Question:

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

Answer:

According to Cauchy-Riemann equations, if the function is regular, the following equations must be held:

\[   \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}   \]

The first equation of the above will give the following:

\[  \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\frac{x}{x^2+(y-1)^2}  \]

\[   = \frac{1\cdot(x^2+(y-1)^2)-x\cdot(2x)}{(x^2+(y-1)^2)^2}   \]

\[    = \frac{-x^2+(y-1)^2}{(x^2+(y-1)^2)^2} = \frac{\partial v}{\partial y}  \]

Therefore, we can obtain \(v\) by integrating the above.

\[  v = \int \frac{-x^2+(y-1)^2}{(x^2+(y-1)^2)^2}dy  \]

\[    = \frac{-(y-1)}{x^2+(y-1)^2}+\varphi(x)  \]

Now, take the derivative in accordance with the second equation.

\[  \frac{\partial}{\partial x}v = \frac{2x(y-1)}{(x^2+(y-1)^2)^2}+\varphi'(x)   \]

\[     = -\frac{\partial}{\partial y}u  \]

Since \(-\frac{\partial u}{\partial y}= \frac{2x(y-1)}{(x^2+(y-1)^2)^2}\), we have

\[  \varphi'(x) = 0 \quad \rightarrow \quad \varphi(x) = C \ \mathrm{(constant)}  \]

Therefore the function \(f(z)\) becomes

\[  f(z) = \frac{x}{x^2+(y-1)^2}+i \left[ \frac{-(y-1)}{x^2+(y-1)^2}+C \right] \]

\[     = \frac{x-iy+i}{x^2+(y-1)^2}+iC  \]

Now, we substitute the following:

\[  x = \frac{z+z’}{2}, \qquad y = \frac{z-z’}{2i}  \]

where \(z=x+iy\) and \(z’=x-iy\).

\[  f(z) = \frac{x-iy+i}{x^2+(y-1)^2}+iC  \]

\[     = \frac{z’+i}{zz’+iz-iz’+1}+iC  \]

\[     = \frac{z’+i}{(z-i)(z’+i)}+iC   \]

The can be a regular function, which is expressed by only \(z\).

\[  f(z) = \frac{1}{z-i}+iC  \]