Complex variables: Regular function

Question:
Consider $f(z)=u(x,y)+iv(x,y)$. When $u(x,y)=\frac{x}{x^2+y^2-2y+1}$, find $f(z)$ which can be a regular function.

Answer:
According to Cauchy-Riemann equations, if the function is regular, the following must be held:
\[
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
\]
Use the first equation.
\begin{eqnarray}
\frac{\partial u}{\partial x} &=& \frac{\partial}{\partial x}\frac{x}{x^2+(y-1)^2}  \\
    &=& \frac{1\cdot(x^2+(y-1)^2)-x\cdot(2x)}{(x^2+(y-1)^2)^2}   \\
    &=& \frac{-x^2+(y-1)^2}{(x^2+(y-1)^2)^2} = \frac{\partial v}{\partial y}
\end{eqnarray}
Therefore,
\begin{eqnarray}
v &=& \int \frac{-x^2+(y-1)^2}{(x^2+(y-1)^2)^2}dy  \\
    &=& \frac{-(y-1)}{x^2+(y-1)^2}+\varphi(x)
\end{eqnarray}
Now, use the second equation.
\begin{eqnarray}
\frac{\partial}{\partial x}v &=& \frac{2x(y-1)}{(x^2+(y-1)^2)^2}+\varphi'(x)
     &=& -\frac{\partial}{\partial y}u
\end{eqnarray}
Since $-\frac{\partial u}{\partial y}= \frac{2x(y-1)}{(x^2+(y-1)^2)^2}$,
\begin{equation}
\varphi'(x) = 0 \quad \rightarrow \quad \varphi(x) = C \ \mathrm{(constant)}
\end{equation}
Therefore the function $f(z)$ becomes
\begin{eqnarray}
f(z) &=& \frac{x}{x^2+(y-1)^2}+i \left[ \frac{-(y-1)}{x^2+(y-1)^2}+C \right] \\
     &=& \frac{x-iy+i}{x^2+(y-1)^2}+iC  \\
\end{eqnarray}
Now, we substitute the following:
\[
x = \frac{z+z'}{2}, \qquad y = \frac{z-z'}{2i}
\]
where $z=x+iy$ and $z'=x-iy$.
\begin{eqnarray}
f(z) &=& \frac{x-iy+i}{x^2+(y-1)^2}+iC  \\
     &=& \frac{z'+i}{zz'+iz-iz'+1}+iC  \\
     &=& \frac{z'+i}{(z-i)(z'+i)}+iC   \\
\end{eqnarray}
Therefore, the function can be expressed by only $z$.
\begin{equation}
f(z) = \frac{1}{z-i}+iC
\end{equation}

Powered by Hirophysics.com