Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

Create An Account

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Evaluate $(-3 + 3i ) ^{ 12}$

Possible Answers:

34,012,224

-531,441

-34,012,224

$17,006,112 \sqrt2 + 17,006,112 \sqrt 2 i$

$-17,006, 112 \sqrt2 + 17,006,112 \sqrt 2 i$

Correct answer:

-34,012,224

Explanation:

First convert the complex number into polar form:

$r = \sqrt{3^2 + (-3)^2 } = \sqrt{9+9} = \sqrt{2 \cdot 9 } = 3 \sqrt2$

$\sin \theta = \frac{3}{3 \sqrt2} = \frac{ 1}{\sqrt2}$

$\theta = \sin ^{-1 } (\frac{1}{\sqrt2}) = \frac{\pi}{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since the real part is negative but the imaginary part is positive, the angle should be in quadrant II, so it is $\frac{ 3 \pi }{4}$

We are evaluating $(3\sqrt2 (\cos \frac{ 3 \pi }{4} + i \sin \frac{ 3 \pi }{4} ) ) ^ {12}$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(3\sqrt2)^{12 } (\cos \frac{ 3 \pi }{4} \cdot 12 + i \sin \frac{3 \pi }{4} \cdot 12 )$ simplify and take the exponent

$34,012,224 (\cos 9\pi + i \sin 9 \pi )$

$9\pi$ is coterminal with $\pi$ since it is an odd multiple of pi

$34, 012, 224 ( \cos \pi + i \sin \pi ) = 34, 012, 224 ( -1 + 0i) = -34,012, 224$

Report an Error

Example Question #6 : Powers And Roots Of Complex Numbers

Use DeMoivre's Theorem to evaluate the expression $(\sqrt3 - 3i ) ^ 6$ .

Possible Answers:

Correct answer:

Explanation:

First convert this complex number to polar form:

$r = \sqrt{ (\sqrt3)^2 + 1^2 } = \sqrt{ 3 + 1 } = \sqrt4 = 2$

$\sin \theta = \frac{ 1}{ 2}$ so $\theta = \frac{\pi }{6} \enspace or \enspace \frac{5 \pi }{6}$

Since this number has positive real and imaginary parts, it is in quadrant I, so the angle is $\frac{ \pi }{6}$

So we are evaluating $(2(\cos \frac{ \pi }{6} + i \sin \frac{ \pi }{6} ))^3$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$2^3 (\cos \frac{3\pi}{6} + i \sin \frac{ 3\pi }{6} ) = 8 ( \cos \frac{\pi}{2} + i \sin \frac{ \pi }{2} ) = 8(0 + i ) = 8i$

Report an Error

Example Question #2 : Powers And Roots Of Complex Numbers

Evaluate: (1-i)^8

Possible Answers:

256

128

16i

Correct answer:

Explanation:

First, convert this complex number to polar form.

$r= \sqrt{ a^2 + b^2 } = \sqrt{1^2 +(-1)^2 }=\sqrt{2}$

$\sin \theta = \frac{-1}{\sqrt2}$

$\theta = \frac{5 \pi }{4} \enspace or \enspace \frac{7 \pi }{4}$

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is $\frac{7\pi}{4}$ .

This gives us $(\sqrt{2} ( \cos \frac{7 \pi }{4} + i \sin \frac{ 7 \pi }{4} )) ^ 8$

To evaluate, use DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(\sqrt2 )^8 ( \cos \frac{ 7 \pi }{4} \cdot 8 + i \sin \frac{ 7 \pi }{4} \cdot 8 )$ simplifying

$2 ^ 4 ( \cos 14 \pi + i \sin 14 \pi )$ , $14 \pi$ is coterminal with since it is an even multiple of $\pi$

$2^ 4 (\cos 0 + i \sin 0 ) = 16(1 + 0i) = 16$

Report an Error

Example Question #1 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

(2 + 2i )^9

Possible Answers:

11,585.238 i

512

11,585.238

8,192 + 8,192 i

Correct answer:

8,192 + 8,192 i

Explanation:

First, convert the complex number to polar form:

$r = \sqrt{ 2^2 + 2^2 } = \sqrt{4 + 4 } = \sqrt{2 \cdot 4 } = 2 \sqrt 2$

$\sin \theta = \frac{ 2 }{ 2 \sqrt 2 } = \frac{ 1 }{ \sqrt2 }$

$\theta = \sin ^{-1 } (\frac{ 1}{ \sqrt 2 }) = \frac{ \pi }{4} \enspace or \enspace \frac{ 3 \pi }{4}$

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is $\frac{ \pi }{4}$

This means we're evaluating

$(2 \sqrt2 (\cos \frac{ \pi }{4} + i \sin \frac{ \pi }{4} ))^ 9$

Using DeMoivre's Theorem:

DeMoivre's Theorem is

$(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$

We apply it to our situation to get.

$(2 \sqrt2 )^9 (\cos \frac{\pi }{4} \cdot 9 + i \sin \frac{ \pi }{4} \cdot 9 )$

First, evaluate $(2\sqrt2)^9$ . We can split this into $2^9 \cdot (\sqrt2)^9$ which is equivalent to $2^9 \cdot (\sqrt2)^8 \cdot \sqrt2 = 2^9 \cdot 2^4 \cdot \sqrt 2$

[We can re-write the middle exponent since $\sqrt 2$ is equivalent to $2^{\frac{1}{2}}$ ]

This comes to $8,192 \sqrt 2$

Evaluating sine and cosine at $\frac{\pi }{4} \cdot 9 = \frac{ 9 \pi }{4}$ is equivalent to evaluating them at $\frac{ \pi }{4}$ since $\frac{ 9 \pi }{4} - 2 \pi = \frac{9 \pi }{4 } - \frac{ 8 \pi }{4} = \frac{ \pi }{4}$

This means our expression can be written as:

$8,192 \sqrt2 ( \cos \frac{\pi }{4} + i \sin \frac{ \pi }{4} ) = 8,192 \sqrt2 (\frac{1}{\sqrt2} + i \frac{ 1}{\sqrt2}) = 8,192 + 8,192i$

Report an Error

Example Question #1 : Find The Roots Of Complex Numbers

Evaluate ${}(1+i)^{2n}$ , where is a natural number and is the complex number ${}i^{2}=-1$ .

Possible Answers:

${}2^{n+1}i^{n}$

${}-2^{n}i^{n}$

${}-2^{n+1}i^{n}$

${}2^{n}i^{n}$

${}2^{n}i^{n+1}$

Correct answer:

${}2^{n}i^{n}$

Explanation:

Note that,

${}(1+i)^{2n}=((1+i)^2)^{n}}= (2i)^{n}= 2^{n}i^{n}$

Report an Error

Example Question #1 : Find The Roots Of Complex Numbers

What is the length of

$({1-i})^{3}$ ?

Possible Answers:

$2-\sqrt{2}$

$2+\sqrt{2}$

$2\sqrt{2}$

$\sqrt{2}$

Correct answer:

$2\sqrt{2}$

Explanation:

We have

$|1-i|=\sqrt{2}$ .

Hence,

$|(1-i)^3|=|1-i|^{3}=(\sqrt2)^3=2\sqrt{2}$ .

Report an Error

Example Question #2 : Find The Roots Of Complex Numbers

Solve for $\small z^3=1$ (there may be more than one solution).

Possible Answers:

$\small z=1$

$\small z=1, \frac{-1+\sqrt{3}}{2}, \frac{-1-\sqrt{3}}{2}$

$\small z=-1$

$\small z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

Correct answer:

$\small z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

Explanation:

Solving that equation is equivalent to solving the roots of the polynomial $\small f(z)=z^3-1$ .

Clearly, one of roots is 1.

Thus, we can factor the polynomial as $\small f(z)=(z-1)(z^2+z+1)$

so that we solve for the roots of $\small z^2+z+1$ .

Using the quadratic equation, we solve for roots, which are $\small z=\frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$ .

This means the solutions to $\small z^3=1$ are

$z=1, \frac{-1+i\sqrt{3}}{2}, \frac{-1-i\sqrt{3}}{2}$

Report an Error

Example Question #4 : Find The Roots Of Complex Numbers

Recall that $z=rcis(\theta)$ is just shorthand for $z=r(cos(\theta)+isin(\theta))$ when dealing with complex numbers in polar form.

Express $\frac{1}{(\sqrt{3}-i)^4}$ in polar form.

Possible Answers:

$2cis\left(\frac{\pi}{3}\right)$

$2cis\left(\frac{2\pi}{3}\right)$

$\frac{1}{16}cis\left(\frac{-2\pi}{3}\right)$

$\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$

$2cis\left(\frac{2\pi}{12}\right)$

Correct answer:

$\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$

Explanation:

First we recognize that we are trying to solve $z^{-4}$ where $z=\sqrt{3}-i$ .

Then we want to convert into polar form using,

$arg(z)=\theta =arctan(\frac{b}{a})$ and $\left | z\right |=r=\sqrt{a^2+b^2}$ .

Then since De Moivre's theorem states,

$z^n=r^ncis(n\theta)$ if is an integer, we can say

$z^{-4}=2^{-4}cis(-4*\frac{-\pi}{6})=\frac{1}{16}cis\left(\frac{2\pi}{3}\right)$ .

Report an Error

Example Question #1 : Find The Roots Of Complex Numbers

Solve for z^3-z^2+z=0 (there may be more than one solution).

Possible Answers:

$0, 1, \frac{1}{2}$

$0, \frac{1}{2}\pm\frac{\sqrt{3}}{2}$

$0, \frac{1}{2}\pm\frac{\sqrt{3i}}{2}$

$0, \frac{1}{2}\pm\sqrt{3i}$

Correct answer:

$0, \frac{1}{2}\pm\frac{\sqrt{3i}}{2}$

Explanation:

To solve for the roots, just set equal to zero and solve for z using the quadratic formula ( $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ) : $z^3-z^2+z\rightarrow z(z^2-z+1)$ and now setting both and z^2-z+1 equal to zero we end up with the answers z=0 and $z=\frac{1\pm 3i}{2}$

Report an Error

Example Question #5 : Find The Roots Of Complex Numbers

Compute (3+3i)^5

Possible Answers:

972i

972

-972-972i

972-972i

972+972i

Correct answer:

-972-972i

Explanation:

To solve this question, you must first derive a few values and convert the equation into exponential form: $re^{\pi*z}$ : $r=\sqrt{9+9}=3\sqrt{2}\rightarrow tan(\theta)=\frac{3}{3}=1\rightarrow Arg(z)=\frac{\pi}{4}$

Now plug back into the original equation and solve: $(3+3i)^5=(3\sqrt{2})^5\cdot e^{\frac{i5\pi}{4}=972\sqrt{2}(cos(\frac{5\pi}{4})+isin(\frac{5\pi}{4}))=972\sqrt{2}(\frac {-\sqrt{2}{2}}-\frac ({\sqrt{2}\cdot i}{2})=-972-972i$

Report an Error

View Pre-Calculus Tutors

Tracy
Certified Tutor

Clarion University of Pennsylvania, Bachelor of Science, Physics.

View Pre-Calculus Tutors

Yaakoub
Certified Tutor

CUNY Queens College, Bachelor of Science, Finance. Vaughn College of Aeronautics and Technology, Bachelor of Science, Electri...

View Pre-Calculus Tutors

Emmanuel
Certified Tutor

Baylor University, Bachelor of Science, Biology, General.

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Biology Tutors in Seattle, Physics Tutors in Los Angeles, Physics Tutors in Chicago, GRE Tutors in San Diego, French Tutors in Seattle, MCAT Tutors in Atlanta, ISEE Tutors in Dallas Fort Worth, GMAT Tutors in Denver, MCAT Tutors in Philadelphia, GRE Tutors in Houston

Popular Courses & Classes

MCAT Courses & Classes in Boston, MCAT Courses & Classes in Philadelphia, SAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Washington DC, Spanish Courses & Classes in San Diego, GMAT Courses & Classes in Chicago, GRE Courses & Classes in Houston, MCAT Courses & Classes in Chicago, GRE Courses & Classes in Atlanta

Popular Test Prep

GRE Test Prep in Chicago, ISEE Test Prep in Atlanta, GMAT Test Prep in Seattle, ISEE Test Prep in Boston, SSAT Test Prep in Denver, GRE Test Prep in Houston, SAT Test Prep in Denver, GMAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Seattle, GMAT Test Prep in Chicago

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #6 : Powers And Roots Of Complex Numbers

Example Question #2 : Powers And Roots Of Complex Numbers

Example Question #1 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #2 : Find The Roots Of Complex Numbers

Example Question #4 : Find The Roots Of Complex Numbers

Express $\frac{1}{(\sqrt{3}-i)^4}$ in polar form.

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #5 : Find The Roots Of Complex Numbers

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback:

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Precalculus : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #3 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #6 : Powers And Roots Of Complex Numbers

Example Question #2 : Powers And Roots Of Complex Numbers

Example Question #1 : Evaluate Powers Of Complex Numbers Using De Moivre's Theorem

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #2 : Find The Roots Of Complex Numbers

Example Question #4 : Find The Roots Of Complex Numbers

Express in polar form.

Example Question #1 : Find The Roots Of Complex Numbers

Example Question #5 : Find The Roots Of Complex Numbers

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Express $\frac{1}{(\sqrt{3}-i)^4}$ in polar form.