evaluate-ln-1-

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

evaluate \ln (- 1)

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

w e h a v e - 1 = e^{i (2 k + 1) π} \forall k \in Z \Rightarrow

l n (- 1) = i (2 k + 1) π b u t w e t a k e l n (- 1) = i π

f o r p r i c i p a l d e t e r m i n a t i o n o f t h e c o m p l e x

l o g a r i t h m e .

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

e^{θ i} = \cos θ + i \times \sin θ

\cos θ = - 1

θ = (2 n - 1) π; n \in Z

\Rightarrow \ln (- 1) = (2 n - 1) π; n \in Z

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

l n (- x) = l n (x) + i Π

l n (- 1) = l n (1) + i Π = i Π

L O G (A + i B) a n d l n (A + i B)

L o g (A + i B) = l n (A + i B) + 2 n Π i

= \frac{1}{2} l n (A^{2} + B^{2}) + {i t a n}^{- 1} (\frac{B}{A}) + 2 n Π i