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Precalculus : Products and Quotients of Complex Numbers in Polar Form

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Example Questions

Example Question #463 : Pre Calculus

Divide.

$\frac{4-i^3}{5+i}$

Possible Answers:

$\frac{7}{13}-\frac{21i}{6}$

$\frac{26}{21}+\frac{26}{i}$

$\frac{21}{26}+\frac{i}{26}$

$\frac{21}{26}-\frac{i}{26}$

Correct answer:

$\frac{21}{26}+\frac{i}{26}$

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 .

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)$

Now, distribute and simplify.

$\left(\frac{4-i^3}{5+i}\right)\left(\frac{5-i}{5-i}\right)=\frac{20-4i-5i^3+i^4}{25-i^2}$

Recall that $i^2=-1, i^3=-i,\text{ and }i^4=1$

$\frac{20-4i-5i^3+i^4}{25-i^2}=\frac{20-4i-5(-i)+1}{25-(-1)}=\frac{21+i}{26}=\frac{21}{26}+\frac{i}{26}$

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Example Question #464 : Pre Calculus

Divide

$\frac{5-8i^2}{3+i}$

Possible Answers:

$\frac{39}{10}+\frac{13i}{10}$

$\frac{39}{10}-\frac{13i}{10}$

$\frac{1}{2}+\frac{13i}{10}$

$\frac{10}{39}-\frac{10}{13i}$

Correct answer:

$\frac{39}{10}-\frac{13i}{10}$

Explanation:

Start by simplifying the fraction.

Recall that i^2=-1

$\frac{5-8i^2}{3+i}=\frac{5-8(-1)}{3+i}=\frac{13}{3+i}$

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 .

$\left(\frac{13}{3+i}\right)\left(\frac{3-i}{3-i}\right)$

Now, distribute and simplify.

$\left(\frac{13}{3+i}\right)\left(\frac{3-i}{3-i}\right)=\frac{39-13i}{9-i^2}$

Recall that i^2=-1

$\frac{39-13i}{9-(-1)}=\frac{39-13i}{10}=\frac{39}{10}-\frac{13i}{10}$

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Example Question #465 : Pre Calculus

Simplify $\frac{3+5i}{2-4i}$ into a number of the form a+bi .

Possible Answers:

3+i

None of the other answers.

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

-1+5i

$-\frac{7}{10}+\frac{11}{10}i$

Correct answer:

$-\frac{7}{10}+\frac{11}{10}i$

Explanation:

We have

$\frac{3+5i}{2-4i}$

Multiply by the complex conjugate of the denominator.

The complex conjugate is the denominator with the sign changed:

$\frac{3+5i}{2-4i}\times \frac{2+4i}{2+4i}$

Multiply fractions

$\frac{(3+5i)(2+4i)}{(2-4i)(2+4i)}$

FOIL the numerator and denominator

$\frac{6+12i+10i+20i^2}{4+8i-8i-16i^2}$

Apply the rule of i^2 = -1 :

$\frac{6+12i+10i-20}{4+8i-8i+16}$

Simplify.

$\frac{-14+22i}{20}$

Simplify further using the addition of fractions rule, then factor the i out of the 2nd fraction.

$-\frac{7}{10}+\frac{11}{10}i$

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Example Question #21 : Polar Coordinates And Complex Numbers

Divide:

$\frac{5-i}{3+6i}$

Possible Answers:

$\frac{1}{15}+\frac{11i}{15}$

$15+\frac{15}{11}i$

$\frac{3-11i}{5}$

$\frac{4}{5}-\frac{5i}{8}$

Correct answer:

$\frac{3-11i}{5}$

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 3-6i .

$\left ( \frac{5-i}{3+6i} \right )\left ( \frac{3-6i}{3-6i} \right )$

Now, distribute and simplify.

$\left ( \frac{5-i}{3+6i} \right )\left ( \frac{3-6i}{3-6i} \right )=\frac{15-30i-3i+6i^{2}}{9-6i^{2}}$

Recall that i^2=-1

$\frac{15-33i-6}{9+6}$

Then combine like terms:

$\frac{9-33i}{15}$

Then since each term is a multiple of you can simplify:

$\frac{3-11i}{5}$

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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 #463 : Pre Calculus

Example Question #464 : Pre Calculus

Example Question #465 : Pre Calculus

Example Question #21 : Polar Coordinates And Complex Numbers

All Precalculus Resources

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