frac{1 + 7i}{(2 - i)^2} = | vedclass

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(A) Given expression: $z = \frac{1 + 7i}{(2 - i)^2}$First,expand the denominator: $(2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$Now,simplify the fra

Vedclass · Accepted Answer

$\sqrt{2} \left( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right)$

Vedclass · Answer

(A) Given expression: $z = \frac{1 + 7i}{(2 - i)^2}$First,expand the denominator: $(2 - i)^2 = 4 - 4i + i^2 = 4 - 4i - 1 = 3 - 4i$Now,simplify the fraction: $z = \frac{1 + 7i}{3 - 4i}$Multiply the numerator and denominator by the conjugate of the denominator $(3 + 4i)$: $z = \frac{(1 + 7i)(3 + 4i)}{(3 - 4i)(3 + 4i)} = \frac{3 + 4i + 21i + 28i^2}{3^2 + 4^2} = \frac{3 + 25i - 28}{9 + 16} = \frac{-25 + 25i}{25} = -1 + i$To convert to polar form $r(\cos 	heta + i\sin 	heta)$,find the modulus $r$ and argument $	heta$: $r = |z| = \sqrt{(-1)^2 + (1)^2} = \sqrt{2}$Since $z$ is in the second quadrant (real part negative,imaginary part positive),$	heta = \pi - 	an^{-1}\left| \frac{1}{-1} ight| = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$Therefore,$z = \sqrt{2} \left( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} ight)$

$\frac{1 + 7i}{(2 - i)^2} = $

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