Question: 5] c)ii) show that $ilog(\frac{x-i}{x+i}) = \pi - 2tan^{-1}x$

0

0

we have $log(x+i) = ½ log(x^2+1) + itan^{-1}\frac{1}{x}$

and $log(x-i) =½ log(x^2+1) –itan^{-1}\frac{1}{x}$

$\frac{log((x-i)}{(x+i))} = log(x-i) - log(x+i)$

$ = -2itan^{-1}\frac{1}{x} = -2i(π/2–tan^{-1}x )$

$∴log\frac{((x-i)}{(x+i))} = -i(π - 2 tan^{-1}x )$

$∴i log(\frac{(x-i)}{(x+i)}) = (π - 2tan^{-1}x )$ .

Please log in to add an answer.

Site

Use of this site constitutes acceptance of our User
Agreement
and Privacy
Policy.