Find the Quotient of Complex Numbers - Pre-Calculus

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Let , . Find a simple form of $\frac{a}{b}$ .

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We remark first that:

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

and we know that :

.

This means that:

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

What is ?

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Since ,

the problem becomes,

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

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

Write

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

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

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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}$i$

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

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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

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$

Divide:

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

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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.

Recall that

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

Divide:

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

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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.

Recall that

Divide.

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

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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.

Recall that

Divide:

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

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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:

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

Divide.

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

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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.

Recall that

Divide.

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

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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.

Recall that

Divide.

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

Tap to see back →

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.

Recall that

Divide.

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

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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.

Recall that

Divide.

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

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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.

Recall that

Divide.

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

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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.

Recall that

Divide.

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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 .

Now, distribute and simplify.

Recall that

Divide

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Start by simplifying the fraction.

Recall that

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.

Recall that

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

Let , . Find a simple form of $\frac{a}{b}$ .

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We remark first that:

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

and we know that :

.

This means that:

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

What is ?

Tap to see back →

Since ,

the problem becomes,

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

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

Write

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

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

Tap to see back →

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}$i$

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

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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

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$