Further Applications of Trigonometry

Polar Form of Complex Numbers

Learning Objectives

In this section, you will:

  • 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 Complex Numbers. 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+bi\,is similar to plotting a real number, except that the horizontal axis represents the real part of the number,\,a,\,and the vertical axis represents the imaginary part of the number,\,bi.

How To

Given a complex number\,a+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\,a\,units in the horizontal direction and\,b\,units in the vertical direction.

Plotting a Complex Number in the Complex Plane

Plot the complex number \,2-3i\,in the complex plane.

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From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See (Figure).

Plot of 2-3i in the complex plane (2 along the real axis, -3 along the imaginary axis).
Figure 1.

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Try It

Plot the point\,1+5i\,in the complex plane.

[reveal-answer q=”fs-id1165134167302″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134167302″]Plot of 1+5i in the complex plane (1 along the real axis, 5 along the imaginary axis).[/hidden-answer]

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|.\,It measures the distance from the origin to a point in the plane. For example, the graph of\,z=2+4i,\,in (Figure), shows\,|z|.

Plot of 2+4i in the complex plane and its magnitude, |z| = rad 20.
Figure 2.

Absolute Value of a Complex Number

Given\,z=x+yi,\,a complex number, the absolute value of\,z\,is defined as

|z|=\sqrt{{x}^{2}+{y}^{2}}

It is the distance from the origin to the point\,\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,\,\left(0,\text{ }0\right).

Finding the Absolute Value of a Complex Number with a Radical

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

[reveal-answer q=”fs-id1165133024224″]Show Solution[/reveal-answer]
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Using the formula, we have

\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}

See (Figure).

Plot of z=(rad5 - i) in the complex plane and its magnitude rad6.
Figure 3.

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Try It

Find the absolute value of the complex number\,z=12-5i.

[reveal-answer q=”fs-id1165137771102″]Show Solution[/reveal-answer]
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13

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Finding the Absolute Value of a Complex Number

Given\,z=3-4i,\,find\,|z|.

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Using the formula, we have

\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\,z\,is 5. See (Figure).

Plot of (3-4i) in the complex plane and its magnitude |z| =5.
Figure 4.

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Try It

Given\,z=1-7i,\,find\,|z|.

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|z|=\sqrt{50}=5\sqrt{2}

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Writing Complex Numbers in Polar Form

The polar form of a complex number expresses a number in terms of an angle\,\theta \,and its distance from the origin\,r.\,Given a complex number in rectangular form expressed as\,z=x+yi,\,we use the same conversion formulas as we do to write the number in trigonometric form:

\begin{array}{l}\,x=r\mathrm{cos}\,\theta \hfill \\ \,y=r\mathrm{sin}\,\theta \hfill \\ \,\,r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}

We review these relationships in (Figure).

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\,\left(x,y\right).\,The modulus, then, is the same as\,r,\,the radius in polar form. We use\,\theta \,to indicate the angle of direction (just as with polar coordinates). Substituting, we have

\begin{array}{l}z=x+yi\hfill \\ z=r\mathrm{cos}\,\theta +\left(r\mathrm{sin}\,\theta \right)i\hfill \\ z=r\left(\mathrm{cos}\,\theta +i\mathrm{sin}\,\theta \right)\hfill \end{array}

Polar Form of a Complex Number

Writing a complex number in polar form involves the following conversion formulas:

\begin{array}{l}\hfill \\ x=r\mathrm{cos}\,\theta \hfill \\ y=r\mathrm{sin}\,\theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}

Making a direct substitution, we have

\begin{array}{l}z=x+yi\hfill \\ z=\left(r\mathrm{cos}\,\theta \right)+i\left(r\mathrm{sin}\,\theta \right)\hfill \\ z=r\left(\mathrm{cos}\,\theta +i\mathrm{sin}\,\theta \right)\hfill \end{array}

where\,r\,is the modulus and \theta is the argument. We often use the abbreviation\,r\text{cis}\,\theta \,to represent\,r\left(\mathrm{cos}\,\theta +i\mathrm{sin}\,\theta \right).

Expressing a Complex Number Using Polar Coordinates

Express the complex number\,4i\,using polar coordinates.

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On the complex plane, the number\,z=4i\,is the same as\,z=0+4i.\,Writing it in polar form, we have to calculate\,r\,first.

\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\,x.\,If\,x=r\mathrm{cos}\,\theta ,\,and\,x=0,\,then\,\theta =\frac{\pi }{2}.\,In polar coordinates, the complex number\,z=0+4i\,can be written as\,z=4\left(\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)\right)\,or\,4\text{cis}\left(\,\frac{\pi }{2}\right).\,See (Figure).

Plot of z=4i in the complex plane, also shows that the in polar coordinate it would be (4,pi/2).
Figure 6.

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Try It

Express\,z=3i\, as \,r\,\text{cis}\,\theta \, in polar form.

[reveal-answer q=”fs-id1165137397894″]Show Solution[/reveal-answer]
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z=3\left(\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)\right)

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Finding the Polar Form of a Complex Number

Find the polar form of\,-4+4i.

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First, find the value of\,r.

\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\,\theta \,using the formula:

\begin{array}{l}\mathrm{cos}\,\theta =\frac{x}{r}\hfill \\ \mathrm{cos}\,\theta =\frac{-4}{4\sqrt{2}}\hfill \\ \mathrm{cos}\,\theta =-\frac{1}{\sqrt{2}}\hfill \\ \,\,\,\,\,\,\,\,\,\theta ={\mathrm{cos}}^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4}\hfill \end{array}

Thus, the solution is\,4\sqrt{2}\text{cis}\left(\frac{3\pi }{4}\right).[/hidden-answer]

Try It

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

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z=2\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right)

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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(\mathrm{cos}\,\theta +i\mathrm{sin}\,\theta \right),\,first evaluate the trigonometric functions\,\mathrm{cos}\,\theta \,and\,\mathrm{sin}\,\theta .\,Then, multiply through by\,r.

Converting from Polar to Rectangular Form

Convert the polar form of the given complex number to rectangular form:

z=12\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right)
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We begin by evaluating the trigonometric expressions.

\mathrm{cos}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\,\text{and}\,\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}\,

After substitution, the complex number is

z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)

We apply the distributive property:

\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\,6\sqrt{3}+6i.[/hidden-answer]

Finding the Rectangular Form of a Complex Number

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

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If\,\mathrm{tan}\,\theta =\frac{5}{12},\,and\,\mathrm{tan}\,\theta =\frac{y}{x},\,we first determine\,r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{.} We then find\,\mathrm{cos}\,\theta =\frac{x}{r}\,and\,\mathrm{sin}\,\theta =\frac{y}{r}.

\begin{array}{l}z=13\left(\mathrm{cos}\,\theta +i\mathrm{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+5i.

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Try It

Convert the complex number to rectangular form:

z=4\left(\mathrm{cos}\frac{11\pi }{6}+i\mathrm{sin}\frac{11\pi }{6}\right)
[reveal-answer q=”fs-id1165137444688″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165137444688″]

z=2\sqrt{3}-2i

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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.

Products of Complex Numbers in Polar Form

If\,{z}_{1}={r}_{1}\left(\mathrm{cos}\,{\theta }_{1}+i\mathrm{sin}\,{\theta }_{1}\right)\,and\,{z}_{2}={r}_{2}\left(\mathrm{cos}\,{\theta }_{2}+i\mathrm{sin}\,{\theta }_{2}\right), then the product of these numbers is given as:

\begin{array}{l}\hfill \\ \begin{array}{l}{z}_{1}{z}_{2}={r}_{1}{r}_{2}\left[\mathrm{cos}\left({\theta }_{1}+{\theta }_{2}\right)+i\mathrm{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.

Finding the Product of Two Complex Numbers in Polar Form

Find the product of\,{z}_{1}{z}_{2},\,given\,{z}_{1}=4\left(\mathrm{cos}\left(80°\right)+i\mathrm{sin}\left(80°\right)\right)\,and\,{z}_{2}=2\left(\mathrm{cos}\left(145°\right)+i\mathrm{sin}\left(145°\right)\right).

[reveal-answer q=”fs-id1165135369384″]Show Solution[/reveal-answer]
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Follow the formula

\begin{array}{l}{z}_{1}{z}_{2}=4\cdot 2\left[\mathrm{cos}\left(80°+145°\right)+i\mathrm{sin}\left(80°+145°\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\mathrm{cos}\left(225°\right)+i\mathrm{sin}\left(225°\right)\right]\hfill \\ {z}_{1}{z}_{2}=8\left[\mathrm{cos}\left(\frac{5\pi }{4}\right)+i\mathrm{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}[/hidden-answer]

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.

Quotients of Complex Numbers in Polar Form

If\,{z}_{1}={r}_{1}\left(\mathrm{cos}\,{\theta }_{1}+i\mathrm{sin}\,{\theta }_{1}\right)\,and\,{z}_{2}={r}_{2}\left(\mathrm{cos}\,{\theta }_{2}+i\mathrm{sin}\,{\theta }_{2}\right), then the quotient of these numbers is

\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{{r}_{1}}{{r}_{2}}\left[\mathrm{cos}\left({\theta }_{1}-{\theta }_{2}\right)+i\mathrm{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\,\frac{{r}_{1}}{{r}_{2}}.
  2. Find\,{\theta }_{1}-{\theta }_{2}.
  3. Substitute the results into the formula:\,z=r\left(\mathrm{cos}\,\theta +i\mathrm{sin}\,\theta \right).\,Replace\,r\,with\,\frac{{r}_{1}}{{r}_{2}},\,and replace\,\theta \,with\,{\theta }_{1}-{\theta }_{2}.
  4. Calculate the new trigonometric expressions and multiply through by\,r.

Finding the Quotient of Two Complex Numbers

Find the quotient of\,{z}_{1}=2\left(\mathrm{cos}\left(213°\right)+i\mathrm{sin}\left(213°\right)\right)\,and\,{z}_{2}=4\left(\mathrm{cos}\left(33°\right)+i\mathrm{sin}\left(33°\right)\right).

[reveal-answer q=”fs-id1165133318744″]Show Solution[/reveal-answer]
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Using the formula, we have

\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}\left[\mathrm{cos}\left(213°-33°\right)+i\mathrm{sin}\left(213°-33°\right)\right]\hfill \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}\left[\mathrm{cos}\left(180°\right)+i\mathrm{sin}\left(180°\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}[/hidden-answer]

Try It

Find the product and the quotient of\,{z}_{1}=2\sqrt{3}\left(\mathrm{cos}\left(150°\right)+i\mathrm{sin}\left(150°\right)\right)\,and\,{z}_{2}=2\left(\mathrm{cos}\left(30°\right)+i\mathrm{sin}\left(30°\right)\right).

[reveal-answer q=”fs-id1165135707943″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135707943″]

\,{z}_{1}{z}_{2}=-4\sqrt{3};\frac{{z}_{1}}{{z}_{2}}=-\frac{\sqrt{3}}{2}+\frac{3}{2}i\,

[/hidden-answer]

Finding Powers 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,{z}^{n}\,is found by raising the modulus to the\,n\text{th}\,power and multiplying the argument by\,n.\,It is the standard method used in modern mathematics.

De Moivre’s Theorem

If\,z=r\left(\mathrm{cos}\,\theta +i\mathrm{sin}\,\theta \right)\,is a complex number, then

\begin{array}{l}{z}^{n}={r}^{n}\left[\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)\right]\\ {z}^{n}={r}^{n}\text{cis}\left(n\theta \right)\end{array}

where\,n\,
is a positive integer.

Evaluating an Expression Using De Moivre’s Theorem

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

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Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write\,\left(1+i\right)\,in polar form. Let us find\,r.

\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\,\theta .\,Using the formula\,\mathrm{tan}\,\theta =\frac{y}{x}\,gives

\begin{array}{l}\mathrm{tan}\,\theta =\frac{1}{1}\hfill \\ \mathrm{tan}\,\theta =1\hfill \\ \,\,\,\,\,\,\,\,\theta =\frac{\pi }{4}\hfill \end{array}

Use De Moivre’s Theorem to evaluate the expression.

\begin{array}{l}{\left(a+bi\right)}^{n}={r}^{n}\left[\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)\right]\hfill \\ \,\,\,\,{\left(1+i\right)}^{5}={\left(\sqrt{2}\right)}^{5}\left[\mathrm{cos}\left(5\cdot \frac{\pi }{4}\right)+i\mathrm{sin}\left(5\cdot \frac{\pi }{4}\right)\right]\hfill \\ \,\,\,\,{\left(1+i\right)}^{5}=4\sqrt{2}\left[\mathrm{cos}\left(\frac{5\pi }{4}\right)+i\mathrm{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}[/hidden-answer]

Finding Roots of Complex Numbers in Polar Form

To find the nth root of a complex number in polar form, we use the\,n\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\,n\text{th}\,roots of complex numbers in polar form.

The nth Root Theorem

To find the\,n\text{th}\,root of a complex number in polar form, use the formula given as

{z}^{\frac{1}{n}}={r}^{\frac{1}{n}}\left[\mathrm{cos}\left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)+i\mathrm{sin}\left(\frac{\theta }{n}+\frac{2k\pi }{n}\right)\right]

where\,k=0,\,\,1,\,\,2,\,\,3,\,.\,\,.\,\,.\,\,,\,\,n-1.\,We add \,\frac{2k\pi }{n}\,\,to\,\frac{\theta }{n}\,in order to obtain the periodic roots.

Finding the nth Root of a Complex Number

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

[reveal-answer q=”fs-id1165137656965″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165137656965″]

We have

\begin{array}{l}{z}^{\frac{1}{3}}={8}^{\frac{1}{3}}\left[\mathrm{cos}\left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)+i\mathrm{sin}\left(\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3}\right)\right]\hfill \\ {z}^{\frac{1}{3}}=2\left[\mathrm{cos}\left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)+i\mathrm{sin}\left(\frac{2\pi }{9}+\frac{2k\pi }{3}\right)\right]\hfill \end{array}

There will be three roots:\,k=0,\,\,1,\,\,2.\,When\,k=0,\,we have

{z}^{\frac{1}{3}}=2\left(\mathrm{cos}\left(\frac{2\pi }{9}\right)+i\mathrm{sin}\left(\frac{2\pi }{9}\right)\right)

When\,k=1,\,we have

\begin{array}{l}{z}^{\frac{1}{3}}=2\left[\mathrm{cos}\left(\frac{2\pi }{9}+\frac{6\pi }{9}\right)+i\mathrm{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(\mathrm{cos}\left(\frac{8\pi }{9}\right)+i\mathrm{sin}\left(\frac{8\pi }{9}\right)\right)\hfill \end{array}

When\,k=2,\, we have

\begin{array}{ll}{z}^{\frac{1}{3}}=2\left[\mathrm{cos}\left(\frac{2\pi }{9}+\frac{12\pi }{9}\right)+i\mathrm{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(\mathrm{cos}\left(\frac{14\pi }{9}\right)+i\mathrm{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=1,\,the angle simplification is

\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}[/hidden-answer]

Try It

Find the four fourth roots of\,16\left(\mathrm{cos}\left(120°\right)+i\mathrm{sin}\left(120°\right)\right).

[reveal-answer q=”fs-id1165134356866″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134356866″]

{z}_{0}=2\left(\mathrm{cos}\left(30°\right)+i\mathrm{sin}\left(30°\right)\right)

{z}_{1}=2\left(\mathrm{cos}\left(120°\right)+i\mathrm{sin}\left(120°\right)\right)

{z}_{2}=2\left(\mathrm{cos}\left(210°\right)+i\mathrm{sin}\left(210°\right)\right)

{z}_{3}=2\left(\mathrm{cos}\left(300°\right)+i\mathrm{sin}\left(300°\right)\right)

[/hidden-answer]

Access these online resources for additional instruction and practice with polar forms of complex numbers.

Key Concepts

  • Complex numbers in the form\,a+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. See (Figure).
  • The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point:\,|z|=\sqrt{{a}^{2}+{b}^{2}}.\,See (Figure) and (Figure).
  • To write complex numbers in polar form, we use the formulas\,x=r\mathrm{cos}\,\theta ,y=r\mathrm{sin}\,\theta ,\,and \,r=\sqrt{{x}^{2}+{y}^{2}}.\,Then,\,z=r\left(\mathrm{cos}\,\theta +i\mathrm{sin}\,\theta \right).\,See (Figure) and (Figure).
  • To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by\,r.\,See (Figure) and (Figure).
  • 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. See (Figure).
  • 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. See (Figure).
  • To find the power of a complex number\,{z}^{n},\,raise \,r\, to the power \,n, and multiply \,\theta \, by \,n.\,See (Figure).
  • Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. See (Figure).

Section Exercises

Verbal

A complex number is\,a+bi.\,Explain each part.

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a is the real part, b is the imaginary part, and\,i=\sqrt{-1}

[/hidden-answer]

What does the absolute value of a complex number represent?

How is a complex number converted to polar form?

[reveal-answer q=”fs-id1165131926323″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165131926323″]

Polar form converts the real and imaginary part of the complex number in polar form using\,x=r\mathrm{cos}\theta \, and \,y=r\mathrm{sin}\theta .

[/hidden-answer]

How do we find the product of two complex numbers?

What is De Moivre’s Theorem and what is it used for?

[reveal-answer q=”fs-id1165133309255″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165133309255″]

{z}^{n}={r}^{n}\left(\mathrm{cos}\left(n\theta \right)+i\mathrm{sin}\left(n\theta \right)\right)\,It is used to simplify polar form when a number has been raised to a power.

[/hidden-answer]

Algebraic

For the following exercises, find the absolute value of the given complex number.

5+\text{​}3i

-7+\text{​}i

[reveal-answer q=”fs-id1165135499578″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135499578″]

5\sqrt{2}

[/hidden-answer]

-3-3i

\sqrt{2}-6i

[reveal-answer q=”fs-id1165133213891″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165133213891″]

\sqrt{38}

[/hidden-answer]

2i

2.2-3.1i

[reveal-answer q=”fs-id1165134116972″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134116972″]

\sqrt{14.45}

[/hidden-answer]

For the following exercises, write the complex number in polar form.

2+2i

8-4i

[reveal-answer q=”fs-id1165134129997″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134129997″]

4\sqrt{5}\mathrm{cis}\left(333.4°\right)

[/hidden-answer]

-\frac{1}{2}-\frac{1}{2}\text{​}i

\sqrt{3}+i

[reveal-answer q=”fs-id1165131993571″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165131993571″]

2\mathrm{cis}\left(\frac{\pi }{6}\right)

[/hidden-answer]

3i

For the following exercises, convert the complex number from polar to rectangular form.

z=7\mathrm{cis}\left(\frac{\pi }{6}\right)

[reveal-answer q=”fs-id1165135367688″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135367688″]

\frac{7\sqrt{3}}{2}+i\frac{7}{2}

[/hidden-answer]

z=2\mathrm{cis}\left(\frac{\pi }{3}\right)

z=4\mathrm{cis}\left(\frac{7\pi }{6}\right)

[reveal-answer q=”fs-id1165134573209″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134573209″]

-2\sqrt{3}-2i

[/hidden-answer]

z=7\mathrm{cis}\left(25°\right)

z=3\mathrm{cis}\left(240°\right)

[reveal-answer q=”fs-id1165135640462″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135640462″]

-1.5-i\frac{3\sqrt{3}}{2}

[/hidden-answer]

z=\sqrt{2}\mathrm{cis}\left(100°\right)

For the following exercises, find\,{z}_{1}{z}_{2}\,in polar form.

{z}_{1}=2\sqrt{3}\mathrm{cis}\left(116°\right);\,\text{ }{z}_{2}=2\mathrm{cis}\left(82°\right)

[reveal-answer q=”fs-id1165135419732″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135419732″]

4\sqrt{3}\mathrm{cis}\left(198°\right)

[/hidden-answer]

{z}_{1}=\sqrt{2}\mathrm{cis}\left(205°\right);\text{ }{z}_{2}=2\sqrt{2}\mathrm{cis}\left(118°\right)

{z}_{1}=3\mathrm{cis}\left(120°\right);\text{ }{z}_{2}=\frac{1}{4}\mathrm{cis}\left(60°\right)
[reveal-answer q=”fs-id1165135472921″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135472921″]

\frac{3}{4}\mathrm{cis}\left(180°\right)

[/hidden-answer]

{z}_{1}=3\mathrm{cis}\left(\frac{\pi }{4}\right);\text{ }{z}_{2}=5\mathrm{cis}\left(\frac{\pi }{6}\right)

{z}_{1}=\sqrt{5}\mathrm{cis}\left(\frac{5\pi }{8}\right);\text{ }{z}_{2}=\sqrt{15}\mathrm{cis}\left(\frac{\pi }{12}\right)

[reveal-answer q=”fs-id1165133077994″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165133077994″]

5\sqrt{3}\mathrm{cis}\left(\frac{17\pi }{24}\right)

[/hidden-answer]

{z}_{1}=4\mathrm{cis}\left(\frac{\pi }{2}\right);\text{ }{z}_{2}=2\mathrm{cis}\left(\frac{\pi }{4}\right)

For the following exercises, find\,\frac{{z}_{1}}{{z}_{2}}\,in polar form.

{z}_{1}=21\mathrm{cis}\left(135°\right);\text{ }{z}_{2}=3\mathrm{cis}\left(65°\right)

[reveal-answer q=”fs-id1165134174904″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134174904″]

7\mathrm{cis}\left(70°\right)

[/hidden-answer]

{z}_{1}=\sqrt{2}\mathrm{cis}\left(90°\right);\text{ }{z}_{2}=2\mathrm{cis}\left(60°\right)

{z}_{1}=15\mathrm{cis}\left(120°\right);\text{ }{z}_{2}=3\mathrm{cis}\left(40°\right)

[reveal-answer q=”fs-id1165137480080″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165137480080″]

5\mathrm{cis}\left(80°\right)

[/hidden-answer]

{z}_{1}=6\mathrm{cis}\left(\frac{\pi }{3}\right);\text{ }{z}_{2}=2\mathrm{cis}\left(\frac{\pi }{4}\right)

{z}_{1}=5\sqrt{2}\mathrm{cis}\left(\pi \right);\text{ }{z}_{2}=\sqrt{2}\mathrm{cis}\left(\frac{2\pi }{3}\right)

[reveal-answer q=”fs-id1165134388962″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134388962″]

5\mathrm{cis}\left(\frac{\pi }{3}\right)

[/hidden-answer]

{z}_{1}=2\mathrm{cis}\left(\frac{3\pi }{5}\right);\text{ }{z}_{2}=3\mathrm{cis}\left(\frac{\pi }{4}\right)

For the following exercises, find the powers of each complex number in polar form.

Find\,{z}^{3}\,when\,z=5\mathrm{cis}\left(45°\right).

[reveal-answer q=”fs-id1165133349423″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165133349423″]

125\mathrm{cis}\left(135°\right)

[/hidden-answer]

Find\,{z}^{4}\,when\,z=2\mathrm{cis}\left(70°\right).

Find\,{z}^{2}\,when\,z=3\mathrm{cis}\left(120°\right).

[reveal-answer q=”fs-id1165135512799″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135512799″]

9\mathrm{cis}\left(240°\right)

[/hidden-answer]

Find\,{z}^{2}\,when\,z=4\mathrm{cis}\left(\frac{\pi }{4}\right).

Find\,{z}^{4}\,when\,z=\mathrm{cis}\left(\frac{3\pi }{16}\right).

[reveal-answer q=”fs-id1165134177544″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134177544″]

\mathrm{cis}\left(\frac{3\pi }{4}\right)

[/hidden-answer]

Find\,{z}^{3}\,when\,z=3\mathrm{cis}\left(\frac{5\pi }{3}\right).

For the following exercises, evaluate each root.

Evaluate the cube root of\,z\,when\,z=27\mathrm{cis}\left(240°\right).

[reveal-answer q=”fs-id1165135609212″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135609212″]

\,3\mathrm{cis}\left(80°\right),3\mathrm{cis}\left(200°\right),3\mathrm{cis}\left(320°\right)\,

[/hidden-answer]

Evaluate the square root of\,z\,when\,z=16\mathrm{cis}\left(100°\right).

Evaluate the cube root of\,z\,when\,z=32\mathrm{cis}\left(\frac{2\pi }{3}\right).

[reveal-answer q=”fs-id1165135238453″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135238453″]

\,2\sqrt[3]{4}\mathrm{cis}\left(\frac{2\pi }{9}\right),2\sqrt[3]{4}\mathrm{cis}\left(\frac{8\pi }{9}\right),2\sqrt[3]{4}\mathrm{cis}\left(\frac{14\pi }{9}\right)

[/hidden-answer]

Evaluate the square root of\,z\,when\,z=32\text{cis}\left(\pi \right).

Evaluate the cube root of\,z\,when\,z=8\mathrm{cis}\left(\frac{7\pi }{4}\right).

[reveal-answer q=”fs-id1165131907304″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165131907304″]

2\sqrt{2}\mathrm{cis}\left(\frac{7\pi }{8}\right),2\sqrt{2}\mathrm{cis}\left(\frac{15\pi }{8}\right)

[/hidden-answer]

Graphical

For the following exercises, plot the complex number in the complex plane.

2+4i

-3-3i

[reveal-answer q=”fs-id1165133221775″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165133221775″]Plot of -3 -3i in the complex plane (-3 along real axis, -3 along imaginary axis).[/hidden-answer]

5-4i

-1-5i

[reveal-answer q=”fs-id1165133260388″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165133260388″]Plot of -1 -5i in the complex plane (-1 along real axis, -5 along imaginary axis).[/hidden-answer]

3+2i

2i

[reveal-answer q=”fs-id1165135702570″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135702570″]Plot of 2i in the complex plane (0 along the real axis, 2 along the imaginary axis).[/hidden-answer]

-4

6-2i
[reveal-answer q=”fs-id1165135672734″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135672734″]Plot of 6-2i in the complex plane (6 along the real axis, -2 along the imaginary axis).[/hidden-answer]

-2+i

1-4i

[reveal-answer q=”fs-id1165137580704″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165137580704″]Plot of 1-4i in the complex plane (1 along the real axis, -4 along the imaginary axis).[/hidden-answer]

Technology

For the following exercises, find all answers rounded to the nearest hundredth.

Use the rectangular to polar feature on the graphing calculator to change\,5+5i\,to polar form.

Use the rectangular to polar feature on the graphing calculator to change\,3-2i\,
to polar form.

[reveal-answer q=”fs-id1165134329643″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165134329643″]

\,3.61{e}^{-0.59i}\,

[/hidden-answer]

Use the rectangular to polar feature on the graphing calculator to change -3-8i\,
to polar form.

Use the polar to rectangular feature on the graphing calculator to change\,4\mathrm{cis}\left(120°\right)\,to rectangular form.

[reveal-answer q=”fs-id1165135694966″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135694966″]

\,-2+3.46i\,

[/hidden-answer]

Use the polar to rectangular feature on the graphing calculator to change\,2\mathrm{cis}\left(45°\right)\,to rectangular form.

Use the polar to rectangular feature on the graphing calculator to change\,5\mathrm{cis}\left(210°\right)\,to rectangular form.

[reveal-answer q=”fs-id1165135252140″]Show Solution[/reveal-answer]
[hidden-answer a=”fs-id1165135252140″]

\,-4.33-2.50i\,

[/hidden-answer]

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\,n\text{th}\,power or nth roots of a complex number; states that, for a positive integer\,n,{z}^{n}\,is found by raising the modulus to the\,n\text{th}\,power and multiplying the angles by\,n\,
modulus
the absolute value of a complex number, or the distance from the origin to the point\,\left(x,y\right);\,also called the amplitude
polar form of a complex number
a complex number expressed in terms of an angle \theta and its distance from the origin\,r;\,can be found by using conversion formulas\,x=r\mathrm{cos}\,\theta ,\,\,y=r\mathrm{sin}\,\theta ,\,\,and\,r=\sqrt{{x}^{2}+{y}^{2}}

License

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Polar Form of Complex Numbers by OpenStax is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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