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Precalculus : Products and Quotients of Complex Numbers in Polar Form

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

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

Possible Answers:

${}i2^{100}$

${}2^{101}$

${}-2^{50}$

${}-i2^{100}$

${}-2^{100}$

Correct answer:

${}-2^{50}$

Explanation:

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

We also know that:

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

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

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

Since we know that,

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

We have then:

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

Since we know that:

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

We have then:

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

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Example Question #1 : Find The Product Of Complex Numbers

Compute the following sum:

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

Possible Answers:

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

{}-i

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

{}i

Correct answer:

Explanation:

Note that this is a geometric series.

Therefore we have:

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

Note that,

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

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

this shows that the sum is 0.

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

Find the following product.

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

Possible Answers:

${}2^{n-1}$

{}2

${}2i^{n}$

${}-2i^{n}$

${}2^{n}$

Correct answer:

${}2^{n}$

Explanation:

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

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

Therefore,

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We have

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

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

Thus this gives us,

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

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

Evaluate:

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

Possible Answers:

i+4

5i-7

9i-2

i^2

Correct answer:

9i-2

Explanation:

To evaluate this problem we need to FOIL the binomials.

(2+i)(1+4i)

=2(1)+2(4i)+1(i)+4i^2

=2+9i+4i^2

Now recall that i^2=-1

Thus,

=2+9i-4

=9i-2

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

Find the product , if

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

Possible Answers:

10+4i

3-7i

3-3i

3+4i

Correct answer:

10+4i

Explanation:

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

Using this method we get the following,

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

and because i^2=-1

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

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Example Question #6 : Find The Product Of Complex Numbers

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

Possible Answers:

$i+\sqrt{-1}$

Correct answer:

Explanation:

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

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

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

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

The correct answer is:

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

Find the product of the two complex numbers

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

Possible Answers:

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

$\small -49+79i$

$\small 10+17i$

$\small 21+70i$

Correct answer:

$\small -49+79i$

Explanation:

The product is

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

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Example Question #451 : Pre Calculus

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

Possible Answers:

$-i^{n}$

$2i^{n+1}$

$-i^{n+1}$

$i^{n}$

$4i^{n+1}$

Correct answer:

$i^{n}$

Explanation:

We remark first that:

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

and we know that :

(1+i)^2=1+2i-i^2=2i .

This means that:

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

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

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Example Question #1 : Find The Quotient Of Complex Numbers

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

Possible Answers:

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

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

-2i

i^2

Correct answer:

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

Explanation:

Since i^=-1 ,

the problem becomes,

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

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

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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 #1 : Find The Product Of Complex Numbers

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

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

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

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

Example Question #6 : Find The Product Of Complex Numbers

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

Example Question #451 : Pre Calculus

Example Question #1 : Find The Quotient Of Complex Numbers

All Precalculus Resources

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