + y \, i \), can be written in the form\[ z = r \left( \cos\left( \theta \right) + i \, \sin\left( \theta \right) \right) \equiv r \, \mathrm{cis}\left( \theta \right), \nonumber \]where\[\begin{array}{rcl} x & = & r \cos\left( \theta \right) \\ y & = & r \sin\left( \theta \right) \\ r & = & \sqrt{x^2+y^2} \\ \end{array} \nonumber \]\(r\) is the modulus of \( z \), and \(\theta\) is the argument.

Example \(\PageIndex{1}\)

Express the complex number \(4i\) in polar form.

Solution

On the complex plane, the number \(z=4i\) is the same as \(z=0+4i\). To write this in polar form, we must first calculate \(r\).\[\begin{array}{rcl} r & = & \sqrt{x^2+y^2} \\ & = & \sqrt{0^2+4^2} \\ & = & \sqrt{16} \\ & = & 4 \end{array} \nonumber \]Next, we look at \(x\). If \(x=r \cos\left( \theta \right)\) and \(x=0\), then \(\theta=\frac{\pi}{2}\). Therefore, in polar form, the complex number \(z=0+4i\) can be written as\[z = 4\left(\cos\left(\dfrac{\pi}{2}\right)+i \, \sin\left(\dfrac{\pi}{2}\right)\right) = 4 \, \mathrm{cis}\left( \frac{\pi}{2}\right). \nonumber \]See Figure \(\PageIndex{2}\).

Figure \( \PageIndex{ 2 } \)

Checkpoint \(\PageIndex{1}\)

Express \(z=3i\) as \(r \, \mathrm{cis}\left( \theta \right)\) in polar form.

Answer: \(z=3\left(\cos\left(\dfrac{\pi}{2}\right)+i \, \sin\left(\dfrac{\pi}{2}\right)\right)\) or \( z = 3 \, \mathrm{cis}\left( \dfrac{\pi}{2} \right) \)

Example \(\PageIndex{2}\)

Find the polar form of \(−4+4i\).

Solution: We begin by finding the value of \(r\).\[\begin{array}{rcl}
r & = & \sqrt{x^2+y^2} \\
& = & \sqrt{(−4)^2+4^2} \\
& = & \sqrt{32} \\
& = & 4\sqrt{2} \\
\end{array} \nonumber \]We now find the angle \(\theta\).\[\begin{array}{rrcl}
& \cos\left( \theta \right) & = & \dfrac{x}{r} \\
\implies & \cos\left( \theta \right) & = & -\dfrac{4}{4\sqrt{2}} \\
\implies & \cos\left( \theta \right) & = & -\dfrac{1}{\sqrt{2}} \\
\implies & \widehat{\theta} & = & \cos^{−1} \left(\dfrac{1}{\sqrt{2}}\right) \\
\implies & \widehat{\theta} & = & \dfrac{\pi}{4} \\
\end{array} \nonumber \]Using the fact that \( z = -4 + 4i \) is in the second quadrant, we get \( \theta = \frac{3\pi}{4} \). Thus, the polar form is\[ z = 4\sqrt{2} \, \mathrm{cis} \left(\dfrac{3\pi}{4}\right). \nonumber \]

It's very tempting to evaluate the trigonometric functions at \( \frac{3\pi}{4} \) in Example \( \PageIndex{ 2 } \). I would not blame you for wanting to do that because that's what we have been doing this entire course; however, in doing so, you would be reverting back to the standard (recangular) form of the complex number (thereby undoing all of your work to get the number in polar form).

Checkpoint \(\PageIndex{2}\)

Write \(z=\sqrt{3}+i\) in polar form.

Answer: \(z=2\left(\cos\left(\dfrac{\pi}{6}\right)+ i \, \sin\left(\dfrac{\pi}{6}\right)\right) = 2 \mathrm{cis}\left( \dfrac{\pi}{6} \right)\)

Converting a Complex Number from Polar to Rectangular Form

Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given\[z = r\left(\cos\left( \theta \right)+i \, \sin\left( \theta \right)\right), \nonumber \]we first evaluate the trigonometric functions \(\cos\left( \theta \right)\) and \(\sin\left( \theta \right)\), then multiply through by \(r\).

Example \(\PageIndex{3}\)

Convert the polar form of the given complex number to rectangular form:\[ z = 12\left(\cos\left(\dfrac{\pi}{6}\right)+i \, \sin\left(\dfrac{\pi}{6}\right)\right)\nonumber \]

Solution: We begin by evaluating the trigonometric expressions.\[\begin{array}{rclcrcl}
\cos\left(\dfrac{\pi}{6}\right) & = & \dfrac{\sqrt{3}}{2} & \text{ and } & \sin(\dfrac{\pi}{6}) & = & \dfrac{1}{2}\\ \end{array} \nonumber \]After substitution, the complex number is\[ z = 12\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2} \, i\right) = 6\sqrt{3} + 6 i \nonumber \]Hence, the rectangular form of the given point in polar form is \(6\sqrt{3}+6i\).

Checkpoint \(\PageIndex{3}\)

Convert the complex number to rectangular form:\[z = 4\left(\cos \left(\dfrac{11\pi}{6}\right) + i \, \sin \left(\dfrac{11\pi}{6}\right)\right) \nonumber \]

Answer: \(z=2\sqrt{3}−2i\)

Finding Products of Complex Numbers in Polar Form

Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.

Theorem: Products of Complex Numbers in Polar Form

If \(z_1=r_1(\cos\left( \theta_1 \right)+i \, \sin\left( \theta_1 \right))\) and \(z_2=r_2(\cos\left( \theta_2 \right)+i \, \sin\left( \theta_2 \right))\), then the product of these numbers is given as\[ z_1z_2 = r_1r_2 \left[ \cos\left(\theta_1+\theta_2\right) + i \, \sin\left(\theta_1+\theta_2\right) \right] = r_1 r_2 \, \mathrm{cis}\left(\theta_1+\theta_2\right).\nonumber \]

Notice that the product calls for multiplying the moduli and adding the angles. The proof of this theorem is left as a homework exercise.

Example \(\PageIndex{4}\)

Find the product of \(z_1z_2\), given \(z_1 = 4 \left(\cos\left(80^{\circ}\right) + i \, \sin\left(80^{\circ}\right)\right)\) and \(z_2 = 2 \, \mathrm{cis}\left(145^{\circ}\right)\).

Solution

Follow the formula

\[\begin{array}{rcl} z_1 z_2 & = & 4 \cdot 2 \left[ \cos\left(80^{\circ} +145^{\circ}\right) + i \, \sin\left(80^{\circ} +145^{\circ} \right)\right] \\
& = & 8 \left[\cos\left(225^{\circ}\right) + i \, \sin\left(225^{\circ}\right) \right] \\
& = & 8 \left[−\dfrac{\sqrt{2}}{2} + i \, \left(−\dfrac{\sqrt{2}}{2}\right) \right] \\
& = & −4\sqrt{2} − 4i\sqrt{2} \\
\end{array} \nonumber \]

Checkpoint \(\PageIndex{4}\)

Let \( z_1 = -\sqrt{3} + i \) and \( z_2 = 3 + 3 i \).

Find the product of \( z_1 \) and \( z_2 \) in rectangular form (without converting to polar form).
Without converting to polar form, determine the angle \( z_1 z_2 \) makes with the positive \( x \)-axis.
Rewrite \( z_1 \) and \( z_2 \) in polar form.
Find the product of \( z_1 \) and \( z_2 \) in polar form.
Using your answer from part (d), determine the angle \( z_1 z_2 \) makes with the positive \( x \)-axis.

Answers

\( z_1 z_2 =(-3 -3\sqrt{3})+ (3 - 3\sqrt{3}) i\)
\( 195^{ \circ } \)
\( z_1 = 2 \left( \cos\left( 150^{ \circ } \right) + i \, \sin\left( 150^{ \circ } \right) \right) \) and \( z_2 = 3\left( \cos\left( 45^{ \circ} \right) + i \, \sin\left( 45^{ \circ }\right) \right) \)
\( z_1 z_2 = 6 \left( \cos\left( 195^{ \circ } \right) + i \, \sin\left( 195^{ \circ } \right) \right)\)
\( 195^{ \circ } \)

Finding Quotients of Complex Numbers in Polar Form

The quotient of two complex numbers in polar form follows the same patterns set for products.

Theorem: Quotients of Complex Numbers in Polar Form

If \(z_1 = r_1 \left(\cos\left( \theta_1 \right) + i \, \sin\left( \theta_1 \right)\right) \) and \(z_2 = r_2 \left( \cos\left( \theta_2 \right) + i \, \sin\left( \theta_2 \right) \right) \), then the quotient of these numbers is\[\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} \left[ \cos\left(\theta_1−\theta_2\right) + i \, \sin\left(\theta_1−\theta_2\right) \right] = \dfrac{r_1}{r_2} \, \mathrm{cis}\left(\theta_1−\theta_2\right), \nonumber \]where \( z_2 \neq 0\).

The proof of this theorem is left as a homework exercise.

Example \(\PageIndex{5}\)

Find the quotient of \(z_1 = 2\left(\cos(213^{\circ} )+i \, \sin(213^{\circ} )\right)\) and \(z_2=4\left(\cos(33^{\circ} )+i \, \sin(33^{\circ} )\right)\).

Solution: Using the formula, we have\[\begin{array}{rcl}
\dfrac{z_1}{z_2} & = & \dfrac{2}{4}\left[\cos(213^{\circ} −33^{\circ} )+i \, \sin(213^{\circ} −33^{\circ} )\right] \\
& = & \dfrac{1}{2}\left[\cos(180^{\circ} )+i \, \sin(180^{\circ} )\right] \\
& = & \dfrac{1}{2}[−1+0i] \\
& = & −\dfrac{1}{2}+0i \\
& = & −\dfrac{1}{2} \\
\end{array} \nonumber \]

Checkpoint \(\PageIndex{5}\)

Find the product and the quotient of \(z_1=2\sqrt{3}\left(\cos(150^{\circ} )+i \, \sin(150^{\circ} )\right)\) and \(z_2=2\left(\cos(30^{\circ} )+i \, \sin(30^{\circ} )\right)\).

Answer: \(z_1z_2=−4\sqrt{3}\); \(\dfrac{z_1}{z_2}=−\dfrac{\sqrt{3}}{2}+\dfrac{3}{2}i\)

Finding Powers of Complex Numbers in Polar Form

Finding powers of complex numbers is greatly simplified when they are written in polar form. For example, let \( z = 1 + i \). This is a fairly innocuous complex number; however, let's try to compute \( z^5 = \left( 1 + i \right)^5 \).\[ \begin{array}{rcl}
z^5 & = & \left( 1 + i \right)^5 \\
& = & \left( 1 + i \right)^2 \left( 1 + i \right)^2 \left( 1 + i \right) \\
& = & \left( 1 + 2i + i^2 \right) \left( 1 + 2i + i^2 \right) \left( 1 + i \right) \\
& = & \left( 1 + 2i - 1 \right) \left( 1 + 2i - 1 \right) \left( 1 + i \right) \\
& = & \left( 2i \right) \left( 2i \right) \left( 1 + i \right) \\
& = & 4i^2 \left( 1 + i \right) \\
& = & -4 \left( 1 + i \right) \\
& = & -4 - 4i \\
\end{array} \nonumber \]This is only the fifth power of a somewhat simple complex number. Imagine working with the tenth power of something like \( z = -\sqrt{3} + i \).If our complex number is written in polar form, however, something fantastic happens.

Theorem: De Moivre's Theorem

If \(z = r\left(\cos\left( \theta \right) + i \, \sin\left( \theta \right)\right)\) is a complex number and \( n \in \mathbb{N} \), then\[z^n = r^n \left[\cos(n\theta)+ i \, \sin(n\theta) \right] = r^n \, \mathrm{cis}(n\theta). \nonumber \]

The following derivation of why De Moivre's Theorem works requires a proof method called Mathematical Induction. If you have already taken College Algebra, then you have worked with, and performed, proofs using Mathematical Induction;² however, since College Algebra is not a prerequisite for this course, you might not have seen it. As such, I provide the proof only for those who are interested.

Proof of De Moivre's Theorem

Let \(z = r\left(\cos\left( \theta \right) + i \, \sin\left( \theta \right)\right)\) and \( n \in \mathbb{N} \).

CLAIM: \( z^n = r^n \left[\cos(n\theta)+ i \, \sin(n\theta) \right] \).

For \( n = 1 \),\[ \begin{array}{rclcl}
z^1 & = & \left( r \left( \cos\left( \theta \right) + i \, \sin\left( \theta \right) \right) \right)^1 & \quad & \\
& = & r^1 \left( \cos\left( \theta \right) + i \, \sin\left( \theta \right) \right)^1 & \quad & \left( \text{Laws of Exponents} \right) \\
& = & r^1 \left( \cos\left( 1 \cdot \theta \right) + i \, \sin\left( 1 \cdot \theta \right) \right) & \quad & \left( \text{Multiplicative Identity} \right)\\
\end{array} \nonumber \]Thus the claim is true for \( n = 1 \).

Induction Hypothesis: Suppose the claim is true for \( n = k \). That is, suppose \( z^k = r^k \left[\cos(k\theta)+ i \, \sin(k\theta) \right] \).

Assuming this Induction Hypothesis,\[ \begin{array}{rclcl}
z^{k + 1} & = & z^k \cdot z^1 & \quad & \left( \text{Laws of Exponents} \right) \\
& = & r^k \left[\cos(k\theta)+ i \, \sin(k\theta) \right] \cdot z^1 & \quad & \left( \text{Induction Hypothesis} \right) \\
& = & r^k \left[\cos(k\theta)+ i \, \sin(k\theta) \right] \cdot r\left[\cos\left( \theta \right) + i \, \sin\left( \theta \right)\right] & \quad & \left( \text{substituting }z = r\left(\cos\left( \theta \right) + i \, \sin\left( \theta \right)\right) \right) \\
& = & r^{k + 1} \left[\cos(k\theta)+ i \, \sin(k\theta) \right] \left[\cos\left( \theta \right) + i \, \sin\left( \theta \right)\right] & \quad & \left( \text{Commutative Property of Multiplication} \right) \\
& = & r^{k + 1} \left[\cos(k\theta)\cos\left( \theta \right) + i \, \cos(k\theta) \sin\left( \theta \right) + i \, \sin\left( k \theta \right) \cos\left( \theta \right) + i^2 \sin\left( k \theta \right) \sin\left( \theta \right)\right] & \quad & \left( \text{distributing} \right) \\
& = & r^{k + 1} \left[\cos(k\theta)\cos\left( \theta \right) + i \, \cos(k\theta) \sin\left( \theta \right) + i \, \sin\left( k \theta \right) \cos\left( \theta \right) - \sin\left( k \theta \right) \sin\left( \theta \right)\right] & \quad & \left( i^2 = -1 \right) \\
& = & r^{k + 1} \left[\cos(k\theta)\cos\left( \theta \right) - \sin\left( k \theta \right) \sin\left( \theta \right) + i \, \cos(k\theta) \sin\left( \theta \right) + i \, \sin\left( k \theta \right) \cos\left( \theta \right) \right] & \quad & \left( \text{Commutative Property of Addition} \right) \\
& = & r^{k + 1} \left[\left(\cos(k\theta)\cos\left( \theta \right) - \sin\left( k \theta \right) \sin\left( \theta \right)\right) + i \left( \cos(k\theta) \sin\left( \theta \right) + \sin\left( k \theta \right) \cos\left( \theta \right) \right) \right] & \quad & \left( \text{grouping and factoring} \right) \\
& = & r^{k + 1} \left[\cos\left((k+1)\theta \right) + i \sin\left( (k + 1) \theta \right) \right] & \quad & \left( \text{Sum of Angles Identities} \right) \\
\end{array} \nonumber \]Thus the claim is true for \( n = k + 1 \).

Since the claim holds for \( n = 1 \), and subsequently for \( n = k + 1 \) when it is assumed it works for \( n = k \), the claim must be true for all \( n \in \mathbb{N} \).

De Moivre’s Theorem states that, for a positive integer \(n\), \(z^n\) is found by raising the modulus to the \(n^{th}\) power and multiplying the argument by \(n\). It is the standard method used in modern Mathematics.

Example \(\PageIndex{6}\)

Evaluate the expression \({(1+i)}^5\) using De Moivre’s Theorem.

Solution

Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write \((1+i)\) in polar form. Let us find \(r\).\[ \begin{array}{rcl}
r & = & \sqrt{x^2+y^2} \\
& = & \sqrt{{(1)}^2+{(1)}^2} \\
& = & \sqrt{2} \\
\end{array} \nonumber \]Then we find \(\theta\). Using the formula \(\tan \theta=\dfrac{y}{x}\) gives\[\begin{array}{rrcl}
& \tan \left(\theta\right) & = & \dfrac{1}{1} \\
\implies & \tan \left(\theta\right) & = & 1 \\
\implies & \widehat{\theta} & = & \dfrac{\pi}{4} \\
\end{array} \nonumber \]Since \( (1 + i) \) is in the first quadrant, this means \( \theta = \frac{\pi}{4} \).

Now we use De Moivre’s Theorem to evaluate the expression.\[ \begin{array}{rcl}
(1+i)^5 & = & (\sqrt{2})^5 \left[ \cos\left(5 \cdot \dfrac{\pi}{4}\right) + i \, \sin\left(5 \cdot \dfrac{\pi}{4}\right) \right] \\
& = & 4\sqrt{2}\left[ \cos\left(\dfrac{5\pi}{4}\right)+i \, \sin\left(\dfrac{5\pi}{4}\right) \right] \\
& = & 4\sqrt{2}\left[ −\dfrac{\sqrt{2}}{2} + i \, \left(−\dfrac{\sqrt{2}}{2}\right) \right] \\
& = & −4−4i \\
\end{array} \nonumber \]

Checkpoint \(\PageIndex{6}\)

Compute \( \left( -\sqrt{3} + i \right)^{10} \).

Answer: \( 2^{10} \, \mathrm{cis}\left( \dfrac{25 \pi}{3} \right) = 1024 \, \mathrm{cis}\left( \dfrac{\pi}{3}\right) = 1024 \left( \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\, i \right) = 512 + 512 \, i \sqrt{3} \)

Finding Roots of Complex Numbers in Polar Form

DeMoivre's Theorem also works for fractional values of \(n\), so we can compute roots of complex numbers. For example, by applying the theorem with \(n=\frac{1}{2}\), we see that one of the square roots of \(z=r(\cos\left( \theta \right)+i \sin\left( \theta \right))\) is\[ z^{1 / 2} = r^{1 / 2}\left(\cos \left(\dfrac{\theta}{2}\right)+i \, \sin \left(\dfrac{\theta}{2}\right)\right). \nonumber \]Now, every number, whether real or complex, has two square roots. To find the other root, remember that we can add a multiple of \(2 \pi\) to the argument of \(z\), that is, we can also write the polar form of \(z\) as\[z=r(\cos (\theta+2 \pi)+i \, \sin (\theta+2 \pi))\).

The second square root of \(z\) is thus\[r^{1 / 2}\left(\cos \left(\dfrac{\theta}{2}+\pi\right)+i \, \sin \left(\dfrac{\theta}{2}+\pi\right)\right). \nonumber \]As an example, consider \(w=-8+8 i \sqrt{3}\), whose polar form is \(16\left(\cos \left(\frac{2 \pi}{3}\right)+i \,\sin\left( \frac{2 \pi}{3}\right)\right)\), or by adding \(2 \pi\) to the argument, \(16\left(\cos \left(\frac{8 \pi}{3}\right)+i \, \sin \left(\frac{8 \pi}{3}\right)\right)\). The two square roots of \(w\) are\[ \begin{array}{rclcrcl}
z_1 & = & 16^{1 / 2}\left(\cos \left(\dfrac{\pi}{3}\right) + i \, \sin \left(\dfrac{\pi}{3}\right)\right) & \quad &z_2 & = & 16^{1 / 2}\left(\cos\left( \dfrac{4 \pi}{3}\right) + i \, \sin\left(\dfrac{4 \pi}{3}\right)\right) \\
\\
& = & 4\left(\dfrac{1}{2}+i \, \dfrac{\sqrt{3}}{2}\right) & \quad & & = & 4\left(\dfrac{-1}{2}+i \dfrac{-\sqrt{3}}{2}\right) \\
\\
&= & 2+2 i \sqrt{3} & \quad && = & -2-2 i \sqrt{3} \\
\end{array} \nonumber \]You can verify that both of these numbers are square roots of \(w=-8+8 i \sqrt{3}\), for instance,\[ \begin{array}{rcl}
(2+2 i \sqrt{3})^2 & = & 4+2(2)(2 i \sqrt{3})+12 i^2 \\
& = & -8+8 i \sqrt{3} \\
\end{array} \nonumber \]The graphs of \(w\) and its two square roots are shown below.

It is not hard to show that every number has three complex cube roots, four complex fourth roots, and so on.

In general, to find the \(n^{th}\) root of a complex number in polar form, we use the \(n^{th}\) Root Theorem, or we can use De Moivre’s Theorem as we did in our work above. The following theorem is presented without proof.

Theoren: The \(n^{th}\) Root Theorem

If \( z = r \left( \cos\left( \theta \right) + i \, \sin\left( \theta \right) \right) \) is a complex number, the \( n^{th} \) root of \( z \) is\[ z^{\tfrac{1}{n}} = r^{\tfrac{1}{n}}\left[ \cos\left(\dfrac{\theta}{n}+\dfrac{2k\pi}{n}\right)+i \sin\left(\dfrac{\theta}{n}+\dfrac{2k\pi}{n}\right) \right]. \nonumber \]where \(k=0, 1, 2, 3, \ldots, n−1\). We add \(\frac{2k\pi}{n}\) to \(\frac{\theta}{n}\) in order to obtain the periodic roots.

Example \( \PageIndex{ 7 } \)

Find the cube roots of \(z = 8\left(\cos\left(\frac{2\pi}{3}\right)+i \, \sin\left(\frac{2\pi}{3}\right)\right)\).

Solution

We have\[ \begin{array}{rcl}
z^{\frac{1}{3}} & = & 8^{\frac{1}{3}}\left[ \cos\left(\frac{\frac{2\pi}{3}}{3}+\frac{2k\pi}{3}\right)+i \, \sin\left(\frac{\frac{2\pi}{3}}{3}+\frac{2k\pi}{3}\right) \right] \\
\\
& = & 2\left[ \cos\left(\frac{2\pi}{9}+\frac{2k\pi}{3}\right)+i \, \sin\left(\frac{2\pi}{9}+\frac{2k\pi}{3}\right) \right] \\
\end{array} \nonumber \]There will be three roots - one for each of \(k=0\) , \(1\), and \(2\).

When \(k=0\), we have\[ z^{\frac{1}{3}} = 2\left(\cos\left(\dfrac{2\pi}{9}\right)+i \, \sin\left(\dfrac{2\pi}{9}\right)\right).\nonumber \]When \(k=1\), we have\[\begin{array}{rclcl}
z^{\frac{1}{3}} & = & 2\left[ \cos\left(\dfrac{2\pi}{9}+\dfrac{6\pi}{9}\right)+i \, \sin\left(\dfrac{2\pi}{9}+\dfrac{6\pi}{9}\right) \right] & \quad & \left(\text{adding }\dfrac{2(1)\pi}{3} \text{ to each angle}\right) \\
\\
& = & 2\left(\cos\left(\dfrac{8\pi}{9}\right)+i \, \sin\left(\dfrac{8\pi}{9}\right)\right) & & \\
\end{array} \nonumber \]When \(k=2\), we have\[ \begin{array}{rclcl}
z^{\frac{1}{3}} & = & 2\left[ \cos\left(\dfrac{2\pi}{9}+\dfrac{12\pi}{9}\right)+i \, \sin\left(\dfrac{2\pi}{9}+\dfrac{12\pi}{9}\right) \right] & \quad & \left(\text{adding }\dfrac{2(2)\pi}{3} \text{ to each angle}\right) \\
& = & 2\left(\cos\left(\dfrac{14\pi}{9}\right)+i \, \sin\left(\dfrac{14\pi}{9}\right)\right) & & \\
\end{array} \nonumber \]Remember to find the common denominator to simplify fractions in situations like this one. For \(k=1\), the angle simplification is\[ \begin{array}{rcl}
\dfrac{\dfrac{2\pi}{3}}{3}+\dfrac{2(1)\pi}{3} & = & \dfrac{2\pi}{3}(\dfrac{1}{3})+\dfrac{2(1)\pi}{3}\left(\dfrac{3}{3}\right) \\
\\
& = & \dfrac{2\pi}{9}+\dfrac{6\pi}{9} \\
\\
& = & \dfrac{8\pi}{9} \\
\end{array} \nonumber \]

Exercise \(\PageIndex{7}\)

Find the four fourth roots of \(16(\cos(120^{\circ} )+i \, \sin(120^{\circ} ))\).

Answer

\(z_0 = 2(\cos(30^{\circ} )+i \, \sin(30^{\circ} ))\)

\(z_1=2(\cos(120^{\circ} )+i \, \sin(120^{\circ} ))\)

\(z_2=2(\cos(210^{\circ} )+i \, \sin(210^{\circ} ))\)

\(z_3=2(\cos(300^{\circ} )+i \, \sin(300^{\circ} ))\)

Homework

Basic Skills

For Problems 1 - 4, simplify and plot each complex number as a point on the complex plane.

\(1, i, i^2, i^3\) and \(i^4\)
\(-1,-i,-i^2,-i^3\) and \(-i^4\)
\(1+2 i\) and \(i(1+2 i)\)
\(3-4 i\) and \(i(3-4 i)\)

For Problems 5 - 8, write the complex numbers in standard form. Give exact values for your answers.

\(6\left(\cos \dfrac{2 \pi}{3}+i \, \sin \dfrac{2 \pi}{3}\right)\)
\(4\left(\cos \dfrac{7 \pi}{4}+i \, \sin \dfrac{7 \pi}{4}\right)\)
\(\sqrt{2}\left(\cos \dfrac{3 \pi}{4}+i \, \sin \dfrac{3 \pi}{4}\right)\)
\(\dfrac{3}{2}\left(\cos \dfrac{5 \pi}{6}+i \, \sin \dfrac{5 \pi}{6}\right)\)

For Problems 9 - 12, write the complex numbers in standard form. Round your answers to hundredths.

\(5(\cos 5.2+i \, \sin 5.2)\)
\(3(\cos 3.5+i \, \sin 3.5)\)
\(12\left(\cos 115^{\circ}+i \, \sin 115^{\circ}\right)\)
\(20\left(\cos 250^{\circ}+i \, \sin 250^{\circ}\right)\)

For Problems 13 - 16, write the complex numbers in polar form. Give exact values for your answers.

\(3 i\) and \(-3 i\)
\(2+2 i\) and \(2-2 i\)
\(-3-\sqrt{3} i\) and \(3-\sqrt{3} i\)
\(2 \sqrt{3}+2 i\) and \(-2 \sqrt{3}+2 i\)

For Problems17 - 22, write the complex numbers in polar form. Round your answers to hundredths.

\(-4+2 i\) and \(4-2 i\)
\(-3-8 i\) and \(3+8 i\)
\(9-5 i\) and \(9+5 i\)
\(2+6 i\) and \(2-6 i\)
\(3+4 i, 3-4 i,-3+4 i\), and \(-3-4 i\)
\(1+3 i, 1-3 i,-1+3 i\), and \(-1-3 i\)
What can you conclude about the polar forms of \(z\) and \(\bar{z}\)?
What can you conclude about the polar forms of \(z\) and \(-z\)?

For Problems25 - 28, find the product \(z_1 z_2\) and the quotient \(\frac{z_1}{z_2}\).

\(z_1=4\left(\cos \dfrac{4 \pi}{3}+i \, \sin \dfrac{4 \pi}{3}\right)\) and \(z_2=\dfrac{1}{2}\left(\cos \dfrac{5 \pi}{6}+i \, \sin \dfrac{5 \pi}{6}\right)\)
\(z_1=6\left(\cos \dfrac{5 \pi}{8}+i \, \sin \dfrac{5 \pi}{8}\right)\) and\(z_2=\dfrac{3}{2}\left(\cos \dfrac{\pi}{8}+i \, \sin \dfrac{\pi}{8}\right)\)
\(z_1=3\left(\cos \dfrac{3 \pi}{5}+i \, \sin \dfrac{3 \pi}{5}\right)\) and\(z_2=2\left(\cos \dfrac{3 \pi}{10}+i \, \sin \dfrac{3 \pi}{10}\right)\)
\(z_1=4\left(\cos \dfrac{5 \pi}{12}+i \, \sin \dfrac{5 \pi}{12}\right)\) and \(z_2=6\left(\cos \dfrac{3 \pi}{4}+i \, \sin \dfrac{3 \pi}{4}\right)\)

For Problems 29 - 32, convert the complex number to polar form, then find the product \(z_1 z_2\) and the quotient \(\frac{z_1}{z_2}\).

\(z_1=2 i, z_2=4 i\)
\(z_1=-2 i, z_2=3 i\)
\(z_1=2 \sqrt{3}-2 i, \quad z_2=-1+i\)
\(z_1=\sqrt{3}+i, \quad z_2=-1+\sqrt{3}\)

For Problems33 - 38, find the power.

\((2+2 i)^5\)
\((\sqrt{2}-\sqrt{2} i)^6\)
\((-1+\sqrt{3} i)^8\)
\(\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} i\right)^{12}\)
\((\sqrt{3}-i)^{10}\)
\((1-i)^{20}\)

For Problems39 - 42, use DeMoivre’s theorem to find the reciprocal.

\(2-2 i\)
\(3+\sqrt{3} i\)
\(-\sqrt{2}+\sqrt{6} i\)
\(-1-i\)

For Problems43 - 48,
(a) Find the roots and plot them in the complex plane.
(b) Write the roots in standard form.

The square roots of \(9 i\).
The fourth roots of \(-81\).
The fifth roots of 32.
The cube roots of \(i\).
The cube roots of \(4 \sqrt{3}+4 i\).
The square roots of \(-2+2 \sqrt{3} i\).
Show that any complex number of the form \(z=\cos \theta+i \, \sin \theta\) lies on the unit circle in the complex plane.
Show that if \(|z|=1\), then \(\dfrac{1}{z}=\bar{z}\).
1. Find three distinct cube roots of 1.
2. Find four distinct fourth roots of 1.
3. Find five distinct fifth roots of 1.
4. Find six distinct sixth roots of 1.
1. Find the sum of the three distinct cube roots of 1. (Hint: Plot the roots.)
2. Find the sum of the four distinct fourth roots of 1.
3. Find the sum of the five distinct fifth roots of 1.
4. Find the sum of the six distinct sixth roots of 1.
If \(n\) is a positive integer, define\[\omega_k=\cos \dfrac{2 \pi k}{n}+i \, \sin \dfrac{2 \pi k}{n} \nonumber \]for \(k=0,1,2, \cdots, n-1\). Show that \(\left(\omega_k\right)^n=1\). (We call \(\omega_k\) an \(n^{t h}\) root of unity.)
Let \(\omega=\cos \frac{2 \pi}{n}+i \, \sin \frac{2 \pi}{n}\), where \(n\) is a positive integer. Show that the \(n\) distinct \(n^{\text {th }}\) roots of unity are \(\omega, \omega^2, \omega^3, \cdots, \omega^{n-1}\).

For Problems 55 - 60, solve the equation.

\(z^4+4 z^2+8=0\)
\(z^6+4 z^3+8=0\)
\(z^6-8=0\)
\(z^4-9 i=0\)
\(z^4+2 z^2+4=0\)
\(z^4-2 z^2+4=0\)
1. Let \(z=\cos \theta+i \, \sin \theta\). Compute \(z^2\) by expanding the product.
2. Use DeMoivre's theorem to compute \(z^2\).
3. Compare your answers to (a) and (b) to write identities for \(\sin 2\theta\) and \(\cos 2\theta\).
1. Let \(z=\cos \theta+i \, \sin \theta\). Compute \(z^3\) by expanding the product.
2. Use DeMoivre's theorem to compute \(z^3\).
3. Compare your answers to (a) and (b) to write identities for \(\sin 3 \theta\) and \(\cos 3 \theta\).

Problems 63 and 64 show that multiplication by \(i\) results in a rotation of \(90^{\circ}\).

Suppose that \(z=a+b i\) and that the real numbers \(a\) and \(b\) are both nonzero.
1. What is the slope of the segment in the complex plane joining the origin to \(z\)?
2. What is the slope of the segment in the complex plane joining the origin to \(z i\)?
3. What is the product of the slopes of the two segments from parts (a) and (b)? What can you conclude about the angle between the two segments?
Suppose that \(z=a+b i\) and that \(a\) and \(b\) are both real numbers.
1. If \(a \neq 0\) and \(b=0\), then what is the slope of the segment in the complex plane joining the origin to \(z\)? What is the slope of the segment joining the origin to \(z i\)?
2. If \(a=0\) and \(b \neq 0\), then what is the slope of the segment in the complex plane joining the origin to \(z\)? What is the slope of the segment joining the origin to \(z i\)?
3. What can you conclude about the angle between the two segments from parts (a) and (b)?

Proofs

Prove the product rule by following the steps.
1. Suppose \(z_1=a+b i\) and \(z_2=c+d i\). Compute \(z_1 z_2\).
2. Now suppose that \(z_1=r(\cos \alpha+i \, \sin \alpha)\) and \(z_2=R(\cos \beta+i \, \sin \beta)\). Write \(a, b, c\) and \(d\) in terms of \(r, R, \alpha\) and \(\beta\).
3. Substitute your expressions for \(a, b, c\) and \(d\) into your formula for \(z_1 z_2\).
4. Use the laws of sines and cosines to simplify your answer to part (c).
Let \(z_1=r(\cos \alpha+i \, \sin \alpha)\) and \(z_2=R(\cos \beta+i \, \sin \beta)\). Prove the quotient rule as follows: Set \(w=\frac{r}{R}(\cos (\alpha-\beta)+i \, \sin (\alpha-\beta))\), and show that \(z_1=w z_2\).