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Question Number 41198 by mondodotto@gmail.com last updated on 03/Aug/18

$$\mathrm{evaluate}\mathbf{ln}(-1)$$

Commented by math khazana by abdo last updated on 03/Aug/18

$$wehave-1=e^{i(2k+1)\pi }\mathrm{\forall }k\in Z\Rightarrow$$$$ln(-1)=i(2k+1)\pi butwetakeln(-1)=i\pi$$$$forpricipaldeterminationofthecomplex$$$$logarithme.$$

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

$${\mathrm{e}}^{\theta \mathrm{i}}=\cos \theta +\mathrm{i}\times \sin \theta$$$$\cos \theta =-1$$$$\theta =(2n-1)\pi ;n\in \mathbb{Z}$$$$\Rightarrow \ln (-1)=(2n-1)\pi ;n\in \mathbb{Z}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18

$$ln(-x)=ln\left(x\right)+i\mathrm{\Pi }$$$$ln(-1)=ln\left(1\right)+i\mathrm{\Pi }=i\mathrm{\Pi }$$$$\mathit{L}OG(A+iB)andln(A+iB)$$$$Log(A+iB)=ln(A+iB)+2n\mathrm{\Pi }i$$$$=\frac{1}{2}ln(A^2+B^2)+{itan}^{-1}\left(\frac{B}{A}\right)+2n\mathrm{\Pi }i$$

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