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If $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$,then $(x, y)$ is equal to

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

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

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The value of the series $1 + i^2 + i^4 + i^6 + ..... + i^{2n}$ is:

$\frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y}{3-i} = i$. Find the value of $x+y$.

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