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Question Number 48226 by gunawan last updated on 21/Nov/18
calculate  log(−1+(√3) i)^2
$$\mathrm{calculate} \\ $$$$\mathrm{log}\left(−\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{i}\right)^{\mathrm{2}} \\ $$
Answered by Smail last updated on 21/Nov/18
ln((−1+(√3)i)^2 )=2ln(2(−(1/2)+i((√3)/2)))  =2ln2+2ln(cos(((2π)/3))+isin(((2π)/3)))  =2ln2+2ln(e^(i((2π)/3)) )=2ln2+i((4π)/3)
$${ln}\left(\left(−\mathrm{1}+\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} \right)=\mathrm{2}{ln}\left(\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{2}{ln}\mathrm{2}+\mathrm{2}{ln}\left({cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right) \\ $$$$=\mathrm{2}{ln}\mathrm{2}+\mathrm{2}{ln}\left({e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)=\mathrm{2}{ln}\mathrm{2}+{i}\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$

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