Complex Analysis Solutions: Real and Imaginary Parts

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The discussion focuses on finding the real and imaginary parts of two complex functions. For the first function, w = (e^(conjugated(z)))^2, the real part is u = (e^(x^2-y^2))cos(2xy) and the imaginary part is v = -(e^(x^2-y^2))sin(2xy). The second function, w = tg(z), yields the real part u = (sin(x)cos(x))/(cosh^2(y) - sin^2(x)) and the imaginary part v = (sinh(y)cosh(y))/(cosh^2(y) - sin^2(x)). Additionally, there are hints provided for using LaTeX to format mathematical expressions correctly. The thread emphasizes the importance of understanding the transformations involved in complex analysis.
Fabio010
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Find the real part and imaginary part of the following exercises.

1) w = ((e^(conjugated(z)))^2

2) w = tgz



Solutions:

1) u= (e^(x^2-y^2))cos2xy v= -(e^(x^2-y^2))sin2xy

2) u= (sinxcosx)/(ch^2y-sin^2x) v= (shychy)/(ch^2-sin^2x)



-------------------------------------

Attempts:
 
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Fabio010 said:
Find the real part and imaginary part of the following exercises.

1) w = ((e^(conjugated(z)))^2

2) w = tgz



Solutions:

1) u= (e^(x^2-y^2))cos2xy v= -(e^(x^2-y^2))sin2xy

2) u= (sinxcosx)/(ch^2y-sin^2x) v= (shychy)/(ch^2-sin^2x)



-------------------------------------

Attempts:


It won't take you long to learn enough LaTeX to properly write mathematics in this forum...

Hints: putting \,z:=x+iy\,\,,\,\,x,y\in\Bbb R\,:

$$(1)\,\,\left(e^{\overline z}\right)^2=\left(e^{x-iy}\right)^2=e^{2x-2iy}=e^{2x}e^{-2iy}=e^{2x}\left(\cos 2y-i\sin 2y\right)$$

$$(2)\;\;\;\tan z=\frac{\sin z}{\cos z}=\frac{e^{iz}-e^{-iz}}{2i}\cdot\frac{2}{e^{iz}+e^{-iz}}=\frac{1}{i}\frac{e^{2iz}-1}{e^{2iz}+1}=i \frac{1-e^{2iz}}{1+e^{2iz}} $$

and now you can use (1) above

DonAntonio
 

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