if-tan-x-iy-a-ib-determine-x-and-y-interms-of-a-and-b-

Question Number 96377 by mathmax by abdo last updated on 01/Jun/20

if \tan (x + iy) = a + ib determine x and y interms of a and b

Answered by mathmax by abdo last updated on 02/Jun/20

\tan (x + iy) = a + ib \Rightarrow x + iy = \arctan (a + ib) we know

\arctan (z) = \frac{1}{2 i} \ln (\frac{1 + iz}{1 - iz}) \Rightarrow \arctan (a + ib) = \frac{1}{2 i} \ln (\frac{1 + i (a + ib)}{1 - i (a + ib)})

= \frac{1}{2 i} \ln (\frac{1 - b + ia}{1 + b - ia}) we have 1 - b + ia = \sqrt{{(1 - b)}^{2} + a^{2}} e^{iarctan (\frac{a}{1 - b})} \Rightarrow

\ln (1 - b + ia) = \frac{1}{2} \ln {{(1 - b)}^{2} + a^{2}} + iarctan (\frac{a}{1 - b}) also

1 + b - ia = \sqrt{{(1 + b)}^{2} + a^{2}} e^{- iarctan (\frac{a}{1 + b})} \Rightarrow \ln (1 + b - ia) = \frac{1}{2} \ln {{(1 + b)}^{2} + a^{2}}

- iarctan (\frac{a}{1 + b}) \Rightarrow 2 i \arctan (a + ib)

\frac{1}{2} \ln {{(1 - b)}^{2} + a^{2}} + iarctan (\frac{a}{1 - b}) - \frac{1}{2} \ln {{(1 + b)}^{2} + a^{2}} + iarctan (\frac{a}{1 + b})

= \frac{1}{2} \ln (\frac{{(1 - b)}^{2} + a^{2}}{{(1 + b)}^{2} + a^{2}}) + i {\arctan (\frac{a}{1 - b}) + \arctan (\frac{a}{1 + b}) \Rightarrow

\arctan (a + ib) = \frac{1}{4 i} \ln (\frac{{(1 - b)}^{2} + a^{2}}{{(1 + b)}^{2} + a^{2}}) + \frac{1}{2} {\arctan (\frac{a}{1 - b}) + \arctan (\frac{a}{1 + b})} = x + iy

\Rightarrow {\begin{cases} x = \frac{1}{2} {\arctan (\frac{a}{1 - b}) + \arctan (\frac{a}{1 + b})} \\ y = \frac{1}{4} \ln (\frac{a^{2} + b^{2} + 2 b + 1}{a^{2} + b^{2} - 2 b + 1}) \end{cases}

Commented by Ar Brandon last updated on 02/Jun/20

Vous avez des choses tellement intéressantes monsieur Mathmax.��