If x+iy = frac{1+7i}{(2-i)^2},then operatorna | vedclass

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(C) Given $x+iy = \frac{1+7i}{(2-i)^2}$.Expanding the denominator: $(2-i)^2 = 4 + i^2 - 4i = 4 - 1 - 4i = 3 - 4i$.So,$x+iy = \frac{1+7i}{3-4i}$.Multip

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$-1$

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(C) Given $x+iy = \frac{1+7i}{(2-i)^2}$.Expanding the denominator: $(2-i)^2 = 4 + i^2 - 4i = 4 - 1 - 4i = 3 - 4i$.So,$x+iy = \frac{1+7i}{3-4i}$.Multiplying the numerator and denominator by the conjugate $(3+4i)$:$x+iy = \frac{(1+7i)(3+4i)}{(3-4i)(3+4i)} = \frac{3 + 4i + 21i + 28i^2}{9 - 16i^2} = \frac{3 + 25i - 28}{9 + 16} = \frac{-25 + 25i}{25} = -1 + i$.Comparing real and imaginary parts,we get $x = -1$ and $y = 1$.Now,evaluate $\operatorname{cosec}\left(	an^{-1} \frac{y}{x} - \frac{\pi}{4}ight)$:$= \operatorname{cosec}\left(	an^{-1} \left(\frac{1}{-1}ight) - \frac{\pi}{4}ight) = \operatorname{cosec}\left(	an^{-1}(-1) - \frac{\pi}{4}ight)$.Since $	an^{-1}(-1) = -\frac{\pi}{4}$,we have:$= \operatorname{cosec}\left(-\frac{\pi}{4} - \frac{\pi}{4}ight) = \operatorname{cosec}\left(-\frac{\pi}{2}ight) = -1$.

If $x+iy = \frac{1+7i}{(2-i)^2}$,then $\operatorname{cosec}\left(\tan^{-1} \frac{y}{x} - \frac{\pi}{4}\right) = $

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