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Precalculus : Find the Product of Complex Numbers

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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:

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

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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.

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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 $\displaystyle {}2$

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

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

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

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

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,

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

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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,

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

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

Evaluate:

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

Possible Answers:

$\displaystyle 9$

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

i+4 $\displaystyle i+4$

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

i^2 $\displaystyle i^2$

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 $\displaystyle =9i-2$

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Example Question #1 : 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:

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

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

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

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

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 $\displaystyle i^2=-1$

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

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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: $\displaystyle -i$

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

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:

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

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

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

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

Correct answer:

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

Explanation:

The product is

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

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Precalculus : Find the Product of Complex Numbers

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

Example Question #1 : Polar Coordinates And Complex Numbers

Example Question #1 : Find The Product Of Complex Numbers

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

All Precalculus Resources

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