[mathjax]
A complex-valued function \(f(z)\) is said to be a holomorphic function if it is differentiable at every point in its domain.
By the way, let \(f(z)\) be defined as
$$f(z)=u(x,y)+iv(x,y),$$
where \(z\) is a complex variable, \(u(x,y)\) and \(v(x,y)\) are real-number functions, and \(x\) and \(y\) are real variables. The following two equations are collectively called Cauchy-Riemann’s equations.
$$\frac{∂u(x,y)}{∂x}=\frac{∂v(x,y)}{∂y}$$
$$\frac{∂u(x,y)}{∂y}=-\frac{∂v(x,y)}{∂x}$$
You can determine the complex-valued function \(f(z)\) is a holomorphic function when \(u(x, y)\) and \(v(x, y)\) satisfy the Cauchy-Riemann equations above.
Practice
Are the following complex-valued functions \(f(z)\) and \(g(z)\) holomorphic functions?
(i)\(f(z)=z^{3}\)
First, substitute \(z=x+iy\) into \(f(z)=z^{3}\).
\begin{eqnarray}
f(z)&=&z^{3}\\&=&(x+iy)^{3}\\&=&x^{3}+3x^{2}(iy)+3x(iy)^{2}+(iy)^{3}\\&=&x^{3}+3x^{2}(iy)-3xy^{2}-iy^{3}\\&=&(x^{3}-3xy^{2})+i(3x^{2}y-y^{3})･･･(1)
\end{eqnarray}
Compare this equation (1) with \(f(z)=u(x, y)+iv(x, y)\) and you can obtain the following formulas.
$$u(x,y)=x^{3}-3xy^{2}$$
$$v(x,y)=3x^{2}y-y^{3}$$
Calculate the derivative of \ (u (x, y) \) with respect to \(x\) and that of \(v(x, y)\) with respect to \(y\) respectively.
$$\frac{∂u(x,y)}{∂x}=\frac{∂}{∂x} (x^{3}-3xy^{2})=3x^{2}-3y^{2}$$
$$\frac{∂v(x,y)}{∂y}=\frac{∂}{∂y} (3x^{2}y-y^{3})=3x^{2}-3y^{2}$$
Therefore, we get the following equation.
$$\frac{∂u(x,y)}{∂x}=\frac{∂v(x,y)}{∂y}･･･(2)$$
Calculate the derivative of \(u(x, y)\) with respect to \(y\) and that of \(v(x, y)\) with respect to \(x\) respectively.
$$\frac{∂u(x,y)}{∂y}=-6xy$$
$$\frac{∂v(x,y)}{∂x}=6xy$$
Then we get the following equation.
$$\frac{∂u(x,y)}{∂y}=-\frac{∂v(x,y)}{∂x}･･･(3)$$
Equations (2) and (3) are the Cauchy-Riemann’s equations. From the above, \(f(z)=z^{3}\) is a holomorphic function.
(ii)\(g(z)=|z|\)
First, substitute \(z=x+iy\) into \(f(z)=|z|\).
$$f(z)=|x+iy|=\sqrt{x^{2}+y^{2}}･･･(4)$$
Compare this equation (4) with \(f(z)=u(x, y)+iv(x, y)\) and you can obtain the following formulas.
$$u(x,y)=\sqrt{x^{2}+y^{2}}$$
$$v(x,y)=0$$
Then calculate the derivative of \ (u (x, y) \) with respect to \(x\) and that of \(v(x, y)\) with respect to \(y\) respectively.
\begin{eqnarray}
\frac{∂u(x,y)}{∂x}&=&\frac{∂(x^{2}+y^{2})^{\frac{1}{2}}}{∂(x^{2}+y^{2})}･\frac{∂(x^{2}+y^{2})}{∂x}\\&=&\frac{1}{2}(x^{2}+y^{2})^{-\frac{1}{2}}･2x\\&=&\frac{x}{\sqrt{x^{2}+y^{2}}}
\end{eqnarray}
$$\frac{∂u(x,y)}{∂y}=\frac{y}{\sqrt{x^{2}+y^{2}}}$$
\(v(x, y)=0\), so clearly the following holds.
$$\frac{∂v(x,y)}{∂x}=0$$
$$\frac{∂v(x,y)}{∂y}=0$$
To summarize so far, we can say the Cauchy-Riemann’s equations holds if the following two equations are established.
$$\frac{x}{\sqrt{x^{2}+y^{2}}}=0･･･(3)$$
$$\frac{y}{\sqrt{x^{2}+y^{2}}}=0･･･(4)$$
However, these Cauchy-Riemann equations are not possible at the same time.
First, if \(x+y≠0\), it is obviously impossible to simultaneously satisfy both equations. If \(x=0\) and \(y =a \ (constant)\), expression (3) is satisfied, but expression (4) is not satisfied. Then when \(x+y=0\), the denominator of the fraction becomes 0.
In conclusion, \(f(z)=|z|\) does not establish the Cauchy-Riemann equations in any \(z\), and is not a holomorphic function.