# DIVISION OF COMPLEX NUMBERS

Division of complex numbers is achieved by rationalizing with the conjugate of the denominator. That is, we multiply the numerators and denominators by the complex conjugate of the denominator.

Before we proceed, you must remember this property of imaginary number:

$i^2=-1$

With that in mind, let us try some examples.

Example 1

Solve $\frac{4+10i}{8-2i}$

Solution:

By rationalizing the denominator

$\frac{4+10i}{8-2i}\times\frac{8+2i}{8+2i}$

$\frac{4(8+2i)+10i(8+2i)}{8(8-2i)+2i(8-2i)}$

$\frac{32+8i+80i+20i^2}{64-16i+16i-4i^2}$

$\frac{32+88i+20i^2}{64-4i^2}$

If $i^2=-1$, then

$\frac{32+88i+20(-1)}{64-4(-1)}$

$\frac{32-20+88i}{64+4}$

$\frac{12+88i}{68}$

Splitting the fraction

$\frac{12}{68}+\frac{88i}{68}$

$\frac{3}{17}+\frac{22i}{17}$

Example 2

Solve $\frac{4-2i}{6+8i}$

Solution:

First, we rationalize the denominator

$\frac{4-2i}{6+8i}\times\frac{6-8i}{6-8i}$

$\frac{4(6-8i)-2i(6-8i)}{6(6-8i)+8i(6-8i)}$

$\frac{24-32i-12i+16i^2}{36-48i+48i-64i^2}$

Given that $i^2=-1$, therefore

$\frac{24-44i+16(-1)}{36-64(-1)}$

$\frac{24-16-44i}{36+64}$

$\frac{8-44i}{100}$

Splitting the fraction

$\frac{8}{100}-\frac{44i}{100}i$

$\frac{2}{25}+\frac{11}{25}i$

Example 3

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

Solution:

$\frac{3+2i}{5-i}\times\frac{5+i}{5+1}$

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

$\frac{15+3i+10i+2i^2}{25-i^2}$

Recalled that $i^2=-1$

$\frac{15+13i+2(-1)}{25-(-1)}$

$\frac{15-2+13i}{26}$

$\frac{13+13i}{26}$

Splitting the fraction

$\frac{13}{26}+\frac{13}{26}i$

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

Example 4

If $z_1=1-3i$, $z_2=-2+5i$ and $z_3=-3-4i$. Find

i. $\frac{Z_1}{Z_3}$

ii. $\frac{Z_1Z_2}{Z_1+Z-2}$

Solution:

I. By substitution

$\frac{1-3i}{-3-4i}$

Now, let's rationalize the denominator

$\frac{1-3i}{-3-4i}\times\frac{-3+4i}{-3+4i}$

$\frac{1(-3+4i)-3i(-3+4i)}{9-16i^2}$

$\frac{-3+4i+9i-12i^2}{9-16(-1)}$

$\frac{-3+13i+12}{25}$

$\frac{9+13i}{25}$

$\frac{9}{25}+\frac{13}{25}i$

ii. To solve this, we have to evaluate the numerator and denominator separately.

we know that $z_1=1-3i$, $z_2=-2+5i$, hence

$(1-3i)(-2+5i)$

So, let's multiply these complex numbers

$1(-2+5i)-3i(-2+5i)$

$-2+5i+6i-15i^2$

$-2+11i+15$

$13+11i$

Now, to the denominator...

$(1-3i)+(-2+5i)$

So, let add these complex numbers

$(1-2)+(-3i+5i)$

$-1+2i$

Now, let evaluate together

$\frac{13+11i}{1+2i}$

By rationalizing the denominator

$\frac{13+11i}{-1+2i}\times\frac{-1-2i}{-1-2i}$

$\frac{13(-1-2i)+11i(-1-2i)}{5}$

$\frac{-13-26i-11i-22i^2}{5}$

if $i^2=-1$, then

$\frac{-13+22 -26i-11i}{5}$

$\frac{9-37i}{5}$

$\frac{9}{5}-\frac{37}{5}i$

That is all for now. Got questions relating to this? tell me in the comment box.

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