Convert the following complex number into pol | vedclass

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Question

(A) Let $z = \frac{1+7i}{(2-i)^2}$.Expanding the denominator: $(2-i)^2 = 4 + i^2 - 4i = 4 - 1 - 4i = 3 - 4i$.So,$z = \frac{1+7i}{3-4i}$.Multiply numer

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

Vedclass · Answer

(A) Let $z = \frac{1+7i}{(2-i)^2}$.Expanding the denominator: $(2-i)^2 = 4 + i^2 - 4i = 4 - 1 - 4i = 3 - 4i$.So,$z = \frac{1+7i}{3-4i}$.Multiply numerator and denominator by the conjugate $(3+4i)$:$z = \frac{(1+7i)(3+4i)}{(3-4i)(3+4i)} = \frac{3 + 4i + 21i + 28i^2}{3^2 + 4^2} = \frac{3 + 25i - 28}{9 + 16} = \frac{-25 + 25i}{25} = -1 + i$.For polar form $z = r(\cos 	heta + i \sin 	heta)$:$r = \sqrt{(-1)^2 + (1)^2} = \sqrt{1+1} = \sqrt{2}$.Since $x = -1$ and $y = 1$,the complex number lies in the $II$ quadrant.$\cos 	heta = \frac{-1}{\sqrt{2}}$ and $\sin 	heta = \frac{1}{\sqrt{2}}$.Thus,$	heta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.The polar form is $\sqrt{2}\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}ight)$.

Convert the following complex number into polar form: $\frac{1+7i}{(2-i)^2}$

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