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Polar Form of Complex Numbers

Plot complex numbers in the complex plane.

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, \(bi.\)

Example

Try it.

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

Solution

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

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 , shows \(|z|.\)

Example

Try it.

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

Solution

Using the formula, we have

\[\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}} \\ |z|=\sqrt{{(\sqrt{5})}^{2}+{(-1)}^{2}} \\ |z|=\sqrt{5+1} \\ |z|=\sqrt{6}\end{array}\]

See .

Example

Try it.

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

Solution

Using the formula, we have

\[\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}} \\ |z|=\sqrt{{(3)}^{2}+{(-4)}^{2}} \\ |z|=\sqrt{9+16} \\ \begin{array}{l}|z|=\sqrt{25} \\ |z|=5\end{array}\end{array}\]

The absolute value \(z\) is 5. See .

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:

\[\begin{array}{l}x=r\cos \ \theta \\ y=r\sin \ \theta \\ r=\sqrt{{x}^{2}+{y}^{2}}\end{array}\]

We review these relationships in .

We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point \((x,y).\) 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

\[\begin{array}{l}z=x+yi \\ z=r\cos \ \theta +(r\sin \ \theta )i \\ z=r(\cos \ \theta +i\sin \ \theta )\end{array}\]

Example

Try it.

Express the complex number \(4i\) using polar coordinates.

Solution

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.

\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{0}^{2}+{4}^{2}} \\ r=\sqrt{16} \\ r=4\end{array}\]

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

Example

Try it.

Find the polar form of \(-4+4i.\)

Solution

First, find the value of \(r.\)

\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{(-4)}^{2}+({4}^{2})} \\ r=\sqrt{32} \\ r=4\sqrt{2}\end{array}\]

Find the angle \(\theta\) using the formula:

\[\begin{array}{l}\cos \ \theta =\frac{x}{r} \\ \cos \ \theta =\frac{-4}{4\sqrt{2}} \\ \cos \ \theta =-\frac{1}{\sqrt{2}} \\ \theta ={\cos }^{-1}(-\frac{1}{\sqrt{2}})=\frac{3\pi }{4}\end{array}\]

Thus, the solution is \(4\sqrt{2}\text{cis}(\frac{3\pi }{4}).\)

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 \(z=r(\cos \ \theta +i\sin \ \theta ),\) first evaluate the trigonometric functions \(\cos \ \theta\) and \(\sin \ \theta .\) Then, multiply through by \(r.\)

Example

Try it.

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

\[z=12(\cos (\frac{\pi }{6})+i\sin (\frac{\pi }{6}))\]

Solution

We begin by evaluating the trigonometric expressions.

\[\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}\ \text{and}\ \sin (\frac{\pi }{6})=\frac{1}{2}\]

After substitution, the complex number is

\[z=12(\frac{\sqrt{3}}{2}+\frac{1}{2}i)\]

We apply the distributive property:

\[\begin{array}{l}z=12(\frac{\sqrt{3}}{2}+\frac{1}{2}i) \\ =(12)\frac{\sqrt{3}}{2}+(12)\frac{1}{2}i \\ =6\sqrt{3}+6i\end{array}\]

The rectangular form of the given point in complex form is \(6\sqrt{3}+6i.\)

Example

Try it.

Find the rectangular form of the complex number given \(r=13\) and \(\tan \ \theta =\frac{5}{12}.\) Assume the number is in the first quadrant.

Solution

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

\[\begin{array}{l}z=13(\cos \ \theta +i\sin \ \theta ) \\ =13(\frac{12}{13}+\frac{5}{13}i) \\ =12+5i\end{array}\]

The rectangular form of the given number in complex form is \(12+5i.\)

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.

Example

Try it.

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

Solution

Follow the formula

\[\begin{array}{l}{z}_{1}{z}_{2}=4⋅2[\cos (80^{\circ}+145^{\circ})+i\sin (80^{\circ}+145^{\circ})] \\ {z}_{1}{z}_{2}=8[\cos (225^{\circ})+i\sin (225^{\circ})] \\ {z}_{1}{z}_{2}=8[\cos (\frac{5\pi }{4})+i\sin (\frac{5\pi }{4})] \\ {z}_{1}{z}_{2}=8[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})] \\ {z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2}\end{array}\]

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.

Example

Try it.

Find the quotient of \({z}_{1}=2(\cos (213^{\circ})+i\sin (213^{\circ}))\) and \({z}_{2}=4(\cos (33^{\circ})+i\sin (33^{\circ})).\)

Solution

Using the formula, we have

\[\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}[\cos (213^{\circ}-33^{\circ})+i\sin (213^{\circ}-33^{\circ})] \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}[\cos (180^{\circ})+i\sin (180^{\circ})] \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}[-1+0i] \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}\end{array}\]

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.

Example

Try it.

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 \(r.\)

\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{(1)}^{2}+{(1)}^{2}} \\ r=\sqrt{2}\end{array}\]

Then we find \(\theta .\) Using the formula \(\tan \ \theta =\frac{y}{x}\) gives

\[\begin{array}{l}\tan \ \theta =\frac{1}{1} \\ \tan \ \theta =1 \\ \theta =\frac{\pi }{4}\end{array}\]

Use De Moivre’s Theorem to evaluate the expression.

\[\begin{array}{l}{(a+bi)}^{n}={r}^{n}[\cos (n\theta )+i\sin (n\theta )] \\ {(1+i)}^{5}={(\sqrt{2})}^{5}[\cos (5⋅\frac{\pi }{4})+i\sin (5⋅\frac{\pi }{4})] \\ {(1+i)}^{5}=4\sqrt{2}[\cos (\frac{5\pi }{4})+i\sin (\frac{5\pi }{4})] \\ {(1+i)}^{5}=4\sqrt{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})] \\ {(1+i)}^{5}=-4-4i\end{array}\]

Finding Roots of Complex Numbers in Polar Form

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

Example

Try it.

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

Solution

We have

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

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

\[{z}^{\frac{1}{3}}=2(\cos (\frac{2\pi }{9})+i\sin (\frac{2\pi }{9}))\]

When \(k=1,\) we have

\[\begin{array}{llll}{z}^{\frac{1}{3}}=2[\cos (\frac{2\pi }{9}+\frac{6\pi }{9})+i\sin (\frac{2\pi }{9}+\frac{6\pi }{9})]\begin{array}{llll} & & & \end{array}\text{ Add }\frac{2(1)\pi }{3}\text{ to each angle.} \\ {z}^{\frac{1}{3}}=2(\cos (\frac{8\pi }{9})+i\sin (\frac{8\pi }{9}))\end{array}\]

When \(k=2,\) we have

\[\begin{array}{lllll}{z}^{\frac{1}{3}}=2[\cos (\frac{2\pi }{9}+\frac{12\pi }{9})+i\sin (\frac{2\pi }{9}+\frac{12\pi }{9})]\begin{array}{llll} & & & \end{array} & \text{Add }\frac{2(2)\pi }{3}\text{ to each angle.} \\ {z}^{\frac{1}{3}}=2(\cos (\frac{14\pi }{9})+i\sin (\frac{14\pi }{9})) & \end{array}\]

Remember to find the common denominator to simplify fractions in situations like this one. For \(k=1,\) the angle simplification is

\[\begin{array}{l}\frac{\frac{2\pi }{3}}{3}+\frac{2(1)\pi }{3}=\frac{2\pi }{3}(\frac{1}{3})+\frac{2(1)\pi }{3}(\frac{3}{3}) \\ =\frac{2\pi }{9}+\frac{6\pi }{9} \\ =\frac{8\pi }{9}\end{array}\]

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 .
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 and .
To write complex numbers in polar form, we use the formulas \(x=r\cos \ \theta ,y=r\sin \ \theta ,\) and \(r=\sqrt{{x}^{2}+{y}^{2}}.\) Then, \(z=r(\cos \ \theta +i\sin \ \theta ).\) See and .
To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by \(r.\) See and .
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 .
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 .
To find the power of a complex number \({z}^{n},\) raise \(r\) to the power \(n,\) and multiply \(\theta\) by \(n.\) See .
Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. See .

Plotting Complex Numbers in the Complex Plane

Example

Try it.

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

Solution

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

Finding the Absolute Value of a Complex Number

Example

Try it.

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

Solution

Using the formula, we have

\[\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}} \\ |z|=\sqrt{{(\sqrt{5})}^{2}+{(-1)}^{2}} \\ |z|=\sqrt{5+1} \\ |z|=\sqrt{6}\end{array}\]

See .

Example

Try it.

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

Solution

Using the formula, we have

\[\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}} \\ |z|=\sqrt{{(3)}^{2}+{(-4)}^{2}} \\ |z|=\sqrt{9+16} \\ \begin{array}{l}|z|=\sqrt{25} \\ |z|=5\end{array}\end{array}\]

The absolute value \(z\) is 5. See .

Writing Complex Numbers in Polar Form

\[\begin{array}{l}x=r\cos \ \theta \\ y=r\sin \ \theta \\ r=\sqrt{{x}^{2}+{y}^{2}}\end{array}\]

We review these relationships in .

\[\begin{array}{l}z=x+yi \\ z=r\cos \ \theta +(r\sin \ \theta )i \\ z=r(\cos \ \theta +i\sin \ \theta )\end{array}\]

Example

Try it.

Express the complex number \(4i\) using polar coordinates.

Solution

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.

\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{0}^{2}+{4}^{2}} \\ r=\sqrt{16} \\ r=4\end{array}\]

Example

Try it.

Find the polar form of \(-4+4i.\)

Solution

First, find the value of \(r.\)

\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{(-4)}^{2}+({4}^{2})} \\ r=\sqrt{32} \\ r=4\sqrt{2}\end{array}\]

Find the angle \(\theta\) using the formula:

\[\begin{array}{l}\cos \ \theta =\frac{x}{r} \\ \cos \ \theta =\frac{-4}{4\sqrt{2}} \\ \cos \ \theta =-\frac{1}{\sqrt{2}} \\ \theta ={\cos }^{-1}(-\frac{1}{\sqrt{2}})=\frac{3\pi }{4}\end{array}\]

Thus, the solution is \(4\sqrt{2}\text{cis}(\frac{3\pi }{4}).\)

Converting a Complex Number from Polar to Rectangular Form

Example

Try it.

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

\[z=12(\cos (\frac{\pi }{6})+i\sin (\frac{\pi }{6}))\]

Solution

We begin by evaluating the trigonometric expressions.

\[\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}\ \text{and}\ \sin (\frac{\pi }{6})=\frac{1}{2}\]

After substitution, the complex number is

\[z=12(\frac{\sqrt{3}}{2}+\frac{1}{2}i)\]

We apply the distributive property:

\[\begin{array}{l}z=12(\frac{\sqrt{3}}{2}+\frac{1}{2}i) \\ =(12)\frac{\sqrt{3}}{2}+(12)\frac{1}{2}i \\ =6\sqrt{3}+6i\end{array}\]

The rectangular form of the given point in complex form is \(6\sqrt{3}+6i.\)

Example

Try it.

Find the rectangular form of the complex number given \(r=13\) and \(\tan \ \theta =\frac{5}{12}.\) Assume the number is in the first quadrant.

Solution

\[\begin{array}{l}z=13(\cos \ \theta +i\sin \ \theta ) \\ =13(\frac{12}{13}+\frac{5}{13}i) \\ =12+5i\end{array}\]

The rectangular form of the given number in complex form is \(12+5i.\)

Finding Products of Complex Numbers in Polar Form

Example

Try it.

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

Solution

Follow the formula

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.

Example

Try it.

Find the quotient of \({z}_{1}=2(\cos (213^{\circ})+i\sin (213^{\circ}))\) and \({z}_{2}=4(\cos (33^{\circ})+i\sin (33^{\circ})).\)

Solution

Using the formula, we have

Finding Powers of Complex Numbers in Polar Form

Example

Try it.

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 \(r.\)

\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{(1)}^{2}+{(1)}^{2}} \\ r=\sqrt{2}\end{array}\]

Then we find \(\theta .\) Using the formula \(\tan \ \theta =\frac{y}{x}\) gives

\[\begin{array}{l}\tan \ \theta =\frac{1}{1} \\ \tan \ \theta =1 \\ \theta =\frac{\pi }{4}\end{array}\]

Use De Moivre’s Theorem to evaluate the expression.

Finding Roots of Complex Numbers in Polar Form

Example

Try it.

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

Solution

We have

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

\[{z}^{\frac{1}{3}}=2(\cos (\frac{2\pi }{9})+i\sin (\frac{2\pi }{9}))\]

When \(k=1,\) we have

When \(k=2,\) we have

Remember to find the common denominator to simplify fractions in situations like this one. For \(k=1,\) the angle simplification is

\[\begin{array}{l}\frac{\frac{2\pi }{3}}{3}+\frac{2(1)\pi }{3}=\frac{2\pi }{3}(\frac{1}{3})+\frac{2(1)\pi }{3}(\frac{3}{3}) \\ =\frac{2\pi }{9}+\frac{6\pi }{9} \\ =\frac{8\pi }{9}\end{array}\]

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 .
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 and .
To write complex numbers in polar form, we use the formulas \(x=r\cos \ \theta ,y=r\sin \ \theta ,\) and \(r=\sqrt{{x}^{2}+{y}^{2}}.\) Then, \(z=r(\cos \ \theta +i\sin \ \theta ).\) See and .
To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by \(r.\) See and .
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 .
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 .
To find the power of a complex number \({z}^{n},\) raise \(r\) to the power \(n,\) and multiply \(\theta\) by \(n.\) See .
Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

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

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See .
Plot the point \(1+5i\) in the complex plane.
Find the absolute value of \(z=\sqrt{5}-i.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Using the formula, we have
\[\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}} \\ |z|=\sqrt{{(\sqrt{5})}^{2}+{(-1)}^{2}} \\ |z|=\sqrt{5+1} \\ |z|=\sqrt{6}\end{array}\]
See .
Find the absolute value of the complex number \(z=12-5i.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

13
Given \(z=3-4i,\) find \(|z|.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Using the formula, we have
\[\begin{array}{l}|z|=\sqrt{{x}^{2}+{y}^{2}} \\ |z|=\sqrt{{(3)}^{2}+{(-4)}^{2}} \\ |z|=\sqrt{9+16} \\ \begin{array}{l}|z|=\sqrt{25} \\ |z|=5\end{array}\end{array}\]
The absolute value \(z\) is 5. See .
Given \(z=1-7i,\) find \(|z|.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(|z|=\sqrt{50}=5\sqrt{2}\)
Express the complex number \(4i\) using polar coordinates.

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

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.
\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{0}^{2}+{4}^{2}} \\ r=\sqrt{16} \\ r=4\end{array}\]
Next, we look at \(x.\) If \(x=r\cos \ \theta ,\) and \(x=0,\) then \(\theta =\frac{\pi }{2}.\) In polar coordinates, the complex number \(z=0+4i\) can be written as \(z=4(\cos (\frac{\pi }{2})+i\sin (\frac{\pi }{2}))\) or \(4\text{cis}(\ \frac{\pi }{2}).\) See .
Express \(z=3i\) as \(r\ \text{cis}\ \theta\) in polar form.

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(z=3(\cos (\frac{\pi }{2})+i\sin (\frac{\pi }{2}))\)
Find the polar form of \(-4+4i.\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

First, find the value of \(r.\)
\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{(-4)}^{2}+({4}^{2})} \\ r=\sqrt{32} \\ r=4\sqrt{2}\end{array}\]
Find the angle \(\theta\) using the formula:
\[\begin{array}{l}\cos \ \theta =\frac{x}{r} \\ \cos \ \theta =\frac{-4}{4\sqrt{2}} \\ \cos \ \theta =-\frac{1}{\sqrt{2}} \\ \theta ={\cos }^{-1}(-\frac{1}{\sqrt{2}})=\frac{3\pi }{4}\end{array}\]
Thus, the solution is \(4\sqrt{2}\text{cis}(\frac{3\pi }{4}).\)
Write \(z=\sqrt{3}+i\) in polar form.

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(z=2(\cos (\frac{\pi }{6})+i\sin (\frac{\pi }{6}))\)
Convert the polar form of the given complex number to rectangular form:
\[z=12(\cos (\frac{\pi }{6})+i\sin (\frac{\pi }{6}))\]

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

We begin by evaluating the trigonometric expressions.
\[\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}\ \text{and}\ \sin (\frac{\pi }{6})=\frac{1}{2}\]
After substitution, the complex number is
\[z=12(\frac{\sqrt{3}}{2}+\frac{1}{2}i)\]
We apply the distributive property:
\[\begin{array}{l}z=12(\frac{\sqrt{3}}{2}+\frac{1}{2}i) \\ =(12)\frac{\sqrt{3}}{2}+(12)\frac{1}{2}i \\ =6\sqrt{3}+6i\end{array}\]
The rectangular form of the given point in complex form is \(6\sqrt{3}+6i.\)
Find the rectangular form of the complex number given \(r=13\) and \(\tan \ \theta =\frac{5}{12}.\) Assume the number is in the first quadrant.

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

If \(\tan \ \theta =\frac{5}{12},\) and \(\tan \ \theta =\frac{y}{x},\) we first confirm \(r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{. }\) We then find \(\cos \ \theta =\frac{x}{r}\) and \(\sin \ \theta =\frac{y}{r}.\)
\[\begin{array}{l}z=13(\cos \ \theta +i\sin \ \theta ) \\ =13(\frac{12}{13}+\frac{5}{13}i) \\ =12+5i\end{array}\]
The rectangular form of the given number in complex form is \(12+5i.\)
Convert the complex number to rectangular form:
\[z=4(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6})\]

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(z=2\sqrt{3}-2i\)
Find the product of \({z}_{1}{z}_{2},\) given \({z}_{1}=4(\cos (80^{\circ})+i\sin (80^{\circ}))\) and \({z}_{2}=2(\cos (145^{\circ})+i\sin (145^{\circ})).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Follow the formula
\[\begin{array}{l}{z}_{1}{z}_{2}=4⋅2[\cos (80^{\circ}+145^{\circ})+i\sin (80^{\circ}+145^{\circ})] \\ {z}_{1}{z}_{2}=8[\cos (225^{\circ})+i\sin (225^{\circ})] \\ {z}_{1}{z}_{2}=8[\cos (\frac{5\pi }{4})+i\sin (\frac{5\pi }{4})] \\ {z}_{1}{z}_{2}=8[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})] \\ {z}_{1}{z}_{2}=-4\sqrt{2}-4i\sqrt{2}\end{array}\]
Find the quotient of \({z}_{1}=2(\cos (213^{\circ})+i\sin (213^{\circ}))\) and \({z}_{2}=4(\cos (33^{\circ})+i\sin (33^{\circ})).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Using the formula, we have
\[\begin{array}{l}\frac{{z}_{1}}{{z}_{2}}=\frac{2}{4}[\cos (213^{\circ}-33^{\circ})+i\sin (213^{\circ}-33^{\circ})] \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}[\cos (180^{\circ})+i\sin (180^{\circ})] \\ \frac{{z}_{1}}{{z}_{2}}=\frac{1}{2}[-1+0i] \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}+0i \\ \frac{{z}_{1}}{{z}_{2}}=-\frac{1}{2}\end{array}\]
Find the product and the quotient of \({z}_{1}=2\sqrt{3}(\cos (150^{\circ})+i\sin (150^{\circ}))\) and \({z}_{2}=2(\cos (30^{\circ})+i\sin (30^{\circ})).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({z}_{1}{z}_{2}=-4\sqrt{3};\frac{{z}_{1}}{{z}_{2}}=-\frac{\sqrt{3}}{2}+\frac{3}{2}i\)
Evaluate the expression \({(1+i)}^{5}\) using De Moivre’s Theorem.

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

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 \(r.\)
\[\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}} \\ r=\sqrt{{(1)}^{2}+{(1)}^{2}} \\ r=\sqrt{2}\end{array}\]
Then we find \(\theta .\) Using the formula \(\tan \ \theta =\frac{y}{x}\) gives
\[\begin{array}{l}\tan \ \theta =\frac{1}{1} \\ \tan \ \theta =1 \\ \theta =\frac{\pi }{4}\end{array}\]
Use De Moivre’s Theorem to evaluate the expression.
\[\begin{array}{l}{(a+bi)}^{n}={r}^{n}[\cos (n\theta )+i\sin (n\theta )] \\ {(1+i)}^{5}={(\sqrt{2})}^{5}[\cos (5⋅\frac{\pi }{4})+i\sin (5⋅\frac{\pi }{4})] \\ {(1+i)}^{5}=4\sqrt{2}[\cos (\frac{5\pi }{4})+i\sin (\frac{5\pi }{4})] \\ {(1+i)}^{5}=4\sqrt{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})] \\ {(1+i)}^{5}=-4-4i\end{array}\]
Evaluate the cube roots of \(z=8(\cos (\frac{2\pi }{3})+i\sin (\frac{2\pi }{3})).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

We have
\[\begin{array}{l}{z}^{\frac{1}{3}}={8}^{\frac{1}{3}}[\cos (\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3})+i\sin (\frac{\frac{2\pi }{3}}{3}+\frac{2k\pi }{3})] \\ {z}^{\frac{1}{3}}=2[\cos (\frac{2\pi }{9}+\frac{2k\pi }{3})+i\sin (\frac{2\pi }{9}+\frac{2k\pi }{3})]\end{array}\]
There will be three roots: \(k=0,\ 1,\ 2.\) When \(k=0,\) we have
\[{z}^{\frac{1}{3}}=2(\cos (\frac{2\pi }{9})+i\sin (\frac{2\pi }{9}))\]
When \(k=1,\) we have
\[\begin{array}{llll}{z}^{\frac{1}{3}}=2[\cos (\frac{2\pi }{9}+\frac{6\pi }{9})+i\sin (\frac{2\pi }{9}+\frac{6\pi }{9})]\begin{array}{llll} & & & \end{array}\text{ Add }\frac{2(1)\pi }{3}\text{ to each angle.} \\ {z}^{\frac{1}{3}}=2(\cos (\frac{8\pi }{9})+i\sin (\frac{8\pi }{9}))\end{array}\]
When \(k=2,\) we have
\[\begin{array}{lllll}{z}^{\frac{1}{3}}=2[\cos (\frac{2\pi }{9}+\frac{12\pi }{9})+i\sin (\frac{2\pi }{9}+\frac{12\pi }{9})]\begin{array}{llll} & & & \end{array} & \text{Add }\frac{2(2)\pi }{3}\text{ to each angle.} \\ {z}^{\frac{1}{3}}=2(\cos (\frac{14\pi }{9})+i\sin (\frac{14\pi }{9})) & \end{array}\]
Remember to find the common denominator to simplify fractions in situations like this one. For \(k=1,\) the angle simplification is
\[\begin{array}{l}\frac{\frac{2\pi }{3}}{3}+\frac{2(1)\pi }{3}=\frac{2\pi }{3}(\frac{1}{3})+\frac{2(1)\pi }{3}(\frac{3}{3}) \\ =\frac{2\pi }{9}+\frac{6\pi }{9} \\ =\frac{8\pi }{9}\end{array}\]
Find the four fourth roots of \(16(\cos (120^{\circ})+i\sin (120^{\circ})).\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({z}_{0}=2(\cos (30^{\circ})+i\sin (30^{\circ}))\)
\({z}_{1}=2(\cos (120^{\circ})+i\sin (120^{\circ}))\)
\({z}_{2}=2(\cos (210^{\circ})+i\sin (210^{\circ}))\)
\({z}_{3}=2(\cos (300^{\circ})+i\sin (300^{\circ}))\)
A complex number is \(a+bi.\) Explain each part.

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

a is the real part, b is the imaginary part, and \(i=\sqrt{-1}\)
What does the absolute value of a complex number represent?
How is a complex number converted to polar form?

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

Polar form converts the real and imaginary part of the complex number in polar form using \(x=r\cos \theta\) and \(y=r\sin \theta .\)
How do we find the product of two complex numbers?
What is De Moivre’s Theorem and what is it used for?

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\({z}^{n}={r}^{n}(\cos (n\theta )+i\sin (n\theta ))\) It is used to simplify polar form when a number has been raised to a power.
\(5+\text{}3i\)
\(-7+\text{}i\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(5\sqrt{2}\)
\(\sqrt{2}-6i\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\sqrt{38}\)
\(2.2-3.1i\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\sqrt{14.45}\)
\(-\frac{1}{2}-\frac{1}{2}\text{}i\)
\(\sqrt{3}+i\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(2\text{cis}(\frac{\pi }{6})\)
\(z=7\text{cis}(\frac{\pi }{6})\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{7\sqrt{3}}{2}+i\frac{7}{2}\)
\(z=2\text{cis}(\frac{\pi }{3})\)
\(z=4\text{cis}(\frac{7\pi }{6})\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(-2\sqrt{3}-2i\)
\(z=7\text{cis}(25^{\circ})\)
\(z=3\text{cis}(240^{\circ})\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(-1.5-i\frac{3\sqrt{3}}{2}\)
\(z=\sqrt{2}\text{cis}(100^{\circ})\)
\({z}_{1}=2\sqrt{3}\text{cis}(116^{\circ});\ {z}_{2}=2\text{cis}(82^{\circ})\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(4\sqrt{3}\text{cis}(198^{\circ})\)
\({z}_{1}=\sqrt{2}\text{cis}(205^{\circ});\ {z}_{2}=2\sqrt{2}\text{cis}(118^{\circ})\)
\({z}_{1}=3\text{cis}(120^{\circ});\ {z}_{2}=\frac{1}{4}\text{cis}(60^{\circ})\)

ଉତ୍ତରକୁ ଖୋଲନ୍ତୁ

\(\frac{3}{4}\text{cis}(180^{\circ})\)
\({z}_{1}=3\text{cis}(\frac{\pi }{4});\ {z}_{2}=5\text{cis}(\frac{\pi }{6})\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

imaginary unit
i² = −1.

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\neq

not equal
The two sides are different.

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

How to: Polar Form of Complex Numbers

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.

Questions people ask

Why radians instead of degrees?

A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.

Why does sin x = 1/2 have infinitely many solutions?

Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.

How do I remember the exact values?

Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.

ନିଜେ ଚେଷ୍ଟାକରନ୍ତୁ

Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ଅଧିକ Trigonometry

The unit circle Trigonometric equations Trigonometric identities Degrees and radians Right-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosines Graphs of sine, cosine and tangent Inverse trigonometric functions