Find the Quotient of Complex Numbers - Pre-Calculus

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Question

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

Answer

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 :

.

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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Question

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

Answer

Since ,

the problem becomes,

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

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

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Question

Write

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

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

Answer

The correct answer is

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

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

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

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

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Question

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

Answer

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 :

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

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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Question

Divide:

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

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

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Question

Divide:

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

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

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Question

Divide.

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

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

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Question

Divide:

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

Answer

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 .

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

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

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

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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Question

Divide.

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

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

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Question

Divide.

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

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

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Question

Divide.

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

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

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Question

Divide.

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

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

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Question

Divide.

$\frac{6-2i}{3-i}$

Answer

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 .

$\left(\frac{6-2i}{3-i}\right)\left(\frac{3+i}{3+i}\right)$

Now, distribute and simplify.

$\left(\frac{6-2i}{3-i}\right)\left(\frac{3+i}{3+i}\right)=\frac{18-6i+6i-2i^2}{3^2-i^2}$

Recall that

$\frac{18-2i^2}{3^2-i^2}=\frac{18-2(-1)}{9-(-1)}=\frac{20}{10}=2$

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Question

Divide.

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

Answer

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 .

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

Now, distribute and simplify.

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

Recall that

$\frac{-5i-i^2}{-i^2}=\frac{-5i-(-1)}{-(-1)}=\frac{1-5i}{1}=1-5i$

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Question

Divide.

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

Answer

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 .

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

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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Question

Divide

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

Answer

Start by simplifying the fraction.

Recall that

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

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 .

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

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

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

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Question

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

Answer

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 :

.

This means that:

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

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

Compare your answer with the correct one above

Question

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

Answer

Since ,

the problem becomes,

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

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

Compare your answer with the correct one above

Question

Write

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

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

Answer

The correct answer is

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

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

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

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

Compare your answer with the correct one above

Question

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

Answer

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 :

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

Simplify.

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

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

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

Compare your answer with the correct one above

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