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Question Number 146702 by deleteduser12 last updated on 13/Nov/23

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$$$$$${ne}^{i\theta }=\frac{3+3i}{2+3i}+\frac{1+5i}{1-2i}$$$${ne}^{i\theta }=\frac{3(1+i)(2-3i)}{(2+3i)(2-3i)}+\frac{(1+5i)(1+2i)}{(1-2i)(1+2i)}$$$${ne}^{i\theta }=\frac{3(5-i)}{13}+\frac{(-9+7i)}{5}$$$${ne}^{i\theta }=\frac{15(5-i)}{65}+\frac{13(-9+7i)}{65}$$$${ne}^{i\theta }=\frac{75-15i}{65}+\frac{-117+91i}{65}$$$${ne}^{i\theta }=\frac{-42+76i}{65}$$$$n=\mid \frac{-42+76i}{65}\mid =\frac{\sqrt{{42}^2+{76}^2}}{65}=\frac{2\sqrt{1885}}{65}$$$$\theta =\arg \left(\frac{-42+76i}{65}\right)=\arctan \left(\frac{76}{-42}\right)=-\arctan \left(\frac{38}{21}\right)$$

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