Question

Represent in a+bi form

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

Answer 1

We have,

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

By rationalizing we get

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

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

$=\frac{1+3i+2i+6i^2}{1-3i^2}$

$=\frac{1+5i-6}{1-3}$ Since $[i^2=-1]$

$=\frac{5-5i}{2}$