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Separate into real and imaginary parts of tanh-1 (x+iy).
2 Answers
written 3.9 years ago by |
Given:tanh−1(x+iy)
tanh−1x=12log[(1+z)(1−z)]
tanh−1(x+iy)=12log[(1+x+iy)(1−x−iy)]
=12log[(1+x+iy)−log(1−x−iy)]
=12[12log[(1+x)2+y2]+i.tan−1(y1+x)]
=−12[12log[(1−x)2+y2]−i.tan−1(y1−x)]
=12[12log[(1+x)2+y2](1−x)2+y2]+i[tan−1y1+x+tan−1y1−x]
=12[12log[(1+x)2+y2](1−x)2+y2]+i.tan−1[y1+x+y1−x/1−y1+x×y1−x]
=12[12log((1+x)2+y2(1−x)2+y2)+i.tan−1(2y1−x2−y2)]
Separate real and imaginary parts,
Real parts: =14log((1+x)2+y2(1−x)2+y2)
Imaginary parts: 12tan−1(2y1−x2−y2)