Complex Analysis - find v given u

[hotvette]

Homework Statement

Let f = u + iv be analytic. Find v for given u.

Relevant Equations

\begin{align*}
u &= \frac{x}{x^2+y^2} = x (x^2+y^2)^{-1} \\
\rule{0mm}{18pt} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y}
\qquad\qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
\end{align*}

Solution Attempt:

\begin{align}
\frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} = (x^2+y^2)^{-1} -x (x^2+y^2)^{-2} (2x)
= (x^2+y^2)^{-1} - 2x^2 (x^2+y^2)^{-2}
\\
\rule{0mm}{18pt} \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} = -x (x^2+y^2)^{-2} (2y) = -2xy(x^2+y^2)^{-2}
\\
\rule{0mm}{18pt} \frac{\partial v}{\partial x} &= 2xy(x^2+y^2)^{-2}
\\
\rule{0mm}{18pt} v &= -y(x^2+y^2)^{-1} + g(y)
\\
\rule{0mm}{18pt} \frac{\partial v}{\partial x} &= y(x^2+y^2)^{-2} (2x)
= 2xy(x^2+y^2)^{-2} \qquad \checkmark
\\
\rule{0mm}{18pt} \frac{\partial v}{\partial y} &= -(x^2+y^2)^{-1} + 2y^2 (x^2+y^2)^{-2} + g'(y)
\end{align}
Equating (1) and (6):
\begin{equation}
-(x^2+y^2)^{-1} + 2y^2 (x^2+y^2)^{-2} + g'(y) = (x^2+y^2)^{-1} - 2x^2 (x^2+y^2)^{-2}
\end{equation}

I get stuck at the last step. I can't see any function of y that would satisfy the equation. What am I doing wrong?

 

[dRic2]

$g'(y)=0$?

 

[FactChecker]

I don't know if this is in the spirit of what is expected for solving the problem, but:
$\frac {x}{x^2+y^2}$ is the real part of $\frac {1}{x+iy}$ which is analytic except at $z=0$.
The real and imaginary parts of $\frac {1}{x+iy}$ are easily found by multiplying both numerator and denominator by the conjugate, $x-iy$.

 

[PeroK]

Also, it would have been better if you had simplified:
$$\frac{\partial u}{\partial x} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$$

 

[FactChecker]

I get stuck at the last step. I can't see any function of y that would satisfy the equation. What am I doing wrong?

Oh no! You did everything right and stopped a little too soon. Try $g(y) \equiv 0$.

 

[FactChecker]

Also, it would have been better if you had simplified:
$$\frac{\partial u}{\partial x} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$$

I agree. It is hard to recognize things as done in the OP.
When taking the derivative of a quotient, I like to use the quotient rule rhyme:
If the quotient rule you wish to know,
It's low D-high less high D-low.
Draw a line and down below
The denominator squared must go.

 

[hotvette]

Drat, I see it now. Thanks!

\begin{align*}
\frac{y^2-x^2}{(x^2+y^2)^2} &= \frac{y^2-x^2}{(x^2+y^2)^2} + g'(y) \\
\rule{0mm}{18pt} g'(y) & = 0 \\
\therefore v &= \frac{-y}{x^2+y^2} + C
\end{align*}