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Precalculus : Find the Quotient of Complex Numbers

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Example Questions

Example Question #2 : Polar Coordinates And Complex Numbers

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:

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

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

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

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

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

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:

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

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

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Example Question #1 : Find The Quotient Of Complex Numbers

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

Possible Answers:

i^2 $\displaystyle i^2$

-2i $\displaystyle -2i$

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

$\displaystyle -i$

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

Correct answer:

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

Explanation:

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

the problem becomes,

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

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

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Example Question #1 : Find The Quotient Of Complex Numbers

Write

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

in the form $\small a+bi$ $\displaystyle \small a+bi$ for some real numbers $\small a$ $\displaystyle \small a$ and $\small b$ $\displaystyle \small b$.

Possible Answers:

$\small \small \frac{3}{2}-\frac{1}{2}i$ $\displaystyle \small \small \frac{3}{2}-\frac{1}{2}i$

$\small \small \frac{3}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ $\displaystyle \small \small \frac{3}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$

$\small \frac{3}{2}+\frac{1}{2}i$ $\displaystyle \small \frac{3}{2}+\frac{1}{2}i$

$\small 1+2i$ $\displaystyle \small 1+2i$

Correct answer:

$\small \frac{3}{2}+\frac{1}{2}i$ $\displaystyle \small \frac{3}{2}+\frac{1}{2}i$

Explanation:

The correct answer is

$\small \frac{3}{2}+\frac{1}{2}i$ $\displaystyle \small \frac{3}{2}+\frac{1}{2}i$

Using simple algebra and multiplying the expression by the complex conjugate of the denominator, we get:

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

$\small =\frac{1-i+2i-2i^2}{2}=\frac{1+i-(-2)}{2}=\frac{3+i}{2}=\frac{3}{2}+\frac{1}{2}i$ $\displaystyle \small =\frac{1-i+2i-2i^2}{2}=\frac{1+i-(-2)}{2}=\frac{3+i}{2}=\frac{3}{2}+\frac{1}{2}i$

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Example Question #1 : Find The Quotient Of Complex Numbers

Divide:

$\frac{4+5i}{5-i}$ $\displaystyle \frac{4+5i}{5-i}$

Possible Answers:

$\frac{5}{8}+\frac{29i}{24}$ $\displaystyle \frac{5}{8}+\frac{29i}{24}$

$\frac{15}{26}-\frac{29i}{26}$ $\displaystyle \frac{15}{26}-\frac{29i}{26}$

$\frac{15}{26}+\frac{29i}{26}$ $\displaystyle \frac{15}{26}+\frac{29i}{26}$

$\frac{29}{26}+\frac{15i}{26}$ $\displaystyle \frac{29}{26}+\frac{15i}{26}$

Correct answer:

$\frac{15}{26}+\frac{29i}{26}$ $\displaystyle \frac{15}{26}+\frac{29i}{26}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 5+i $\displaystyle 5+i$.

$\left(\frac{4+5i}{5-i}\right)\left(\frac{5+i}{5+i}\right)$ $\displaystyle \left(\frac{4+5i}{5-i}\right)\left(\frac{5+i}{5+i}\right)$

Now, distribute and simplify.

$\left(\frac{4+5i}{5-i}\right)\left(\frac{5+i}{5+i}\right)=\frac{20+29i+5i^2}{25-i^2}$ $\displaystyle \left(\frac{4+5i}{5-i}\right)\left(\frac{5+i}{5+i}\right)=\frac{20+29i+5i^2}{25-i^2}$

Recall that i^2=-1 $\displaystyle i^2=-1$

$\frac{20+29i+5(-1)}{25-(-1)}=\frac{15+29i}{26}=\frac{15}{26}+\frac{29i}{26}$ $\displaystyle \frac{20+29i+5(-1)}{25-(-1)}=\frac{15+29i}{26}=\frac{15}{26}+\frac{29i}{26}$

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Example Question #5 : Find The Quotient Of Complex Numbers

Divide:

$\frac{7+i}{2-i}$ $\displaystyle \frac{7+i}{2-i}$

Possible Answers:

$\frac{13}{5}+\frac{9i}{5}$ $\displaystyle \frac{13}{5}+\frac{9i}{5}$

$\frac{13}{5}-\frac{9i}{5}$ $\displaystyle \frac{13}{5}-\frac{9i}{5}$

$\frac{15}{2}+3i$ $\displaystyle \frac{15}{2}+3i$

$\frac{5}{13}+\frac{5}{9i}$ $\displaystyle \frac{5}{13}+\frac{5}{9i}$

Correct answer:

$\frac{13}{5}+\frac{9i}{5}$ $\displaystyle \frac{13}{5}+\frac{9i}{5}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 2+i $\displaystyle 2+i$.

$\left(\frac{7+i}{2-i}\right)\left(\frac{2+i}{2+i}\right)$ $\displaystyle \left(\frac{7+i}{2-i}\right)\left(\frac{2+i}{2+i}\right)$

Now, distribute and simplify.

$\left(\frac{7+i}{2-i}\right)\left(\frac{2+i}{2+i}\right)=\frac{14+9i+i^2}{4-i^2}$ $\displaystyle \left(\frac{7+i}{2-i}\right)\left(\frac{2+i}{2+i}\right)=\frac{14+9i+i^2}{4-i^2}$

Recall that i^2=-1 $\displaystyle i^2=-1$

$\frac{14+9i+i^2}{4-i^2}=\frac{14+9i+(-1)}{4-(-1)}=\frac{13+9i}{5}=\frac{13}{5}+\frac{9i}{5}$ $\displaystyle \frac{14+9i+i^2}{4-i^2}=\frac{14+9i+(-1)}{4-(-1)}=\frac{13+9i}{5}=\frac{13}{5}+\frac{9i}{5}$

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Example Question #1 : Find The Quotient Of Complex Numbers

Divide.

$\frac{2-9i}{3+i}$ $\displaystyle \frac{2-9i}{3+i}$

Possible Answers:

$\frac{3}{4}+\frac{29i}{4}$ $\displaystyle \frac{3}{4}+\frac{29i}{4}$

$-\frac{3}{10}-\frac{29i}{10}$ $\displaystyle -\frac{3}{10}-\frac{29i}{10}$

$-\frac{10}{3}-\frac{10}{29i}$ $\displaystyle -\frac{10}{3}-\frac{10}{29i}$

$\frac{3}{10}+\frac{29i}{10}$ $\displaystyle \frac{3}{10}+\frac{29i}{10}$

Correct answer:

$-\frac{3}{10}-\frac{29i}{10}$ $\displaystyle -\frac{3}{10}-\frac{29i}{10}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 3-i $\displaystyle 3-i$.

$\left(\frac{2-9i}{3+i}\right)\left(\frac{3-i}{3-i}\right)$ $\displaystyle \left(\frac{2-9i}{3+i}\right)\left(\frac{3-i}{3-i}\right)$

Now, distribute and simplify.

$\left(\frac{2-9i}{3+i}\right)\left(\frac{3-i}{3-i}\right)=\frac{6-29i+9i^2}{9-i^2}$ $\displaystyle \left(\frac{2-9i}{3+i}\right)\left(\frac{3-i}{3-i}\right)=\frac{6-29i+9i^2}{9-i^2}$

Recall that i^2=-1 $\displaystyle i^2=-1$

$\frac{6-29i+9i^2}{9-i^2}=\frac{6-29i+9(-1)}{9-(-1)}=\frac{-3-29i}{10}=-\frac{3}{10}-\frac{29i}{10}$ $\displaystyle \frac{6-29i+9i^2}{9-i^2}=\frac{6-29i+9(-1)}{9-(-1)}=\frac{-3-29i}{10}=-\frac{3}{10}-\frac{29i}{10}$

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Example Question #7 : Find The Quotient Of Complex Numbers

Divide.

$\frac{4+3i}{6-i}$ $\displaystyle \frac{4+3i}{6-i}$

Possible Answers:

$\frac{9}{7}+\frac{27}{22i}$ $\displaystyle \frac{9}{7}+\frac{27}{22i}$

$\frac{21}{37}+\frac{22i}{37}$ $\displaystyle \frac{21}{37}+\frac{22i}{37}$

$\frac{7}{9}-\frac{22i}{27}$ $\displaystyle \frac{7}{9}-\frac{22i}{27}$

$\frac{2}{3}+\frac{i}{9}$ $\displaystyle \frac{2}{3}+\frac{i}{9}$

Correct answer:

$\frac{21}{37}+\frac{22i}{37}$ $\displaystyle \frac{21}{37}+\frac{22i}{37}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 6+i $\displaystyle 6+i$.

$\left(\frac{4+3i}{6-i}\right)\left(\frac{6+i}{6+i}\right)$ $\displaystyle \left(\frac{4+3i}{6-i}\right)\left(\frac{6+i}{6+i}\right)$

Now, distribute and simplify.

$\left(\frac{4+3i}{6-i}\right)\left(\frac{6+i}{6+i}\right)=\frac{24+22i+3i^2}{36-i^2}$ $\displaystyle \left(\frac{4+3i}{6-i}\right)\left(\frac{6+i}{6+i}\right)=\frac{24+22i+3i^2}{36-i^2}$

Recall that i^2=-1 $\displaystyle i^2=-1$

$\frac{24+22i+3i^2}{36-i^2}=\frac{24+22i+3(-1)}{36-(-1)}=\frac{21+22i}{37}$ $\displaystyle \frac{24+22i+3i^2}{36-i^2}=\frac{24+22i+3(-1)}{36-(-1)}=\frac{21+22i}{37}$

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Example Question #1 : Find The Quotient Of Complex Numbers

Divide.

$\frac{5-2i}{3-8i}$ $\displaystyle \frac{5-2i}{3-8i}$

Possible Answers:

$\frac{31}{73}+\frac{34i}{73}$ $\displaystyle \frac{31}{73}+\frac{34i}{73}$

$\frac{31}{73}-\frac{34i}{73}$ $\displaystyle \frac{31}{73}-\frac{34i}{73}$

$\frac{21}{29}+\frac{6i}{29}$ $\displaystyle \frac{21}{29}+\frac{6i}{29}$

$\frac{73}{31}-\frac{73}{64i}$ $\displaystyle \frac{73}{31}-\frac{73}{64i}$

Correct answer:

$\frac{31}{73}+\frac{34i}{73}$ $\displaystyle \frac{31}{73}+\frac{34i}{73}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 3+8i $\displaystyle 3+8i$.

$\left(\frac{5-2i}{3-8i}\right)\left(\frac{3+8i}{3+8i}\right)$ $\displaystyle \left(\frac{5-2i}{3-8i}\right)\left(\frac{3+8i}{3+8i}\right)$

Now, distribute and simplify.

$\left(\frac{5-2i}{3-8i}\right)\left(\frac{3+8i}{3+8i}\right)=\frac{15+34i-16i^2}{9-64i^2}$ $\displaystyle \left(\frac{5-2i}{3-8i}\right)\left(\frac{3+8i}{3+8i}\right)=\frac{15+34i-16i^2}{9-64i^2}$

Recall that i^2=-1 $\displaystyle i^2=-1$

$\frac{15+34i-16i^2}{9-64i^2}=\frac{15+34i-16(-1)}{9-64(-1)}=\frac{31+34i}{73}=\frac{31}{73}+\frac{34i}{73}$ $\displaystyle \frac{15+34i-16i^2}{9-64i^2}=\frac{15+34i-16(-1)}{9-64(-1)}=\frac{31+34i}{73}=\frac{31}{73}+\frac{34i}{73}$

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Example Question #1 : Find The Quotient Of Complex Numbers

Divide.

$\frac{5+i}{3+i}$ $\displaystyle \frac{5+i}{3+i}$

Possible Answers:

$\frac{8}{5}+\frac{i}{5}$ $\displaystyle \frac{8}{5}+\frac{i}{5}$

$\frac{3}{8}-\frac{i}{4}$ $\displaystyle \frac{3}{8}-\frac{i}{4}$

$\frac{5}{8}+\frac{i}{5}$ $\displaystyle \frac{5}{8}+\frac{i}{5}$

$\frac{8}{5}-\frac{i}{5}$ $\displaystyle \frac{8}{5}-\frac{i}{5}$

Correct answer:

$\frac{8}{5}-\frac{i}{5}$ $\displaystyle \frac{8}{5}-\frac{i}{5}$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 3-i $\displaystyle 3-i$.

$\left(\frac{5+i}{3+i}\right)\left(\frac{3-i}{3-i}\right)$ $\displaystyle \left(\frac{5+i}{3+i}\right)\left(\frac{3-i}{3-i}\right)$

Now, distribute and simplify.

$\left(\frac{5+i}{3+i}\right)\left(\frac{3-i}{3-i}\right)=\frac{15-2i-i^2}{9-i^2}$ $\displaystyle \left(\frac{5+i}{3+i}\right)\left(\frac{3-i}{3-i}\right)=\frac{15-2i-i^2}{9-i^2}$

Recall that i^2=-1 $\displaystyle i^2=-1$

$\frac{15-2i-i^2}{9-i^2}=\frac{15-2i-(-1)}{9-(-1)}=\frac{16-2i}{10}=\frac{8}{5}-\frac{i}{5}$ $\displaystyle \frac{15-2i-i^2}{9-i^2}=\frac{15-2i-(-1)}{9-(-1)}=\frac{16-2i}{10}=\frac{8}{5}-\frac{i}{5}$

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Example Question #10 : Find The Quotient Of Complex Numbers

Divide.

$\frac{-9+i}{1-i}$ $\displaystyle \frac{-9+i}{1-i}$

Possible Answers:

-5-4i $\displaystyle -5-4i$

-2+3i $\displaystyle -2+3i$

-4-5i $\displaystyle -4-5i$

5+4i $\displaystyle 5+4i$

Correct answer:

-5-4i $\displaystyle -5-4i$

Explanation:

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator.

To find the conjugate, just change the sign in the denominator. The conjugate used will be 1-i $\displaystyle 1-i$.

$\left(\frac{-9+i}{1-i}\right)\left(\frac{1+i}{1+i}\right)$ $\displaystyle \left(\frac{-9+i}{1-i}\right)\left(\frac{1+i}{1+i}\right)$

Now, distribute and simplify.

$\left(\frac{-9+i}{1-i}\right)\left(\frac{1+i}{1+i}\right)=\frac{-9-8i+i^2}{1-i^2}$ $\displaystyle \left(\frac{-9+i}{1-i}\right)\left(\frac{1+i}{1+i}\right)=\frac{-9-8i+i^2}{1-i^2}$

Recall that i^2=-1 $\displaystyle i^2=-1$

$\frac{-9-8i+i^2}{1-i^2}=\frac{-9-8i+(-1)}{1-(-1)}=\frac{-10-8i}{2}=-5-4i$ $\displaystyle \frac{-9-8i+i^2}{1-i^2}=\frac{-9-8i+(-1)}{1-(-1)}=\frac{-10-8i}{2}=-5-4i$

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Precalculus : Find the Quotient of Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #2 : Polar Coordinates And Complex Numbers

Example Question #1 : Find The Quotient Of Complex Numbers

Example Question #1 : Find The Quotient Of Complex Numbers

Example Question #1 : Find The Quotient Of Complex Numbers

Example Question #5 : Find The Quotient Of Complex Numbers

Example Question #1 : Find The Quotient Of Complex Numbers

Example Question #7 : Find The Quotient Of Complex Numbers

Example Question #1 : Find The Quotient Of Complex Numbers

Example Question #1 : Find The Quotient Of Complex Numbers

Example Question #10 : Find The Quotient Of Complex Numbers

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