Finding the Conjugate Function of u for f to be Analytic

[Dustinsfl]

\(u = 2x(1 - y)\) I want to find v such that \(f = u +iv\) is analytic. The hint is find the conjugate function of u.

I am not sure if what I did was finding the conjugate function of u thoug.
\[
u_x = 2(1 - y) = v_y
\]
so
\[
v = 2y - y^2 + g(x) \Rightarrow v_x = g'(x)
\]
and
\[
u_y = -2x = -g'(x)\Rightarrow g(x) = x^2
\]
Therefore, \(v = 2y - y^2 + x^2\) which makes f analytic. Is that v the conjugate function of u?

 

[Euge]

\(u = 2x(1 - y)\) I want to find v such that \(f = u +iv\) is analytic. The hint is find the conjugate function of u.

I am not sure if what I did was finding the conjugate function of u thoug.
\[
u_x = 2(1 - y) = v_y
\]
so
\[
v = 2y - y^2 + g(x) \Rightarrow v_x = g'(x)
\]
and
\[
u_y = -2x = -g'(x)\Rightarrow g(x) = x^2
\]
Therefore, \(v = 2y - y^2 + x^2\) which makes f analytic. Is that v the conjugate function of u?

What you have is right, but I have one minor comment to make. At the last step, you have $u_y = -2x = -g'(x)\Rightarrow g(x) = x^2$. When you solve for $g$, you should have $g(x) = x^2 + C$, where $C$ is an arbitrary constant. But then you can choose $C = 0$, giving the particular conjugate function $v = 2y - y^2 + x^2$.

 

[Opalg]

I have another minor comment to make. In a problem like this, a good way to check whether your answer is correct is to see whether $u+iv$ can be expressed as a (differentiable) function of $z=x+iy$.

In this case, you started with $u = 2x(1-y)$ and you found that $v = 2y - y^2 + x^2$. Then $$\begin{aligned} u+iv &= (2x - 2xy) + i(2y - y^2 + x^2) \\ &= 2(x+iy) + i(x^2 - y^2 + 2ixy) \\ &= 2z + iz^2.\end{aligned}$$ So $u$ and $v$ are the real and imaginary parts of the analytic function $2z+iz^2$.

As Euge has pointed out, you should really add a constant of integration to the expression for $v$, and the analytic function of $z$ will then be $2z+i(z^2 + C),$ where $C$ is a real constant.