Complex Analysis Solutions: Real and Imaginary Parts

[Fabio010]

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:

 

[DonAntonio]

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