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If $ tan \Big[ log(x+iy) \Big] \;=\; a+ib , $ prove that $ tan[log(x^2+y^2)] \;=\; \dfrac{2a}{1-a^2-b^2} \; \; \; where \; a^2+b^2 \ne 1 $
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written 9.4 years ago by
teamques10
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t$an \Big[ log(x+iy) \Big] \;=\; a+ib \\ \; \\ \; \\ \therefore log(x+iy) \;=\; tan^{-1} (a+ib) \; \; \; \; \ldots (i) \\ \; \\ \; \\ log(x-iy) \;=\; tan^{-1} (a-ib) \; \; \; \; \ldots (ii) \\ \; \\ \; \\ $
Adding equations (i) \& (ii)
$ log(x+iy) …
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