Originally Posted by

**Ulysses** Hi there. I have two doubts on this exercise. The exercise gives me the real part u(x,y), and I have to find from it the imaginary part, such that the function I get accomplishes the given condition, and its an analytic function.

$\displaystyle u(x,y)=x^2-y^2+2x+y,f(i)=0$

The first doubt I got I think its quiet silly. I don't know how I should evaluate f(i). I hoped to have there two coordinates, for x and y. But I have just i. I don't know if I should took that as a complex number expressed in its polar form, does it mean that x=0 and y=i?

On the other hand I've tried to find v(x,y), the real part it asks me for.

So I used the Cauchy-Riemann conditions.

$\displaystyle u_x=2x+2=v_y \rightarrow v(x,y)=2xy+2y+\Phi(x)$

$\displaystyle -u_y=v_x=2y-1=2y+\Phi'(x)\rightarrow \Phi'(x)=2y-2x-1 \rightarrow \Phi(x)=2xy-x^2-x+K$

So there is something wrong in there, the function I get for Phi depends on y too. Whats wrong with this? I couldn't find the analytic function under this procedure. Did I do something wrong or is it that it doesn't exist an analytic function for the given u(x,y)?

Bye and thanks.

PD: Ok, I realized about the first part, it was really silly, it just means f(z)=f(i), z=i.