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


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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
	$tan(x+iy) =a+ib ⇒x+iy =arctan(a+ib) we know arctan(z) =(1/(2i))ln(((1+iz)/(1−iz))) ⇒arctan(a+ib) =(1/(2i))ln(((1+i(a+ib))/(1−i(a+ib)))) =(1/(2i))ln(((1−b+ia)/(1+b−ia))) we have 1−b+ia =(√((1−b)^2 +a^2 ))e^(iarctan((a/(1−b)))) ⇒ ln(1−b+ia) =(1/2)ln{(1−b)^2 +a^2 } +iarctan((a/(1−b))) also 1+b−ia =(√((1+b)^2 +a^2 ))e^(−iarctan((a/(1+b)))) ⇒ln(1+b−ia) =(1/2)ln{(1+b)^2 +a^2 } −iarctan((a/(1+b))) ⇒2i arctan(a+ib) (1/2)ln{(1−b)^2 +a^2 }+iarctan((a/(1−b)))−(1/2)ln{(1+b)^2 +a^2 }+iarctan((a/(1+b))) =(1/2)ln((((1−b)^2 +a^2 )/((1+b)^2 +a^2 ))) +i{ arctan((a/(1−b)))+arctan((a/(1+b))) ⇒ arctan(a+ib) =(1/(4i))ln((((1−b)^2 +a^2 )/((1+b)^2 +a^2 )))+(1/2){arctan((a/(1−b)))+arctan((a/(1+b)))}=x+iy ⇒ { ((x =(1/2){ arctan((a/(1−b))) +arctan((a/(1+b)))})),((y =(1/4)ln(((a^2 +b^2 +2b+1)/(a^2 +b^2 −2b+1))))) :}$
	$\tan (x + iy) = a + ib \Rightarrow x + iy = \arctan (a + ib) we know$ $\arctan (z) = \frac{1}{2 i} \ln (\frac{1 + iz}{1 - iz}) \Rightarrow \arctan (a + ib) = \frac{1}{2 i} \ln (\frac{1 + i (a + ib)}{1 - i (a + ib)})$ $= \frac{1}{2 i} \ln (\frac{1 - b + ia}{1 + b - ia}) we have 1 - b + ia = \sqrt{{(1 - b)}^{2} + a^{2}} e^{iarctan (\frac{a}{1 - b})} \Rightarrow$ $\ln (1 - b + ia) = \frac{1}{2} \ln {{(1 - b)}^{2} + a^{2}} + iarctan (\frac{a}{1 - b}) also$ $1 + b - ia = \sqrt{{(1 + b)}^{2} + a^{2}} e^{- iarctan (\frac{a}{1 + b})} \Rightarrow \ln (1 + b - ia) = \frac{1}{2} \ln {{(1 + b)}^{2} + a^{2}}$ $- iarctan (\frac{a}{1 + b}) \Rightarrow 2 i \arctan (a + ib)$ $\frac{1}{2} \ln {{(1 - b)}^{2} + a^{2}} + iarctan (\frac{a}{1 - b}) - \frac{1}{2} \ln {{(1 + b)}^{2} + a^{2}} + iarctan (\frac{a}{1 + b})$ $= \frac{1}{2} \ln (\frac{{(1 - b)}^{2} + a^{2}}{{(1 + b)}^{2} + a^{2}}) + i {\arctan (\frac{a}{1 - b}) + \arctan (\frac{a}{1 + b}) \Rightarrow$ $\arctan (a + ib) = \frac{1}{4 i} \ln (\frac{{(1 - b)}^{2} + a^{2}}{{(1 + b)}^{2} + a^{2}}) + \frac{1}{2} {\arctan (\frac{a}{1 - b}) + \arctan (\frac{a}{1 + b})} = x + iy$ $\Rightarrow {\begin{cases} x = \frac{1}{2} {\arctan (\frac{a}{1 - b}) + \arctan (\frac{a}{1 + b})} \\ y = \frac{1}{4} \ln (\frac{a^{2} + b^{2} + 2 b + 1}{a^{2} + b^{2} - 2 b + 1}) \end{cases}$
	Commented by Ar Brandon last updated on 02/Jun/20
	Vous avez des choses tellement intéressantes monsieur Mathmax.��
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