Question-146702

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

Answered by Olaf_Thorendsen last updated on 15/Jul/21

ne^(iθ) = ((3+3i)/(2+3i))+((1+5i)/(1−2i)) ne^(iθ) = ((3(1+i)(2−3i))/((2+3i)(2−3i)))+(((1+5i)(1+2i))/((1−2i)(1+2i))) ne^(iθ) = ((3(5−i))/(13))+(((−9+7i))/5) ne^(iθ) = ((15(5−i))/(65))+((13(−9+7i))/(65)) ne^(iθ) = ((75−15i)/(65))+((−117+91i)/(65)) ne^(iθ) = ((−42+76i)/(65)) n = ∣((−42+76i)/(65))∣ = ((√(42^2 +76^2 ))/(65)) = ((2(√(1885)))/(65)) θ = arg(((−42+76i)/(65))) = arctan(((76)/(−42))) = −arctan(((38)/(21)))

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

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