Question Number 142383 by liki last updated on 30/May/21
Commented by liki last updated on 30/May/21
$$\mathrm{help}\:\mathrm{me}\:\mathrm{qn}\:\mathrm{no}\:\mathrm{4} \\ $$
Answered by physicstutes last updated on 31/May/21
$$\mathrm{4}\left(\mathrm{a}\right)\:{f}\left({z}\right)\:=\:\frac{\mathrm{7}−{z}}{\mathrm{1}−{z}^{\mathrm{2}} },\:{z}\:=\:\mathrm{1}\:+\:\mathrm{2}{i} \\ $$$${f}\left({z}\right)\:=\:\frac{\mathrm{7}−\mathrm{1}−\mathrm{2}{i}}{\mathrm{1}−\left(\mathrm{1}−\mathrm{4}+\mathrm{4}{i}\right)}\:=\:\frac{\mathrm{6}−\mathrm{2}{i}}{\mathrm{4}−\mathrm{4}{i}}\: \\ $$$$\mid{f}\left({z}\right)\mid\:=\:\frac{\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}{\:\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}\:=\:\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\: \\ $$$$\mathrm{2}\mid{f}\left({z}\right)\mid\:=\:\sqrt{\mathrm{5}} \\ $$$$\mid{z}\mid\:=\:\sqrt{\mathrm{1}+\mathrm{2}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{5}} \\ $$$$\mathrm{Hence}\:\:\mid{z}\mid\:=\:\mathrm{2}\mid{f}\left({z}\right)\mid. \\ $$