Separate into real and imaginary parts of tanh-1 (x+iy).

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written 3.0 years ago by

teamques10 ★ 65k

$Given: \tanh^{-1} (x+iy) $

$\tanh^{-1} x= \dfrac 1 2 \log \bigg[\dfrac {(1+z)}{(1-z)}\bigg]$

$\tanh^{-1}( x+iy)= \dfrac 1 2 \log \bigg[\dfrac {(1+x+iy)}{(1-x-iy)}\bigg]$

$= \dfrac 1 2 \log [ {(1+x+iy)}-\log{(1-x-iy)}]$

$= \dfrac 1 2 \bigg[\dfrac 1 2 \log [ {(1+x)^2+y^2]}+i .\tan^{-1}\bigg(\dfrac {y} {1+x}\bigg)\bigg]$

$= -\dfrac 1 2 \bigg[\dfrac 1 2 \log [ {(1-x)^2+y^2]}-i …

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