# Thread: Complex Analysis: Analytic Function

1. ## Complex Analysis: Analytic Function

Here is my question: Let $f=u+iv$ be analytic and suppose that $f(i)=1$. If $u=2x^2+2x+1-2y^2$, find $v$.

My proof so far follows:

By differentiating with respect to x and y, we see that $u_x=4x+2$, $u_y=-4y$, $v_x=4y$, and $v_y=4x+2$. In order to move back to v, I will integrate on $v_x$ and $v_y$ as follows:
$4xy+2y+g(x)+c=v$
$4xy+g(y_+c=v$
$\rightarrow 4xy+2y+0+c=v$

To solve for c, $f=u+iv$ and $f(i)=1$, so $\cdots$

and now this is the really dumb part... I am not sure if I am supposed to substitute this literally as u + i*v, using the u as given and the v that I have derived, substituting (0,1) for i, or do something else entirely. I feel that I am nearly at the solution, but I am losing it here in something that should be quite simple. I will appreciate your help. Thanks.

2. ## Re: Complex Analysis: Analytic Function

Originally Posted by tarheelborn
Here is my question: Let $f=u+iv$ be analytic and suppose that $f(i)=1$. If $u=2x^2+2x+1-2y^2$, find $v$...
The condition $f(i)=1$ means that $u(0,1)=1$ and if is $u=2x^2+2x+1-2y^2$ then $u(0,1)=-1$... contradiction!...

Kind regards

$\chi$ $\sigma$

3. ## Re: Complex Analysis: Analytic Function

But don't I have to put the v in there somewhere, too, since that's what I've had to find? And I'm trying to solve for c? Since f=u+iv to start with, don't they go together? That's where I am getting confused. I have discussed this problem with classmates and nobody mentioned coming up with a contradiction. I know that we are supposed to solve for c and I really thought that the answer was either 2 or 2i, but for the life of me, I can't get there.