Thank you for reporting, we will resolve it shortly
Q.
If $z=1+i$, then the multiplicative inverse of $z^{2}$ is (where, $i=\sqrt{-1}$ )
Complex Numbers and Quadratic Equations
Solution:
Let $z= 1+i $ then $z^{2}=(1+i)^{2} $
$ =1^{2}+i^{2}+2 \cdot 1 \cdot i=1+i^{2}+2 i $
$=1-1+2 i=2 i $
Now $2 i \times-\frac{i}{2}$
$ \Rightarrow - i ^{2}=1$
Hence, $-\frac{i}{2}$ is multiplicative inverse of $z^{2}$.