Question

Simplify the following :
$\frac{1+2i}{1-(1-i{)}^2}$

Answer 1

We know that $\frac{1+2i}{1-(-2i)}$ = 1

$i^4=1,i^{4n}=1,i^{4n+1}=i.$

$i^{4n+2}=i^2=-1,i^{4n+3}=i^3=-1$

$(1+i{)}^2=1+i^2+2i=2i$

$\therefore (1-i{)}^2=conj.=-2i,1+i^2=0,1-i^2=2$

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