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Question Number 130742 by mohammad17 last updated on 28/Jan/21

Answered by Dwaipayan Shikari last updated on 28/Jan/21

z=9+3i  =(√(9^2 +3^2 )) e^(itan^(−1) (1/3)) =3(√(10))e^(itan^(−1) (1/3))   z=7+2i=(√(53)) e^(itan^(−1) (2/7))   z=2+6i=2(√(10)) e^(itan^(−1) 3)

$${z}=\mathrm{9}+\mathrm{3}{i}\:\:=\sqrt{\mathrm{9}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\:{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{3}\sqrt{\mathrm{10}}{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${z}=\mathrm{7}+\mathrm{2}{i}=\sqrt{\mathrm{53}}\:{e}^{{itan}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{7}}} \\ $$$${z}=\mathrm{2}+\mathrm{6}{i}=\mathrm{2}\sqrt{\mathrm{10}}\:{e}^{{itan}^{−\mathrm{1}} \mathrm{3}} \:\: \\ $$

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