Question

If $z=1+i$, then the multiplicative inverse of $z^2$ is (where, $i=\sqrt{-1}$ )

• A. $2i$
• B. $1-i$
• C. $-\frac{i}{2}$
• D. $\frac{i}{2}$

Answer 1

Answer: (C)

Let $z=1+i$ then $z^2=(1+i{)}^2$
$=1^2+i^2+2\cdot 1\cdot i=1+i^2+2i$
$=1-1+2i=2i$
Now $2i\times -\frac{i}{2}$
$\Rightarrow -i^2=1$
Hence, $-\frac{i}{2}$ is multiplicative inverse of $z^2$.