- #1

- 996

- 5

- 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?

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