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 $z$.

$$f(z) = \frac{1}{z-i}+iC$$