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If $z=1+i$, then what is the inverse of $z^{2}$ ?
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The correct answer is:
$-\frac{1}{2}$
Given that $z=1+i$ $\Rightarrow z^{2}=(1+i)^{2}=1+i^{2}+2 i=1-1+2 i=2 i$
Inverse of $z^{2}=\frac{1}{2 i}=-\frac{i^{2}}{2 \mathrm{i}}\left[\right.$ Since $\left.i^{2}=-1 ;-\mathrm{i}^{2}=1\right]$
$=-\frac{\mathrm{i}}{2}$
Inverse of $z^{2}=\frac{1}{2 i}=-\frac{i^{2}}{2 \mathrm{i}}\left[\right.$ Since $\left.i^{2}=-1 ;-\mathrm{i}^{2}=1\right]$
$=-\frac{\mathrm{i}}{2}$
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