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

Marks:- 4

Mumbai Unversity>FE>Sem1>Applied Maths1

m1(81) • 32 views
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modified 8 weeks ago  • written 8 weeks ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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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 )$

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

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

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written 8 weeks ago by gravatar for Mayank Aggarwal Mayank Aggarwal0
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