1. ## Complex Variables

consider the complex function

f(z) = e ^(2z^(2)+2)

let f(x+iy)=u(x,y)+iv(x,y)
calculate u(x,y) and v(x,y)

after substituing z=x+iy I get

e^(2x^(2)-2y^(2)+2) . e^(4xiy)

how do i get rid of e.
if i have done it wrong then plz tell me whats the mistake.plz help

2. to change it to e I think you change it in the form of e^itheta = cos(theta)+isin(theta)

3. so what do u get as u and v

4. Im not sure myself how to follow on, Im hoping someone else could help.

5. ## i have got it.lemme help u ronaldo

f(z) = e^(2z² + 2)

f(x + iy)
= e^(2(x + iy)² + 2)
= e^(2(x² + 2ixy + i²y²) + 2)
= e^(2(x² + 2ixy - y²) + 2)
= e^(2x² + 4ixy - 2y² + 2)
= e^(4ixy)e^(2x² - 2y² + 2)
= [cos(4xy) + isin(4xy)]e^(2x² - 2y² + 2)
= cos(4xy)e^(2x² - 2y² + 2) + isin(4xy)e^(2x² - 2y² + 2)

So:
u(x,y) = cos(4xy)e^(2x² - 2y² + 2)
v(x,y) = sin(4xy)e^(2x² - 2y² + 2)

6. is z^2=x^2+y^2 or x^2-y^2?

7. Originally Posted by usman206
consider the complex function f(z) = e ^(2z^(2)+2)
You notation is very hard to follow. If this is it, then
$\begin{gathered}
2z^2 + 2 = \left( {2x^2 - 2y^2 + 2} \right) + i\left( {4xy} \right) \hfill \\
e^{2z^2 + 2} = e^{\left( {2x^2 - 2y^2 + 2} \right)} \left[ {\cos (4xy) + i\sin (4xy)} \right] \hfill \\
\end{gathered}$