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

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written 3.9 years ago by

teamques10 ★ 69k

modified 3.1 years ago by

v2ishu789 • 40

engineering mathematics

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written 3.1 years ago by

v2ishu789 • 40

modified 3.1 years ago by

sagarkolekar ★ 11k

enter image description here Solution:

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written 3.9 years ago by

teamques10 ★ 69k

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

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

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

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

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

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

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