Question

Derive the following complex in the form of $A+iB$.

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

Answer 1

Consider the given complex number $\frac{1+2i+3i^2}{1-2i+3i^2}$ ,

$\because i^2=-1$

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

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

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

By rationalization ,we get

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

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

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

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

Rationalise again ,we get

$\frac{1}{i}=\frac{1}{i}\times \frac{i}{i}=\frac{i}{i^2}$

$\frac{i}{(-1)}=-i$

$=0-i$

$\frac{1+2i+3i^2}{1-2i+3i^2}=0-i$

$A+iB=0-i$

Hence, this is the required answer.