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.
Plotting Complex Numbers in the Complex Plane
Plotting a complex number [latex]a+bi[/latex] is similar to plotting a real number, except that the horizontal axis represents the real part of the number, [latex]a[/latex], and the vertical axis represents the imaginary part of the number, [latex]bi[/latex].How To: Given a complex number [latex]a+bi[/latex], plot it in the complex plane.
- Label the horizontal axis as the real axis and the vertical axis as the imaginary axis.
- Plot the point in the complex plane by moving [latex]a[/latex] units in the horizontal direction and [latex]b[/latex] units in the vertical direction.
Example 1: Plotting a Complex Number in the Complex Plane
Plot the complex number [latex]2 - 3i[/latex] in the complex plane.Solution
From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. Figure 1
Try It 1
Plot the point [latex]1+5i[/latex] in the complex plane. SolutionFinding 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 [latex]|z|[/latex]. It measures the distance from the origin to a point in the plane. For example, the graph of [latex]z=2+4i[/latex], in Figure 2, shows [latex]|z|[/latex]. Figure 2
A General Note: Absolute Value of a Complex Number
Given [latex]z=x+yi[/latex], a complex number, the absolute value of [latex]z[/latex] is defined as[latex]|z|=\sqrt{{x}^{2}+{y}^{2}}[/latex]
It is the distance from the origin to the point [latex]\left(x,y\right)[/latex].
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, [latex]\left(0,\text{ }0\right)[/latex].
Example 2: Finding the Absolute Value of a Complex Number with a Radical
Find the absolute value of [latex]z=\sqrt{5}-i[/latex].Solution
Using the formula, we have[latex]\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}[/latex]
Figure 3
Try It 2
Find the absolute value of the complex number [latex]z=12 - 5i[/latex]. SolutionExample 3: Finding the Absolute Value of a Complex Number
Given [latex]z=3 - 4i[/latex], find [latex]|z|[/latex].Solution
Using the formula, we have[latex]\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}[/latex]
The absolute value [latex]z[/latex] is 5.
Figure 4
Try It 3
Given [latex]z=1 - 7i[/latex], find [latex]|z|[/latex]. SolutionWriting Complex Numbers in Polar Form
The polar form of a complex number expresses a number in terms of an angle [latex]\theta [/latex] and its distance from the origin [latex]r[/latex]. Given a complex number in rectangular form expressed as [latex]z=x+yi[/latex], we use the same conversion formulas as we do to write the number in trigonometric form:[latex]\begin{array}{l}x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}[/latex]
We review these relationships in Figure 5.
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 [latex]\left(x,y\right)[/latex]. The modulus, then, is the same as [latex]r[/latex], the radius in polar form. We use [latex]\theta [/latex] to indicate the angle of direction (just as with polar coordinates). Substituting, we have
[latex]\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}[/latex]
A General Note: Polar Form of a Complex Number
Writing a complex number in polar form involves the following conversion formulas:[latex]\begin{array}{l}\hfill \\ x=r\cos \theta \hfill \\ y=r\sin \theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}[/latex]
Making a direct substitution, we have
[latex]\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}[/latex]
where [latex]r[/latex] is the modulus and [latex]\theta [/latex] is the argument. We often use the abbreviation [latex]r\text{cis}\theta [/latex] to represent [latex]r\left(\cos \theta +i\sin \theta \right)[/latex].
Example 4: Expressing a Complex Number Using Polar Coordinates
Express the complex number [latex]4i[/latex] using polar coordinates.Solution
On the complex plane, the number [latex]z=4i[/latex] is the same as [latex]z=0+4i[/latex]. Writing it in polar form, we have to calculate [latex]r[/latex] first.[latex]\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}[/latex]
Next, we look at [latex]x[/latex]. If [latex]x=r\cos \theta [/latex], and [latex]x=0[/latex], then [latex]\theta =\frac{\pi }{2}[/latex]. In polar coordinates, the complex number [latex]z=0+4i[/latex] can be written as [latex]z=4\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)[/latex] or [latex]4\text{cis}\left(\frac{\pi }{2}\right)[/latex].
Figure 6
Try It 4
Express [latex]z=3i[/latex] as [latex]r\text{cis}\theta [/latex] in polar form. SolutionExample 5: Finding the Polar Form of a Complex Number
Find the polar form of [latex]-4+4i[/latex].Solution
First, find the value of [latex]r[/latex].[latex]\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}[/latex]
Find the angle [latex]\theta [/latex] using the formula:
[latex]\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}[/latex]
Thus, the solution is [latex]4\sqrt{2}\text{cis}\left(\frac{3\pi }{4}\right)[/latex].
Try It 5
Write [latex]z=\sqrt{3}+i[/latex] in polar form. SolutionConverting 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 [latex]z=r\left(\cos \theta +i\sin \theta \right)[/latex], first evaluate the trigonometric functions [latex]\cos \theta [/latex] and [latex]\sin \theta [/latex]. Then, multiply through by [latex]r[/latex].Example 6: Converting from Polar to Rectangular Form
Convert the polar form of the given complex number to rectangular form:[latex]z=12\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)[/latex]
Solution
We begin by evaluating the trigonometric expressions.[latex]\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\\\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}[/latex]
After substitution, the complex number is
[latex]z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)[/latex]
We apply the distributive property:
[latex]\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}[/latex]
The rectangular form of the given point in complex form is [latex]6\sqrt{3}+6i[/latex].
Example 7: Finding the Rectangular Form of a Complex Number
Find the rectangular form of the complex number given [latex]r=13[/latex] and [latex]\tan \theta =\frac{5}{12}[/latex].Solution
If [latex]\tan \theta =\frac{5}{12}[/latex], and [latex]\tan \theta =\frac{y}{x}[/latex], we first determine [latex]r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{.}[/latex] We then find [latex]\cos \theta =\frac{x}{r}[/latex] and [latex]\sin \theta =\frac{y}{r}[/latex].[latex]\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}[/latex]
The rectangular form of the given number in complex form is [latex]12+5i[/latex].
Try It 6
Convert the complex number to rectangular form:[latex]z=4\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)[/latex]
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 [latex]{z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right)[/latex] and [latex]{z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right)[/latex], then the product of these numbers is given as:[latex]\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}[/latex]
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 [latex]{z}_{1}{z}_{2}[/latex], given [latex]{z}_{1}=4\left(\cos \left(80^\circ \right)+i\sin \left(80^\circ \right)\right)[/latex] and [latex]{z}_{2}=2\left(\cos \left(145^\circ \right)+i\sin \left(145^\circ \right)\right)[/latex].Solution
Follow the formula[latex]\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}[/latex]
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 [latex]{z}_{1}={r}_{1}\left(\cos {\theta }_{1}+i\sin {\theta }_{1}\right)[/latex] and [latex]{z}_{2}={r}_{2}\left(\cos {\theta }_{2}+i\sin {\theta }_{2}\right)[/latex], then the quotient of these numbers is[latex]\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}[/latex]
Notice that the moduli are divided, and the angles are subtracted.
How To: Given two complex numbers in polar form, find the quotient.
- Divide [latex]\frac{{r}_{1}}{{r}_{2}}[/latex].
- Find [latex]{\theta }_{1}-{\theta }_{2}[/latex].
- Substitute the results into the formula: [latex]z=r\left(\cos \theta +i\sin \theta \right)[/latex]. Replace [latex]r[/latex] with [latex]\frac{{r}_{1}}{{r}_{2}}[/latex], and replace [latex]\theta [/latex] with [latex]{\theta }_{1}-{\theta }_{2}[/latex].
- Calculate the new trigonometric expressions and multiply through by [latex]r[/latex].
Example 9: Finding the Quotient of Two Complex Numbers
Find the quotient of [latex]{z}_{1}=2\left(\cos \left(213^\circ \right)+i\sin \left(213^\circ \right)\right)[/latex] and [latex]{z}_{2}=4\left(\cos \left(33^\circ \right)+i\sin \left(33^\circ \right)\right)[/latex].Solution
Using the formula, we have[latex]\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}[/latex]
Try It 7
Find the product and the quotient of [latex]{z}_{1}=2\sqrt{3}\left(\cos \left(150^\circ \right)+i\sin \left(150^\circ \right)\right)[/latex] and [latex]{z}_{2}=2\left(\cos \left(30^\circ \right)+i\sin \left(30^\circ \right)\right)[/latex]. SolutionFinding 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 [latex]n,{z}^{n}[/latex] is found by raising the modulus to the [latex]n\text{th}[/latex] power and multiplying the argument by [latex]n[/latex]. It is the standard method used in modern mathematics.A General Note: De Moivre’s Theorem
If [latex]z=r\left(\cos \theta +i\sin \theta \right)[/latex] is a complex number, then[latex]\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}[/latex]
where [latex]n[/latex] is a positive integer.
Example 10: Evaluating an Expression Using De Moivre’s Theorem
Evaluate the expression [latex]{\left(1+i\right)}^{5}[/latex] using De Moivre’s Theorem.Solution
Since De Moivre’s Theorem applies to complex numbers written in polar form, we must first write [latex]\left(1+i\right)[/latex] in polar form. Let us find [latex]r[/latex].[latex]\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}[/latex]
Then we find [latex]\theta [/latex]. Using the formula [latex]\tan \theta =\frac{y}{x}[/latex] gives
[latex]\begin{array}{l}\tan \theta =\frac{1}{1}\hfill \\ \tan \theta =1\hfill \\ \theta =\frac{\pi }{4}\hfill \end{array}[/latex]
Use De Moivre’s Theorem to evaluate the expression.
[latex]\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}[/latex]
Finding Roots of Complex Numbers in Polar Form
To find the nth root of a complex number in polar form, we use the [latex]n\text{th}[/latex] 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 [latex]n\text{th}[/latex] roots of complex numbers in polar form.A General Note: The nth Root Theorem
To find the [latex]n\text{th}[/latex] root of a complex number in polar form, use the formula given as[latex]{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][/latex]
where [latex]k=0,1,2,3,...,n - 1[/latex]. We add [latex]\frac{2k\pi }{n}[/latex] to [latex]\frac{\theta }{n}[/latex] in order to obtain the periodic roots.
Example 11: Finding the nth Root of a Complex Number
Evaluate the cube roots of [latex]z=8\left(\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right)[/latex].Solution
We have[latex]\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}[/latex]
There will be three roots: [latex]k=0,1,2[/latex]. When [latex]k=0[/latex], we have
[latex]{z}^{\frac{1}{3}}=2\left(\cos \left(\frac{2\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}\right)\right)[/latex]
When [latex]k=1[/latex], we have
[latex]\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}[/latex]
When [latex]k=2[/latex], we have
[latex]\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}[/latex]
Remember to find the common denominator to simplify fractions in situations like this one. For [latex]k=1[/latex], the angle simplification is
[latex]\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}[/latex]
Try It 8
Find the four fourth roots of [latex]16\left(\cos \left(120^\circ \right)+i\sin \left(120^\circ \right)\right)[/latex]. SolutionKey Concepts
- Complex numbers in the form [latex]a+bi[/latex] 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: [latex]|z|=\sqrt{{a}^{2}+{b}^{2}}[/latex].
- To write complex numbers in polar form, we use the formulas [latex]x=r\cos \theta ,y=r\sin \theta [/latex], and [latex]r=\sqrt{{x}^{2}+{y}^{2}}[/latex]. Then, [latex]z=r\left(\cos \theta +i\sin \theta \right)[/latex].
- To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by [latex]r[/latex].
- 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 [latex]{z}^{n}[/latex], raise [latex]r[/latex] to the power [latex]n[/latex], and multiply [latex]\theta [/latex] by [latex]n[/latex].
- 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 [latex]n\text{th}[/latex] power or nth roots of a complex number; states that, for a positive integer [latex]n,{z}^{n}[/latex] is found by raising the modulus to the [latex]n\text{th}[/latex] power and multiplying the angles by [latex]n[/latex]
- modulus
- the absolute value of a complex number, or the distance from the origin to the point [latex]\left(x,y\right)[/latex]; also called the amplitude
- polar form of a complex number
- a complex number expressed in terms of an angle [latex]\theta [/latex] and its distance from the origin [latex]r[/latex]; can be found by using conversion formulas [latex]x=r\cos \theta ,y=r\sin \theta [/latex], and [latex]r=\sqrt{{x}^{2}+{y}^{2}}[/latex]
Section Exercises
1. A complex number is [latex]a+bi[/latex]. 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. [latex]5+3i[/latex] 7. [latex]−7+i[/latex] 8. [latex]−3−3i[/latex] 9. [latex]\sqrt{2}−6i[/latex] 10. [latex]2i[/latex] 11. [latex]2.2−3.1i[/latex] For the following exercises, write the complex number in polar form. 12. [latex]2+2i[/latex] 13. [latex]8−4i[/latex] 14. [latex]−\frac{1}{2}−\frac{1}{2}i[/latex] 15. [latex]\sqrt{3}+i[/latex] 16. [latex]3i[/latex] For the following exercises, convert the complex number from polar to rectangular form. 17. [latex]z=7\text{cis}\left(\frac{\pi}{6}\right)[/latex] 18. [latex]z=2\text{cis}\left(\frac{\pi}{3}\right)[/latex] 19. [latex]z=4\text{cis}\left(\frac{7\pi}{6}\right)[/latex] 20. [latex]z=7\text{cis}\left(25^{\circ}\right)[/latex] 21. [latex]z=3\text{cis}\left(240^{\circ}\right)[/latex] 22. [latex]z=\sqrt{2}\text{cis}\left(100^{\circ}\right)[/latex] For the following exercises, find z1z2 in polar form. 23. [latex]z_{1}=2\sqrt{3}\text{cis}\left(116^{\circ}\right)\text{; }\left(118^{\circ}\right)[/latex] 24. [latex]z_{1}=\sqrt{2}\text{cis}\left(205^{\circ}\right)\text{; }z_{2}=\frac{1}{4}\text{cis}\left(60^{\circ}\right)[/latex] 25. [latex]z_{1}=3\text{cis}\left(120^{\circ}\right)\text{; }z_{2}=\frac{1}{4}\text{cis}\left(60^{\circ}\right)[/latex] 26. [latex]z_{1}=3\text{cis}\left(\frac{5\pi}{4}\right)\text{; }z_{2}=5\text{cis}\left(\frac{\pi}{6}\right)[/latex] 27. [latex]z_{1}=\sqrt{5}\text{cis}\left(\frac{5\pi}{8}\right)\text{; }z_{2}=\sqrt{15}\text{cis}\left(\frac{\pi}{12}\right)[/latex] 28. [latex]z_{1}=4\text{cis}\left(\frac{\pi}{2}\right)\text{; }z_{2}=2\text{cis}\left(\frac{\pi}{4}\right)[/latex] For the following exercises, find [latex]\frac{z_{1}}{z_{2}}[/latex] in polar form. 29. [latex]z_{1}=21\text{cis}\left(135^{\circ}\right)\text{; }z_{2}=3\text{cis}\left(65^{\circ}\right)[/latex] 30. [latex]z_{1}=\sqrt{2}\text{cis}\left(90^{\circ}\right)\text{; }z_{2}=2\text{cis}\left(60^{\circ}\right)[/latex] 31. [latex]z_{1}=15\text{cis}\left(120^{\circ}\right)\text{; }z_{2}=3\text{cis}\left(40^{\circ}\right)[/latex] 32. [latex]z_{1}=6\text{cis}\left(\frac{\pi}{3}\right)\text{; }z_{2}=2\text{cis}\left(\frac{\pi}{4}\right)[/latex] 33. [latex]z_{1}=5\sqrt{2}\text{cis}\left(\pi\right)\text{; }z_{2}=\sqrt{2}\text{cis}\left(\frac{2\pi}{3}\right)[/latex] 34. [latex]z_{1}=2\text{cis}\left(\frac{3\pi}{5}\right)\text{; }z_{2}=3\text{cis}\left(\frac{\pi}{4}\right)[/latex] For the following exercises, find the powers of each complex number in polar form. 35. Find [latex]z^{3}[/latex] when [latex]z=5\text{cis}\left(45^{\circ}\right)[/latex]. 36. Find [latex]z^{4}[/latex] when [latex]z=2\text{cis}\left(70^{\circ}\right)[/latex]. 37. Find [latex]z^{2}[/latex] when [latex]z=3\text{cis}\left(120^{\circ}\right)[/latex]. 38. Find [latex]z^{2}[/latex] when [latex]z=4\text{cis}\left(\frac{\pi}{4}\right)[/latex] 39. Find [latex]z^{4}[/latex] when [latex]z=\text{cis}\left(\frac{3\pi}{16}\right)[/latex]. 40. Find [latex]z^{3}[/latex] when [latex]z=3\text{cis}\left(\frac{5\pi}{3}\right)[/latex]. For the following exercises, evaluate each root. 41. Evaluate the cube root of z when [latex]z=27\text{cis}\left(240^{\circ}\right)[/latex]. 42. Evaluate the square root of z when [latex]z=16\text{cis}\left(100^{\circ}\right)[/latex]. 43. Evaluate the cube root of z when [latex]z=32\text{cis}\left(\frac{2\pi}{3}\right)[/latex]. 44. Evaluate the square root of z when [latex]z=32\text{cis}\left(\pi\right)[/latex]. 45. Evaluate the cube root of z when [latex]z=8\text{cis}\left(\frac{7\pi}{4}\right)[/latex]. For the following exercises, plot the complex number in the complex plane. 46. [latex]2+4i[/latex] 47. [latex]−3−3i[/latex] 48. [latex]5−4i[/latex] 49. [latex]−1−5i[/latex] 50. [latex]3+2i[/latex] 51. [latex]2i[/latex] 52. [latex]−4[/latex] 53. [latex]6−2i[/latex] 54. [latex]−2+i[/latex] 55. [latex]1−4i[/latex] 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 [latex]5+5i[/latex] to polar form. 57. Use the rectangular to polar feature on the graphing calculator to change [latex]3−2i[/latex] 58. Use the rectangular to polar feature on the graphing calculator to change [latex]−3−8i[/latex] 59. Use the polar to rectangular feature on the graphing calculator to change [latex]4\text{cis}\left(120^{\circ}\right)[/latex] to rectangular form. 60. Use the polar to rectangular feature on the graphing calculator to change [latex]2\text{cis}\left(45^{\circ}\right)[/latex] to rectangular form. 61. Use the polar to rectangular feature on the graphing calculator to change [latex]5\text{cis}\left(210^{\circ}\right)[/latex] to rectangular form.Licenses & Attributions
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