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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.