Find an analytic function from its real part

**Ulysses** · September 23rd 2011, 02:43 PM

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.

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

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.

$u_x=2x+2=v_y \rightarrow v(x,y)=2xy+2y+\Phi(x)$
$-u_y=v_x=2y-1=2y+\Phi'(x)\rightarrow \Phi'(x)=2y-2x-1 \rightarrow \Phi(x)=2xy-x^2-x+K$

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.

**chisigma** · September 23rd 2011, 04:17 PM

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.

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

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.

$u_x=2x+2=v_y \rightarrow v(x,y)=2xy+2y+\Phi(x)$
$-u_y=v_x=2y-1=2y+\Phi'(x)\rightarrow \Phi'(x)=2y-2x-1 \rightarrow \Phi(x)=2xy-x^2-x+K$

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.

Your porcedure is correct!... from the condition...

$-u_{y}=2y-1=v_{x}= 2y+\phi^{'}(x)$ (1)

... You derive immediately that is...

$\phi^{'}(x)= -1 \implies \phi(x)=-x+c$ (2)

... so that is...

$v=2 x y +2 y -x + c$ (3)

Taking into account that $f(i)=0$ You obtain in (3) $c=-2$ ...

Kind regards

$\chi$ $\sigma$

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