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Find the multiplicative inverse of the complex number $\;\normalsize -i\;$

$(a)\;3 \normalsize i\qquad(b)\;2 \normalsize i\qquad(c)\;0\qquad(d)\;\normalsize i$

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Answer : $\; \normalsize i$
Explanation :
Let$\;z= \normalsize - i$
Then , $\;\overline{z}=\normalsize i = |z|^{2}=( \normalsize i)^{2}=1$
Therefore , the multiplicative inverse of $\;\normalsize -i\;$ is given by
$z^{-1} =\large\frac{\overline{z}}{|z|^{2}} =\large\frac{\normalsize i}{\normalsize 1} = \normalsize i$
answered Apr 10, 2014 by yamini.v
 

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