A) 2 si
B) 1 - i
C) - i/2
D) i/2
Correct Answer: C
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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