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Question-199642

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Question Number 199642 by mokys last updated on 06/Nov/23
Commented by mokys last updated on 07/Nov/23
whers steps
$${whers}\:{steps} \\ $$
Commented by Frix last updated on 07/Nov/23
e^(−((5π)/(12))i) (√2)
$$\mathrm{e}^{−\frac{\mathrm{5}\pi}{\mathrm{12}}\mathrm{i}} \sqrt{\mathrm{2}} \\ $$
Answered by Frix last updated on 07/Nov/23
z=(((√3)−i)/(1+i))=((((√3)−i)(1−i))/((1+i)(1−i)))=((−1+(√3))/2)−((1+(√3))/2)i ∣z∣=(√((((−1+(√3))/2))^2 +(−((1+(√3))/2))^2 ))=(√2) arg (z) =tan^(−1) ((−((1+(√3))/2))/((−1+(√3))/2)) =−tan^(−1) (2+(√3)) =−((5π)/(12)) z=e^(−((5π)/(12))i) (√2)
$${z}=\frac{\sqrt{\mathrm{3}}−\mathrm{i}}{\mathrm{1}+\mathrm{i}}=\frac{\left(\sqrt{\mathrm{3}}−\mathrm{i}\right)\left(\mathrm{1}−\mathrm{i}\right)}{\left(\mathrm{1}+\mathrm{i}\right)\left(\mathrm{1}−\mathrm{i}\right)}=\frac{−\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\mid{z}\mid=\sqrt{\left(\frac{−\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}} \\ $$$$\mathrm{arg}\:\left({z}\right)\:=\mathrm{tan}^{−\mathrm{1}} \:\frac{−\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}}{\frac{−\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=−\mathrm{tan}^{−\mathrm{1}} \:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:=−\frac{\mathrm{5}\pi}{\mathrm{12}} \\ $$$${z}=\mathrm{e}^{−\frac{\mathrm{5}\pi}{\mathrm{12}}\mathrm{i}} \sqrt{\mathrm{2}} \\ $$

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