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Guias de estudo > Precalculus II

Polar Form of Complex Numbers

Learning Objectives

By the end of this section, you will be able to:
  • Plot complex numbers in the complex plane.
  • Find the absolute value of a complex number.
  • 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. We first encountered complex numbers in Precalculus I. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem.

Plotting Complex Numbers in the Complex Plane

Plotting a complex number a+bia+bi is similar to plotting a real number, except that the horizontal axis represents the real part of the number, aa, and the vertical axis represents the imaginary part of the number, bibi.

How To: Given a complex number a+bia+bi, plot it in the complex plane.

  1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis.
  2. Plot the point in the complex plane by moving aa units in the horizontal direction and bb units in the vertical direction.

Example 1: Plotting a Complex Number in the Complex Plane

Plot the complex number 2−3i2 - 3i in the complex plane.

Solution

From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction.
Plot of 2-3i in the complex plane (2 along the real axis, -3 along the imaginary axis). Figure 1

Try It 1

Plot the point 1+5i1+5i in the complex plane. Solution

 Finding the Absolute Value of a Complex Number

The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude, or ∣z∣|z|. It measures the distance from the origin to a point in the plane. For example, the graph of z=2+4iz=2+4i, in Figure 2, shows ∣z∣|z|.
Plot of 2+4i in the complex plane and its magnitude, |z| = rad 20. Figure 2

A General Note: Absolute Value of a Complex Number

Given z=x+yiz=x+yi, a complex number, the absolute value of zz is defined as
∣z∣=x2+y2|z|=\sqrt{{x}^{2}+{y}^{2}}
It is the distance from the origin to the point (x,y)\left(x,y\right). Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, (0, 0)\left(0,\text{ }0\right).

Example 2: Finding the Absolute Value of a Complex Number with a Radical

Find the absolute value of z=5−iz=\sqrt{5}-i.

Solution

Using the formula, we have
∣z∣=x2+y2∣z∣=(5)2+(−1)2∣z∣=5+1∣z∣=6\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(\sqrt{5}\right)}^{2}+{\left(-1\right)}^{2}}\hfill \\ |z|=\sqrt{5+1}\hfill \\ |z|=\sqrt{6}\hfill \end{array}
Plot of z=(rad5 - i) in the complex plane and its magnitude rad6. Figure 3

Try It 2

Find the absolute value of the complex number z=12−5iz=12 - 5i. Solution

Example 3: Finding the Absolute Value of a Complex Number

Given z=3−4iz=3 - 4i, find ∣z∣|z|.

Solution

Using the formula, we have
∣z∣=x2+y2∣z∣=(3)2+(−4)2∣z∣=9+16∣z∣=25∣z∣=5\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ |z|=\sqrt{{\left(3\right)}^{2}+{\left(-4\right)}^{2}}\hfill \\ |z|=\sqrt{9+16}\hfill \\ \begin{array}{l}|z|=\sqrt{25}\\ |z|=5\end{array}\hfill \end{array}
The absolute value zz is 5.
Plot of (3-4i) in the complex plane and its magnitude |z| =5. Figure 4

Try It 3

Given z=1−7iz=1 - 7i, find ∣z∣|z|. Solution

 Writing Complex Numbers in Polar Form

The polar form of a complex number expresses a number in terms of an angle θ\theta and its distance from the origin rr. Given a complex number in rectangular form expressed as z=x+yiz=x+yi, we use the same conversion formulas as we do to write the number in trigonometric form:
x=rcos⁡θy=rsin⁡θr=x2+y2\begin{array}{l}x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}
We review these relationships in Figure 5.
Triangle plotted in the complex plane (x axis is real, y axis is imaginary). Base is along the x/real axis, height is some y/imaginary value in Q 1, and hypotenuse r extends from origin to that point (x+yi) in Q 1. The angle at the origin is theta. There is an arc going through (x+yi). Figure 5
We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point (x,y)\left(x,y\right). The modulus, then, is the same as rr, the radius in polar form. We use θ\theta to indicate the angle of direction (just as with polar coordinates). Substituting, we have
z=x+yiz=rcos⁡θ+(rsin⁡θ)iz=r(cos⁡θ+isin⁡θ)\begin{array}{l}z=x+yi\hfill \\ z=r\cos \theta +\left(r\sin \theta \right)i\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}

A General Note: Polar Form of a Complex Number

Writing a complex number in polar form involves the following conversion formulas:
x=rcos⁡θy=rsin⁡θr=x2+y2\begin{array}{l}\hfill \\ x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}
Making a direct substitution, we have
z=x+yiz=(rcos⁡θ)+i(rsin⁡θ)z=r(cos⁡θ+isin⁡θ)\begin{array}{l}z=x+yi\hfill \\ z=\left(r\cos \theta \right)+i\left(r\sin \theta \right)\hfill \\ z=r\left(\cos \theta +i\sin \theta \right)\hfill \end{array}
where rr is the modulus and θ\theta is the argument. We often use the abbreviation rcisθr\text{cis}\theta to represent r(cos⁡θ+isin⁡θ)r\left(\cos \theta +i\sin \theta \right).

Example 4: Expressing a Complex Number Using Polar Coordinates

Express the complex number 4i4i using polar coordinates.

Solution

On the complex plane, the number z=4iz=4i is the same as z=0+4iz=0+4i. Writing it in polar form, we have to calculate rr first.
r=x2+y2r=02+42r=16r=4\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{0}^{2}+{4}^{2}}\hfill \\ r=\sqrt{16}\hfill \\ r=4\hfill \end{array}
Next, we look at xx. If x=rcos⁡θx=r\cos \theta , and x=0x=0, then θ=π2\theta =\frac{\pi }{2}. In polar coordinates, the complex number z=0+4iz=0+4i can be written as z=4(cos⁡(π2)+isin⁡(π2))z=4\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right) or 4cis(π2)4\text{cis}\left(\frac{\pi }{2}\right).
Plot of z=4i in the complex plane, also shows that the in polar coordinate it would be (4,pi/2). Figure 6

Try It 4

Express z=3iz=3i as rcisθr\text{cis}\theta in polar form. Solution

Example 5: Finding the Polar Form of a Complex Number

Find the polar form of −4+4i-4+4i.

Solution

First, find the value of rr.
r=x2+y2r=(−4)2+(42)r=32r=42\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(-4\right)}^{2}+\left({4}^{2}\right)}\hfill \\ r=\sqrt{32}\hfill \\ r=4\sqrt{2}\hfill \end{array}
Find the angle θ\theta using the formula:
cos⁡θ=xrcos⁡θ=−442cos⁡θ=−12θ=cos⁡−1(−12)=3π4\begin{array}{l}\cos \theta =\frac{x}{r}\hfill \\ \cos \theta =\frac{-4}{4\sqrt{2}}\hfill \\ \cos \theta =-\frac{1}{\sqrt{2}}\hfill \\ \theta ={\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4}\hfill \end{array}
Thus, the solution is 42cis(3π4)4\sqrt{2}\text{cis}\left(\frac{3\pi }{4}\right).

Try It 5

Write z=3+iz=\sqrt{3}+i in polar form. Solution

 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(cos⁡θ+isin⁡θ)z=r\left(\cos \theta +i\sin \theta \right), first evaluate the trigonometric functions cos⁡θ\cos \theta and sin⁡θ\sin \theta . Then, multiply through by rr.

Example 6: Converting from Polar to Rectangular Form

Convert the polar form of the given complex number to rectangular form:
z=12(cos⁡(π6)+isin⁡(π6))z=12\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)

Solution

We begin by evaluating the trigonometric expressions.
cos⁡(π6)=32sin⁡(π6)=12\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\\\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}
After substitution, the complex number is
z=12(32+12i)z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)
We apply the distributive property:
z=12(32+12i) =(12)32+(12)12i =63+6i\begin{array}{l}z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\hfill \\ \text{ }=\left(12\right)\frac{\sqrt{3}}{2}+\left(12\right)\frac{1}{2}i\hfill \\ \text{ }=6\sqrt{3}+6i\hfill \end{array}
The rectangular form of the given point in complex form is 63+6i6\sqrt{3}+6i.

Example 7: Finding the Rectangular Form of a Complex Number

Find the rectangular form of the complex number given r=13r=13 and tan⁡θ=512\tan \theta =\frac{5}{12}.

Solution

If tan⁡θ=512\tan \theta =\frac{5}{12}, and tan⁡θ=yx\tan \theta =\frac{y}{x}, we first determine r=x2+y2=122+52=13.r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{.} We then find cos⁡θ=xr\cos \theta =\frac{x}{r} and sin⁡θ=yr\sin \theta =\frac{y}{r}.
z=13(cos⁡θ+isin⁡θ)=13(1213+513i)=12+5i\begin{array}{l}z=13\left(\cos \theta +i\sin \theta \right)\hfill \\ =13\left(\frac{12}{13}+\frac{5}{13}i\right)\hfill \\ =12+5i\hfill \end{array}
The rectangular form of the given number in complex form is 12+5i12+5i.

Try It 6

Convert the complex number to rectangular form:
z=4(cos⁡11π6+isin⁡11π6)z=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)
Solution

 Finding Products and Quotients 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.

A General Note: Products of Complex Numbers in Polar Form

If z1=r1(cos⁡θ1+isin⁡θ1){z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right) and z2=r2(cos⁡θ2+isin⁡θ2){z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right), then the product of these numbers is given as:
z1z2=r1r2[cos⁡(θ1+θ2)+isin⁡(θ1+θ2)]z1z2=r1r2cis(θ1+θ2)\begin{array}{l}\hfill \\ \begin{array}{l}{z}_{1}{z}_{2}={r}_{1}{r}_{2}\left[\cos \left({\theta }_{1}+{\theta }_{2}\right)+i\sin \left({\theta }_{1}+{\theta }_{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}={r}_{1}{r}_{2}\text{cis}\left({\theta }_{1}+{\theta }_{2}\right)\hfill \end{array}\hfill \end{array}
Notice that the product calls for multiplying the moduli and adding the angles.

Example 8: Finding the Product of Two Complex Numbers in Polar Form

Find the product of z1z2{z}_{1}{z}_{2}, given z1=4(cos⁡(80∘)+isin⁡(80∘)){z}_{1}=4\left(\cos \left(80^\circ \right)+i\sin \left(80^\circ \right)\right) and z2=2(cos⁡(145∘)+isin⁡(145∘)){z}_{2}=2\left(\cos \left(145^\circ \right)+i\sin \left(145^\circ \right)\right).

Solution

Follow the formula
z1z2=4⋅2[cos⁡(80∘+145∘)+isin⁡(80∘+145∘)]z1z2=8[cos⁡(225∘)+isin⁡(225∘)]z1z2=8[cos⁡(5π4)+isin⁡(5π4)]z1z2=8[−22+i(−22)]z1z2=−42−4i2\begin{array}{l}{z}_{1}{z}_{2}=4\cdot 2\left[\cos \left(80^\circ +145^\circ \right)+i\sin \left(80^\circ +145^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(225^\circ \right)+i\sin \left(225^\circ \right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2}\hfill \end{array}

Finding Quotients of Complex Numbers in Polar Form

The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments.

A General Note: Quotients of Complex Numbers in Polar Form

If z1=r1(cos⁡θ1+isin⁡θ1){z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right) and z2=r2(cos⁡θ2+isin⁡θ2){z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right), then the quotient of these numbers is
z1z2=r1r2[cos⁡(θ1−θ2)+isin⁡(θ1−θ2)],z2≠0z1z2=r1r2cis(θ1−θ2),z2≠0\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\left[\cos \left({\theta }_{1}-{\theta }_{2}\right)+i\sin \left({\theta }_{1}-{\theta }_{2}\right)\right],{z}_{2}\ne 0\\ \frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\text{cis}\left({\theta }_{1}-{\theta }_{2}\right),{z}_{2}\ne 0\end{array}
Notice that the moduli are divided, and the angles are subtracted.

How To: Given two complex numbers in polar form, find the quotient.

  1. Divide r1r2\frac{{r}_{1}}{{r}_{2}}.
  2. Find θ1−θ2{\theta }_{1}-{\theta }_{2}.
  3. Substitute the results into the formula: z=r(cos⁡θ+isin⁡θ)z=r\left(\cos \theta +i\sin \theta \right). Replace rr with r1r2\frac{{r}_{1}}{{r}_{2}}, and replace θ\theta with θ1−θ2{\theta }_{1}-{\theta }_{2}.
  4. Calculate the new trigonometric expressions and multiply through by rr.

Example 9: Finding the Quotient of Two Complex Numbers

Find the quotient of z1=2(cos⁡(213∘)+isin⁡(213∘)){z}_{1}=2\left(\cos \left(213^\circ \right)+i\sin \left(213^\circ \right)\right) and z2=4(cos⁡(33∘)+isin⁡(33∘)){z}_{2}=4\left(\cos \left(33^\circ \right)+i\sin \left(33^\circ \right)\right).

Solution

Using the formula, we have
z1z2=24[cos⁡(213∘−33∘)+isin⁡(213∘−33∘)]z1z2=12[cos⁡(180∘)+isin⁡(180∘)]z1z2=12[−1+0i]z1z2=−12+0iz1z2=−12\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}\left[\cos \left(213^\circ -33^\circ \right)+i\sin \left(213^\circ -33^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[\cos \left(180^\circ \right)+i\sin \left(180^\circ \right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[-1+0i\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i\hfill \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}\hfill \end{array}

Try It 7

Find the product and the quotient of z1=23(cos⁡(150∘)+isin⁡(150∘)){z}_{1}=2\sqrt{3}\left(\cos \left(150^\circ \right)+i\sin \left(150^\circ \right)\right) and z2=2(cos⁡(30∘)+isin⁡(30∘)){z}_{2}=2\left(\cos \left(30^\circ \right)+i\sin \left(30^\circ \right)\right). Solution

 Finding Powers and Roots of Complex Numbers in Polar Form

Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer n,znn,{z}^{n} is found by raising the modulus to the nthn\text{th} power and multiplying the argument by nn. It is the standard method used in modern mathematics.

A General Note: De Moivre’s Theorem

If z=r(cos⁡θ+isin⁡θ)z=r\left(\cos \theta +i\sin \theta \right) is a complex number, then
zn=rn[cos⁡(nθ)+isin⁡(nθ)]zn=rncis(nθ)\begin{array}{l}{z}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\\ {z}^{n}={r}^{n}\text{cis}\left(n\theta \right)\end{array}
where nn is a positive integer.

Example 10: Evaluating an Expression Using De Moivre’s Theorem

Evaluate the expression (1+i)5{\left(1+i\right)}^{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)\left(1+i\right) in polar form. Let us find rr.
r=x2+y2r=(1)2+(1)2r=2\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(1\right)}^{2}+{\left(1\right)}^{2}}\hfill \\ r=\sqrt{2}\hfill \end{array}
Then we find θ\theta . Using the formula tan⁡θ=yx\tan \theta =\frac{y}{x} gives
tan⁡θ=11tan⁡θ=1θ=π4\begin{array}{l}\tan \theta =\frac{1}{1}\hfill \\ \tan \theta =1\hfill \\ \theta =\frac{\pi }{4}\hfill \end{array}
Use De Moivre’s Theorem to evaluate the expression.
(a+bi)n=rn[cos⁡(nθ)+isin⁡(nθ)](1+i)5=(2)5[cos⁡(5⋅π4)+isin⁡(5⋅π4)](1+i)5=42[cos⁡(5π4)+isin⁡(5π4)](1+i)5=42[−22+i(−22)](1+i)5=−4−4i\begin{array}{l}{\left(a+bi\right)}^{n}={r}^{n}\left[\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right]\hfill \\ {\left(1+i\right)}^{5}={\left(\sqrt{2}\right)}^{5}\left[\cos \left(5\cdot \frac{\pi }{4}\right)+i\sin \left(5\cdot \frac{\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=4\sqrt{2}\left[-\frac{\sqrt{2}}{2}+i\left(-\frac{\sqrt{2}}{2}\right)\right]\hfill \\ {\left(1+i\right)}^{5}=-4 - 4i\hfill \end{array}

Finding Roots of Complex Numbers in Polar Form

To find the nth root of a complex number in polar form, we use the nthn\text{th} Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding nthn\text{th} roots of complex numbers in polar form.

A General Note: The nth Root Theorem

To find the nthn\text{th} root of a complex number in polar form, use the formula given as
z1n=r1n[cos⁡(θn+2kπn)+isin⁡(θn+2kπn)]{z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left[\cos \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)+i\sin \left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)\right]
where k=0,1,2,3,...,n−1k=0,1,2,3,...,n - 1. We add 2kπn\frac{2k\pi }{n} to θn\frac{\theta }{n} in order to obtain the periodic roots.

Example 11: Finding the nth Root of a Complex Number

Evaluate the cube roots of z=8(cos⁡(2π3)+isin⁡(2π3))z=8\left(\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right).

Solution

We have
z13=813[cos⁡(2π33+2kπ3)+isin⁡(2π33+2kπ3)]z13=2[cos⁡(2π9+2kπ3)+isin⁡(2π9+2kπ3)]\begin{array}{l}{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]\hfill \\ {z}^{\frac{1}{3}}=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]\hfill \end{array}
There will be three roots: k=0,1,2k=0,1,2. When k=0k=0, we have
z13=2(cos⁡(2π9)+isin⁡(2π9)){z}^{\frac{1}{3}}=2\left(\cos \left(\frac{2\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}\right)\right)
When k=1k=1, we have
z13=2[cos⁡(2π9+6π9)+isin⁡(2π9+6π9)] Add 2(1)π3 to each angle.z13=2(cos⁡(8π9)+isin⁡(8π9))\begin{array}{l}{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)\right]\begin{array}{cccc}& & & \end{array}\text{ Add }\frac{2\left(1\right)\pi }{3}\text{ to each angle.}\hfill \\ {z}^{\frac{1}{3}}=2\left(\cos \left(\frac{8\pi }{9}\right)+i\sin \left(\frac{8\pi }{9}\right)\right)\hfill \end{array}
When k=2k=2, we have
z13=2[cos⁡(2π9+12π9)+isin⁡(2π9+12π9)]Add 2(2)π3 to each angle.z13=2(cos⁡(14π9)+isin⁡(14π9))\begin{array}{ll}{z}^{\frac{1}{3}}=2\left[\cos \left(\frac{2\pi }{9}+\frac{12\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}+\frac{12\pi }{9}\right)\right]\begin{array}{cccc}& & & \end{array}\hfill & \text{Add }\frac{2\left(2\right)\pi }{3}\text{ to each angle.}\hfill \\ {z}^{\frac{1}{3}}=2\left(\cos \left(\frac{14\pi }{9}\right)+i\sin \left(\frac{14\pi }{9}\right)\right)\hfill & \hfill \end{array}
Remember to find the common denominator to simplify fractions in situations like this one. For k=1k=1, the angle simplification is
2π33+2(1)π3=2π3(13)+2(1)π3(33)=2π9+6π9=8π9\begin{array}{l}\frac{\frac{2\pi }{3}}{3}+\frac{2\left(1\right)\pi }{3}=\frac{2\pi }{3}\left(\frac{1}{3}\right)+\frac{2\left(1\right)\pi }{3}\left(\frac{3}{3}\right)\hfill \\ =\frac{2\pi }{9}+\frac{6\pi }{9}\hfill \\ =\frac{8\pi }{9}\hfill \end{array}

Try It 8

Find the four fourth roots of 16(cos⁡(120∘)+isin⁡(120∘))16\left(\cos \left(120^\circ \right)+i\sin \left(120^\circ \right)\right). Solution
 

Key Concepts

  • Complex numbers in the form a+bia+bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label the x-axis as the real axis and the y-axis as the imaginary axis.
  • The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point: ∣z∣=a2+b2|z|=\sqrt{{a}^{2}+{b}^{2}}.
  • To write complex numbers in polar form, we use the formulas x=rcos⁡θ,y=rsin⁡θx=r\cos \theta ,y=r\sin \theta , and r=x2+y2r=\sqrt{{x}^{2}+{y}^{2}}. Then, z=r(cos⁡θ+isin⁡θ)z=r\left(\cos \theta +i\sin \theta \right).
  • To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by rr.
  • To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property.
  • To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles.
  • To find the power of a complex number zn{z}^{n}, raise rr to the power nn, and multiply θ\theta by nn.
  • Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent.

Glossary

argument
the angle associated with a complex number; the angle between the line from the origin to the point and the positive real axis
De Moivre’s Theorem
formula used to find the nthn\text{th} power or nth roots of a complex number; states that, for a positive integer n,znn,{z}^{n} is found by raising the modulus to the nthn\text{th} power and multiplying the angles by nn
modulus
the absolute value of a complex number, or the distance from the origin to the point (x,y)\left(x,y\right); also called the amplitude
polar form of a complex number
a complex number expressed in terms of an angle θ\theta and its distance from the origin rr; can be found by using conversion formulas x=rcos⁡θ,y=rsin⁡θx=r\cos \theta ,y=r\sin \theta , and r=x2+y2r=\sqrt{{x}^{2}+{y}^{2}}

Section Exercises

1. A complex number is a+bia+bi. Explain each part. 2. What does the absolute value of a complex number represent? 3. How is a complex number converted to polar form? 4. How do we find the product of two complex numbers? 5. What is De Moivre's Theorem and what is it used for? For the following exercises, find the absolute value of the given complex number. 6. 5+3i5+3i 7. −7+i−7+i 8. −3−3i−3−3i 9. 2−6i\sqrt{2}−6i 10. 2i2i 11. 2.2−3.1i2.2−3.1i For the following exercises, write the complex number in polar form. 12. 2+2i2+2i 13. 8−4i8−4i 14. −12−12i−\frac{1}{2}−\frac{1}{2}i 15. 3+i\sqrt{3}+i 16. 3i3i For the following exercises, convert the complex number from polar to rectangular form. 17. z=7cis(π6)z=7\text{cis}\left(\frac{\pi}{6}\right) 18. z=2cis(π3)z=2\text{cis}\left(\frac{\pi}{3}\right) 19. z=4cis(7π6)z=4\text{cis}\left(\frac{7\pi}{6}\right) 20. z=7cis(25∘)z=7\text{cis}\left(25^{\circ}\right) 21. z=3cis(240∘)z=3\text{cis}\left(240^{\circ}\right) 22. z=2cis(100∘)z=\sqrt{2}\text{cis}\left(100^{\circ}\right) For the following exercises, find z1z2 in polar form. 23. z1=23cis(116∘); (118∘)z_{1}=2\sqrt{3}\text{cis}\left(116^{\circ}\right)\text{; }\left(118^{\circ}\right) 24. z1=2cis(205∘); z2=14cis(60∘)z_{1}=\sqrt{2}\text{cis}\left(205^{\circ}\right)\text{; }z_{2}=\frac{1}{4}\text{cis}\left(60^{\circ}\right) 25. z1=3cis(120∘); z2=14cis(60∘)z_{1}=3\text{cis}\left(120^{\circ}\right)\text{; }z_{2}=\frac{1}{4}\text{cis}\left(60^{\circ}\right) 26. z1=3cis(5π4); z2=5cis(π6)z_{1}=3\text{cis}\left(\frac{5\pi}{4}\right)\text{; }z_{2}=5\text{cis}\left(\frac{\pi}{6}\right) 27. z1=5cis(5π8); z2=15cis(π12)z_{1}=\sqrt{5}\text{cis}\left(\frac{5\pi}{8}\right)\text{; }z_{2}=\sqrt{15}\text{cis}\left(\frac{\pi}{12}\right) 28. z1=4cis(π2); z2=2cis(π4)z_{1}=4\text{cis}\left(\frac{\pi}{2}\right)\text{; }z_{2}=2\text{cis}\left(\frac{\pi}{4}\right) For the following exercises, find z1z2\frac{z_{1}}{z_{2}} in polar form. 29. z1=21cis(135∘); z2=3cis(65∘)z_{1}=21\text{cis}\left(135^{\circ}\right)\text{; }z_{2}=3\text{cis}\left(65^{\circ}\right) 30. z1=2cis(90∘); z2=2cis(60∘)z_{1}=\sqrt{2}\text{cis}\left(90^{\circ}\right)\text{; }z_{2}=2\text{cis}\left(60^{\circ}\right) 31. z1=15cis(120∘); z2=3cis(40∘)z_{1}=15\text{cis}\left(120^{\circ}\right)\text{; }z_{2}=3\text{cis}\left(40^{\circ}\right) 32. z1=6cis(π3); z2=2cis(π4)z_{1}=6\text{cis}\left(\frac{\pi}{3}\right)\text{; }z_{2}=2\text{cis}\left(\frac{\pi}{4}\right) 33. z1=52cis(π); z2=2cis(2π3)z_{1}=5\sqrt{2}\text{cis}\left(\pi\right)\text{; }z_{2}=\sqrt{2}\text{cis}\left(\frac{2\pi}{3}\right) 34. z1=2cis(3π5); z2=3cis(π4)z_{1}=2\text{cis}\left(\frac{3\pi}{5}\right)\text{; }z_{2}=3\text{cis}\left(\frac{\pi}{4}\right) For the following exercises, find the powers of each complex number in polar form. 35. Find z3z^{3} when z=5cis(45∘)z=5\text{cis}\left(45^{\circ}\right). 36. Find z4z^{4} when z=2cis(70∘)z=2\text{cis}\left(70^{\circ}\right). 37. Find z2z^{2} when z=3cis(120∘)z=3\text{cis}\left(120^{\circ}\right). 38. Find z2z^{2} when z=4cis(π4)z=4\text{cis}\left(\frac{\pi}{4}\right) 39. Find z4z^{4} when z=cis(3π16)z=\text{cis}\left(\frac{3\pi}{16}\right). 40. Find z3z^{3} when z=3cis(5π3)z=3\text{cis}\left(\frac{5\pi}{3}\right). For the following exercises, evaluate each root. 41. Evaluate the cube root of z when z=27cis(240∘)z=27\text{cis}\left(240^{\circ}\right). 42. Evaluate the square root of z when z=16cis(100∘)z=16\text{cis}\left(100^{\circ}\right). 43. Evaluate the cube root of z when z=32cis(2π3)z=32\text{cis}\left(\frac{2\pi}{3}\right). 44. Evaluate the square root of z when z=32cis(π)z=32\text{cis}\left(\pi\right). 45. Evaluate the cube root of z when z=8cis(7π4)z=8\text{cis}\left(\frac{7\pi}{4}\right). For the following exercises, plot the complex number in the complex plane. 46. 2+4i2+4i 47. −3−3i−3−3i 48. 5−4i5−4i 49. −1−5i−1−5i 50. 3+2i3+2i 51. 2i2i 52. −4−4 53. 6−2i6−2i 54. −2+i−2+i 55. 1−4i1−4i For the following exercises, find all answers rounded to the nearest hundredth. 56. Use the rectangular to polar feature on the graphing calculator to change 5+5i5+5i to polar form. 57. Use the rectangular to polar feature on the graphing calculator to change 3−2i3−2i 58. Use the rectangular to polar feature on the graphing calculator to change −3−8i−3−8i 59. Use the polar to rectangular feature on the graphing calculator to change 4cis(120∘)4\text{cis}\left(120^{\circ}\right) to rectangular form. 60. Use the polar to rectangular feature on the graphing calculator to change 2cis(45∘)2\text{cis}\left(45^{\circ}\right) to rectangular form. 61. Use the polar to rectangular feature on the graphing calculator to change 5cis(210∘)5\text{cis}\left(210^{\circ}\right) to rectangular form.

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