Convert into the polar form: (1 + 7i) / (2 - i)2.

To convert into polar form,

\(\mathrm{z}=\frac{1+7 \mathrm{i}}{(2-\mathrm{i})^{2}}\\ =\frac{1+7 i}{4+i^{2}-4 i}\\ =\frac{1+7 \mathrm{i}}{4-1-4 \mathrm{i}}\\ =\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}\\ =\frac{3+4 \mathrm{i}+21 \mathrm{i}+28 \mathrm{i}^{2}}{3^{2}+4^{2}}\\ =\frac{3+4 \mathrm{i}+21 \mathrm{i}-28}{3^{2}+4^{2}}=\frac{-25+25 \mathrm{i}}{25}\\ =-1+i\\ \text \ Let \ r \ \cos \theta=-1 \ and \ r \sin \theta=1\\ \text { On squaring and adding, we obtain}\\ \mathrm{r}^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1\\ \Rightarrow \mathrm{r}^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2\\ \Rightarrow \mathrm{r}^{2}=2 \left( \cos ^{2} \theta+\sin ^{2} \theta=1\right)\\ \Rightarrow \mathrm{r}=\sqrt{2}(\mathrm{As} \mathrm{r}>0)\\ \sqrt{2} \cos \theta=-1 and \sqrt{2} \sin \theta=1\\ \Rightarrow \cos \theta=\frac{-1}{\sqrt{2}} and \sin \theta=\frac{1}{\sqrt{2}}\\ \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}\\ \mathrm{z}=\mathrm{r} \cos \theta+\mathrm{ir} \sin \theta\\ =\sqrt{2} \cos \frac{3 \pi}{4}+\mathrm{i} \sqrt{2} \sin \frac{3 \pi}{4}\\ =\sqrt{2}\left(\cos \frac{3 \pi}{4}+\mathrm{i} \sin \frac{3 \pi}{4}\right)\)

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