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

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" $\frac{\mathbf{鈭�}\mathbf{u}}{\mathbf{鈭�}\mathbf{x}}\mathbf{=}\frac{\mathbf{鈭�}\mathbf{v}}{\mathbf{鈭�}\mathbf{y}}$and $\frac{\mathbf{鈭�}\mathbf{v}}{\mathbf{鈭�}\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{鈭�}\mathbf{u}}{\mathbf{鈭�}\mathbf{y}}$ to be analytic.

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 $v(x,y)=0$.

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

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

Here, $\frac{鈭�u}{鈭�x}鈮�\frac{鈭�v}{鈭�y}$and $\frac{鈭�v}{鈭�x}鈮�-\frac{鈭�u}{鈭�y}$, that is, this function doesn鈥檛 satisfy the Cauchy-Riemann condition

Therefore, the given function is not analytic.