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

My proof so far follows:

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

$\displaystyle 4xy+2y+g(x)+c=v$

$\displaystyle 4xy+g(y_+c=v$

$\displaystyle \rightarrow 4xy+2y+0+c=v$

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

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.