Question

Use the Cauchy-Riemann conditions to find out whether the functions in Problems 1.1 to 1.21 are analytic.

$\left|z\right|$

Short answer

The given function $\left|z\right|$is not analytic.

Step-by-step solution

Given information

The given function is $\left|z\right|$.

Concept of Cauchy-Riemann conditions

For the complex function $\mathit{f}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{+}\mathit{i}\mathit{y}\mathbf{)}\mathbf{=}\mathit{u}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{+}\mathit{i}\mathit{v}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$, where, $\mathit{u}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$is the real part and $\mathit{v}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$is the imaginary part, the conditions are

role="math" localid="1653393713166" $\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{y}}$and $\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{y}}$ to be analytic.

Substitute the value

Substitute $z=x+iy$in $\left|z\right|$as follows:

$\left|z\right|=\left|x+iy\right|$

It is known that, the modulus of a complex number $z=x+iy$is given as,

$\left|z\right|=\sqrt{x^2+y^2}$

So, $\left|z\right|=\left|z+iy\right|$can be further written as follows:

$\left|z\right|=\sqrt{x^2+y^2}+i.0$

Hence, the real part of the given function is $u(x,y)=\sqrt{x^2+y^2}$and the imaginary part is $v(x,y)=0$.

Apply Cauchy-Riemann conditions

Substitute the values of $u$and $v$in $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$and $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}$and simplify.

$$\begin{gathered} \frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(\sqrt{x^2+y^2}\right) \\ =\frac{x}{\sqrt{x^2+y^2}} \\ \frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left(\sqrt{x^2+y^2}\right) \\ ==\frac{y}{\sqrt{x^2+y^2}} \\ \frac{\partial v}{\partial x}=\frac{\partial }{\partial x}\left(0\right) \\ =0 \\ \frac{\partial v}{\partial y}=\frac{\partial }{\partial y}\left(0\right) \\ =0 \end{gathered}$$

Here, $\frac{\partial u}{\partial x}\ne \frac{\partial v}{\partial y}$and $\frac{\partial v}{\partial x}\ne -\frac{\partial u}{\partial y}$, that is, this function doesn’t satisfy the Cauchy-Riemann condition

Therefore, the given function is not analytic.