Question Number 41198 by mondodotto@gmail.com last updated on 03/Aug/18
$$\mathrm{evaluate}\:\boldsymbol{\mathrm{ln}}\left(−\mathrm{1}\right) \\ $$
Commented by math khazana by abdo last updated on 03/Aug/18
$${we}\:\:{have}\:\:−\mathrm{1}\:={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\:\:\:\:\forall{k}\in{Z}\:\Rightarrow \\ $$$${ln}\left(−\mathrm{1}\right)={i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi\:\:\:{but}\:{we}\:{take}\:\:{ln}\left(−\mathrm{1}\right)={i}\pi \\ $$$${for}\:{pricipal}\:{determination}\:{of}\:{the}\:{complex} \\ $$$${logarithme}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18
Answered by MJS last updated on 03/Aug/18
$$\mathrm{e}^{\theta\mathrm{i}} =\mathrm{cos}\:\theta\:+\mathrm{i}×\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\theta\:=−\mathrm{1} \\ $$$$\theta=\left(\mathrm{2}{n}−\mathrm{1}\right)\pi;\:{n}\in\mathbb{Z} \\ $$$$\Rightarrow\:\mathrm{ln}\left(−\mathrm{1}\right)=\left(\mathrm{2}{n}−\mathrm{1}\right)\pi;\:{n}\in\mathbb{Z} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18
$${ln}\left(−{x}\right)={ln}\left({x}\right)+{i}\Pi \\ $$$${ln}\left(−\mathrm{1}\right)={ln}\left(\mathrm{1}\right)+{i}\Pi={i}\Pi \\ $$$$\boldsymbol{{L}}{OG}\left({A}+{iB}\right)\:\:{and}\:{ln}\left({A}+{iB}\right)\: \\ $$$${Log}\left({A}+{iB}\right)={ln}\left({A}+{iB}\right)+\mathrm{2}{n}\Pi{i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} \right)+{itan}^{−\mathrm{1}} \left(\frac{{B}}{{A}}\right)+\mathrm{2}{n}\Pi{i} \\ $$