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Find the multiplicative inverse of the complex number $-i$.

$\sum\limits_{n=1}^{50} i^{(2n-1)!}$ is equal to (where $i = \sqrt{-1}$)

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If $a+ib = \frac{(x+i)^{2}}{2x^{2}+1}$,prove that $a^{2}+b^{2} = \frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$.

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The values of $\theta$,for which $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is real are

${\left( \frac{1 + i}{1 - i} \right)^2} + {\left( \frac{1 - i}{1 + i} \right)^2}$ is equal to

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