We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Products and Quotients of Complex Numbers in Polar Form

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 #1 : Products And Quotients Of Complex Numbers In Polar Form

Find the value of ${}(1+i)^{100}$ $\displaystyle {}(1+i)^{100}$ ,where $\displaystyle i$ the complex number is given by ${}i^2=-1$ $\displaystyle {}i^2=-1$ .

Possible Answers:

${}-i2^{100}$ $\displaystyle {}-i2^{100}$

${}-2^{100}$ $\displaystyle {}-2^{100}$

${}i2^{100}$ $\displaystyle {}i2^{100}$

${}-2^{50}$ $\displaystyle {}-2^{50}$

${}2^{101}$ $\displaystyle {}2^{101}$

Correct answer:

${}-2^{50}$ $\displaystyle {}-2^{50}$

Explanation:

We note that ${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ $\displaystyle {}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ by FOILing.

We also know that:

$(z)^{nm}=(z^{n})^{m}$ $\displaystyle (z)^{nm}=(z^{n})^{m}$

We have by using the above rule: n=2 , m=50

$(1+i)^{100}=((1+i)^{2})^{50}$ $\displaystyle (1+i)^{100}=((1+i)^{2})^{50}$

Since we know that,

${}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$ $\displaystyle {}(1+i)^{2}=1+2i+i^2=1+2i-1=2i$

We have then:

$(1+i)^{100}=((1+i)^{2})^{50}$ $\displaystyle (1+i)^{100}=((1+i)^{2})^{50}$ $=(2i)^{50}$ $\displaystyle =(2i)^{50}$

Since we know that:

$(ab)^{n}=a^{n}b^{n}$ $\displaystyle (ab)^{n}=a^{n}b^{n}$ , we use a=2 ,b=i

We have then:

$2^{50}({i}^2)^{25}=2^{50}(-1)^{25} =-2^{50}}$

Report an Error

Example Question #441 : Pre Calculus

Compute the following sum:

${}1+i+i^2+i^3....+i^{2015}$ $\displaystyle {}1+i+i^2+i^3....+i^{2015}$ . Remember {}i $\displaystyle {}i$ is the complex number satisfying ${}i^{2}=-1$ $\displaystyle {}i^{2}=-1$ .

Possible Answers:

$\displaystyle 0$

{}-i $\displaystyle {}-i$

${}\frac{1}{1-i}$ $\displaystyle {}\frac{1}{1-i}$

{}i $\displaystyle {}i$

${}\frac{i}{i-1}$ $\displaystyle {}\frac{i}{i-1}$

Correct answer:

$\displaystyle 0$

Explanation:

Note that this is a geometric series.

Therefore we have:

${}1+i+i^2....+i^{2015}=\frac{1-i^{2016}}{1-i}$ $\displaystyle {}1+i+i^2....+i^{2015}=\frac{1-i^{2016}}{1-i}$

Note that,

${}i^{2016}$ $\displaystyle {}i^{2016}$ = ${}({i}^{2})^{1008}$ $\displaystyle {}({i}^{2})^{1008}$ and since ${}i^2=-1$ $\displaystyle {}i^2=-1$ we have ${}{-1}^{1008}=1$ $\displaystyle {}{-1}^{1008}=1$ .

$\frac{1-1}{1-i}=\frac{0}{1-i}=0$ $\displaystyle \frac{1-1}{1-i}=\frac{0}{1-i}=0$

this shows that the sum is 0.

Report an Error

Example Question #1 : Polar Coordinates And Complex Numbers

Find the following product.

${}(1-i)^{n}(1+i)^n$ $\displaystyle {}(1-i)^{n}(1+i)^n$

Possible Answers:

${}2^{n-1}$ $\displaystyle {}2^{n-1}$

${}2i^{n}$ $\displaystyle {}2i^{n}$

{}2 $\displaystyle {}2$

${}2^{n}$ $\displaystyle {}2^{n}$

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

Correct answer:

${}2^{n}$ $\displaystyle {}2^{n}$

Explanation:

Note that by FOILing the two binomials we get the following:

${}(1+i)(1-i)=1-i+i-i^2=2$ $\displaystyle {}(1+i)(1-i)=1-i+i-i^2=2$

Therefore,

${}(1-i)^{n}(1-i)^{n}=2^{n}$ $\displaystyle {}(1-i)^{n}(1-i)^{n}=2^{n}$

Report an Error

Example Question #2 : Products And Quotients Of Complex Numbers In Polar Form

Compute the magnitude of ${}(1+i)^{n}$ $\displaystyle {}(1+i)^{n}$ .

Possible Answers:

${}(1-\sqrt{2})^n$ $\displaystyle {}(1-\sqrt{2})^n$

${}(\sqrt{2})^n+1$ $\displaystyle {}(\sqrt{2})^n+1$

${}(\sqrt{2})^{n+1}$ $\displaystyle {}(\sqrt{2})^{n+1}$

${}(\sqrt{2})^n$ $\displaystyle {}(\sqrt{2})^n$

${}(\sqrt{2})^n-1$ $\displaystyle {}(\sqrt{2})^n-1$

Correct answer:

${}(\sqrt{2})^n$ $\displaystyle {}(\sqrt{2})^n$

Explanation:

We have

${}|(1+i)|=\sqrt{2}$ $\displaystyle {}|(1+i)|=\sqrt{2}$ .

We know that ${}|z^n|=|z|^n$ $\displaystyle {}|z^n|=|z|^n$

Thus this gives us,

${}|(1+i)^n|=|1+i|^{n}=(\sqrt{2})^{n}$ $\displaystyle {}|(1+i)^n|=|1+i|^{n}=(\sqrt{2})^{n}$ .

Report an Error

Example Question #2 : Polar Coordinates And Complex Numbers

Evaluate:

$(2+i)\cdot(1+4i)$ $\displaystyle (2+i)\cdot(1+4i)$

Possible Answers:

i^2 $\displaystyle i^2$

9i-2 $\displaystyle 9i-2$

$\displaystyle 9$

5i-7 $\displaystyle 5i-7$

i+4 $\displaystyle i+4$

Correct answer:

9i-2 $\displaystyle 9i-2$

Explanation:

To evaluate this problem we need to FOIL the binomials.

$\displaystyle (2+i)(1+4i)$

$\displaystyle =2(1)+2(4i)+1(i)+4i^2$

$\displaystyle =2+9i+4i^2$

Now recall that i^2=-1 $\displaystyle i^2=-1$

Thus,

=2+9i-4 $\displaystyle =2+9i-4$

=9i-2 $\displaystyle =9i-2$

Report an Error

Example Question #3 : Polar Coordinates And Complex Numbers

Find the product $\displaystyle ab$ , if

$a=3+7i\:\:\:b=1-i$ $\displaystyle a=3+7i\:\:\:b=1-i$ .

Possible Answers:

3+4i $\displaystyle 3+4i$

3-3i $\displaystyle 3-3i$

3-7i $\displaystyle 3-7i$

10+4i $\displaystyle 10+4i$

Correct answer:

10+4i $\displaystyle 10+4i$

Explanation:

To find the product $\displaystyle ab$ , FOIL the complex numbers. FOIL stands for the multiplication of the Firsts, Outers, Inners, and Lasts.

Using this method we get the following,

$\displaystyle ab=(3+7i)(1-i)=3-3i+7i-7i^2=3+4i-7i^2$

and because i^2=-1 $\displaystyle i^2=-1$

$\displaystyle =3+4i-7(-1)=3+4i+7=10+4i$ .

Report an Error

Example Question #1 : Find The Product Of Complex Numbers

Simplify: $i^4 \times i^3$ $\displaystyle i^4 \times i^3$

Possible Answers:

$\displaystyle 1$

$\displaystyle i$

$i+\sqrt{-1}$ $\displaystyle i+\sqrt{-1}$

$\displaystyle -i$

$\displaystyle -1$

Correct answer:

$\displaystyle -i$

Explanation:

The expression $i^4 \times i^3$ $\displaystyle i^4 \times i^3$ can be rewritten as:

$i^4 \times i^3 = i^2\times i^2\times i^2\times i$ $\displaystyle i^4 \times i^3 = i^2\times i^2\times i^2\times i$

Since $i=\sqrt{-1}$ $\displaystyle i=\sqrt{-1}$ , the value of $\displaystyle i^2 = -1$ .

$i^2\times i^2\times i^2\times i = (-1) (-1) (-1)(i) = -i$ $\displaystyle i^2\times i^2\times i^2\times i = (-1) (-1) (-1)(i) = -i$

The correct answer is: $\displaystyle -i$

Report an Error

Example Question #4 : Polar Coordinates And Complex Numbers

Find the product of the two complex numbers

$\small 3+7i$ $\displaystyle \small 3+7i$ and $\small 7+10i$ $\displaystyle \small 7+10i$

Possible Answers:

$\small -49+79i$ $\displaystyle \small -49+79i$

$\small 21+79i+70i^2$ $\displaystyle \small 21+79i+70i^2$

$\small 21+70i$ $\displaystyle \small 21+70i$

$\small 10+17i$ $\displaystyle \small 10+17i$

Correct answer:

$\small -49+79i$ $\displaystyle \small -49+79i$

Explanation:

The product is

$\small (3+7i)(7+10i)=21+30i+49i+70i^2=21-70+79i=-49+79i$ $\displaystyle \small (3+7i)(7+10i)=21+30i+49i+70i^2=21-70+79i=-49+79i$

Report an Error

Example Question #3 : Products And Quotients Of Complex Numbers In Polar Form

Let $a=(1+i)^{n}$ $\displaystyle a=(1+i)^{n}$ , $\displaystyle b=(1-i)^n$ . Find a simple form of $\frac{a}{b}$ $\displaystyle \frac{a}{b}$ .

Possible Answers:

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

$i^{n}$ $\displaystyle i^{n}$

$4i^{n+1}$ $\displaystyle 4i^{n+1}$

$-i^{n}$ $\displaystyle -i^{n}$

$-i^{n+1}$ $\displaystyle -i^{n+1}$

Correct answer:

$i^{n}$ $\displaystyle i^{n}$

Explanation:

We remark first that:

$\frac{1+i}{1-i}=\frac{1+i}{1-i} \frac{1+i}{1+i}$ $\displaystyle \frac{1+i}{1-i}=\frac{1+i}{1-i} \frac{1+i}{1+i}$ $=\frac{(1+i)^2}{2}$ $\displaystyle =\frac{(1+i)^2}{2}$

and we know that :

$\displaystyle (1+i)^2=1+2i-i^2=2i$ .

This means that:

$\frac{1+i}{1-i}=\frac{2i}{2}=i$ $\displaystyle \frac{1+i}{1-i}=\frac{2i}{2}=i$

$\left(\frac{1+i}{1-i}\right)^n=i^{n}$ $\displaystyle \left(\frac{1+i}{1-i}\right)^n=i^{n}$

Report an Error

Example Question #1 : Find The Quotient Of Complex Numbers

What is $\frac{(i^2)}{1-i}$ $\displaystyle \frac{(i^2)}{1-i}$ ?

Possible Answers:

$\displaystyle -i$

-2i $\displaystyle -2i$

$1+\frac{i}{i^2}$ $\displaystyle 1+\frac{i}{i^2}$

$-1+\frac{1}{1-i}$ $\displaystyle -1+\frac{1}{1-i}$

i^2 $\displaystyle i^2$

Correct answer:

$-1+\frac{1}{1-i}$ $\displaystyle -1+\frac{1}{1-i}$

Explanation:

Since i^=-1 $\displaystyle i^=-1$ ,

the problem becomes,

$\frac{-1}{1-i}$ $\displaystyle \frac{-1}{1-i}$

$=-1+\frac{1}{1-i}$ $\displaystyle =-1+\frac{1}{1-i}$

Report an Error

View Pre-Calculus Tutors

David
Certified Tutor

The University of Texas at El Paso, Bachelor of Science, Computational Mathematics. Indiana State University, Master of Scien...

View Pre-Calculus Tutors

Lucille
Certified Tutor

View Pre-Calculus Tutors

Devin
Certified Tutor

Columbia University in the City of New York, Bachelor of Science, Chemical Engineering.

All Precalculus Resources

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

Popular Subjects

Reading Tutors in Atlanta, ISEE Tutors in San Diego, LSAT Tutors in New York City, Reading Tutors in San Diego, LSAT Tutors in Atlanta, SSAT Tutors in Seattle, GRE Tutors in Washington DC, French Tutors in Washington DC, GRE Tutors in Boston, SSAT Tutors in Phoenix

Popular Courses & Classes

Spanish Courses & Classes in Miami, ACT Courses & Classes in San Diego, Spanish Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Boston, ISEE Courses & Classes in Atlanta, ACT Courses & Classes in Denver, MCAT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in San Diego, MCAT Courses & Classes in Atlanta, Spanish Courses & Classes in Los Angeles

Popular Test Prep

GMAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Seattle, SSAT Test Prep in Philadelphia, ACT Test Prep in Miami, ACT Test Prep in Denver, MCAT Test Prep in Dallas Fort Worth, GRE Test Prep in Miami, ACT Test Prep in Boston, SSAT Test Prep in Miami, SAT Test Prep in Miami

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 : Products and Quotients of Complex Numbers in Polar Form

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Products And Quotients Of Complex Numbers In Polar Form

Example Question #441 : Pre Calculus

Example Question #1 : Polar Coordinates And Complex Numbers

Example Question #2 : Products And Quotients Of Complex Numbers In Polar Form

Example Question #2 : Polar Coordinates And Complex Numbers

Example Question #3 : Polar Coordinates And Complex Numbers

Example Question #1 : Find The Product Of Complex Numbers

Example Question #4 : Polar Coordinates And Complex Numbers

Example Question #3 : Products And Quotients Of Complex Numbers In Polar Form

Example Question #1 : Find The Quotient Of Complex Numbers

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: