• # question_answer If $z=1+i,$ then the multiplicative inverse of z2 is (where i = $\sqrt{-1}$) [Karnataka CET 1999] A) 2 si B) 1 - i C) - i/2 D) i/2

Solution :

Given $z=1+i$ and $i=\sqrt{-1}.$ Squaring both sides, we get ${{z}^{2}}={{(1+i)}^{2}}=1+2i+{{i}^{2}}=1+2i-1$ or ${{z}^{2}}=2i.$ Since it is multiplicative identity, therefore multiplicative inverse of ${{z}^{2}}=\frac{1}{2i}\times \frac{i}{i}=\frac{i}{2{{i}^{2}}}=-\frac{i}{2}.$

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