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Question Number 136897 by mohammad17 last updated on 27/Mar/21

Answered by Olaf last updated on 27/Mar/21

$$$$$$f\left(z\right)=e^{i\frac{\pi }{4}}z+(1-2i)$$$$f\left(z\right)-z_0=e^{i\frac{\pi }{4}}(z-z_0)+(e^{i\frac{\pi }{4}}-1)z_0+(1-2i)$$$$\mathrm{We}\mathrm{choose}{z}_0\mathrm{such}\mathrm{as}:(e^{i\frac{\pi }{4}}-1)z_0+(1-2i)=0$$$$\Rightarrow z_0=-\frac{1-2i}{e^{i\frac{\pi }{4}}-1}=-\frac{1-2i}{\frac{1}{\sqrt{2}}-1+\frac{i}{\sqrt{2}}}=-\frac{\sqrt{2}+(2-\sqrt{2})i}{2-\sqrt{2}}$$$$z_0=-\frac{\frac{1}{\sqrt{2}}-1-\sqrt{2}+(2-\sqrt{2}-\frac{1}{\sqrt{2}})i}{2-\sqrt{2}}=(\frac{3}{2}+\sqrt{2})-\frac{\sqrt{2}-1}{2}i$$$$f\left(z\right)-z_0=e^{i\frac{\pi }{4}}(z-z_0)$$$$\mathrm{The}\mathrm{transformation}\mathrm{W}\mathrm{is}\mathrm{the}\mathrm{rotation}$$$$\mathrm{angle}\frac{\pi }{4}\mathrm{around}\mathrm{the}\mathrm{point}{z}_0.$$$$\mathrm{The}\mathrm{area}\mathcal{A}=\{0⩽x⩽2,0⩽y⩽1\}\mathrm{is}\mathrm{a}$$$$\mathrm{rectangle}\mathrm{O},\mathrm{A}\left(2\right),\mathrm{B}(2+i),\mathrm{C}\left(i\right)$$$$\mathrm{The}\mathrm{image}{\mathcal{A}}^{\prime }\mathrm{of}\mathcal{A}\mathrm{by}\mathrm{W}\mathrm{is}\mathrm{another}$$$$\mathrm{rectangle}{\mathrm{O}}^{\prime }{\mathrm{A}}^{\prime }{\mathrm{B}}^{\prime }{\mathrm{C}}^{\prime }$$$${\mathrm{O}}^{\prime }=f\left(\mathrm{O}\right)=1-2i$$$${\mathrm{A}}^{\prime }=f\left(\mathrm{A}\right)=2\times \frac{1}{\sqrt{2}}(1+i)+(1-2i)=\sqrt{2}+1+(\sqrt{2}-2)i$$$${\mathrm{B}}^{\prime }=f\left(\mathrm{B}\right)=(2+i)\frac{1}{\sqrt{2}}(1+i)+(1-2i)=1+\frac{1}{\sqrt{2}}+(\frac{3}{\sqrt{2}}-2)i$$$${\mathrm{C}}^{\prime }=f\left(\mathrm{C}\right)=i\times \frac{1}{\sqrt{2}}(1+i)+(1-2i)=1-\frac{1}{\sqrt{2}}+(\frac{1}{\sqrt{2}}-2)i$$

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