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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.