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Question Number 142383 by liki last updated on 30/May/21

Commented by liki last updated on 30/May/21

$$\mathrm{help}\mathrm{me}\mathrm{qn}\mathrm{no}4$$

Answered by physicstutes last updated on 31/May/21

$$4\left(\mathrm{a}\right)f\left(z\right)=\frac{7-z}{1-z^2},z=1+2i$$$$f\left(z\right)=\frac{7-1-2i}{1-(1-4+4i)}=\frac{6-2i}{4-4i}$$$$\mid f\left(z\right)\mid =\frac{\sqrt{6^2+2^2}}{\sqrt{4^2+4^2}}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$$$$2\mid f\left(z\right)\mid =\sqrt{5}$$$$\mid z\mid =\sqrt{1+2^2}=\sqrt{5}$$$$\mathrm{Hence}\mid z\mid =2\mid f\left(z\right)\mid .$$

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