Question

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

$\overline{\mathbf{z}}$

Short answer

The given function$\overline{z}$ is not analytic.

Step-by-step solution

Given information

The given function is $\overline{z}$.

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

$\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{x}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{v}}{\mathbf{\partial }\mathbf{y}}$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 $\overline{z}$gives:

$\overline{z}=\overline{x+iy}$

It is known that, the conjugate of $z=x+iy$is given by $\overline{z}=x+iy$. So, the above equation is written as follows:

$\overline{z}=x+iy$

The above equation is in the form of $f\left(z\right)=f(x+iy)=u(x,y)+iv(x,y)$such that role="math" localid="1653391064229" $u(x,y)=x$and $v(x,y)=-y$.

Hence, the real part of the given function is$u(x,y)=x$ and the imaginary part is$v(x,y)=-y$.

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.

$$\begin{gathered} \frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left(x\right) \\ =1 \\ \frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\left(y\right) \\ =0 \\ \frac{\partial v}{\partial x}=\frac{\partial }{\partial x}(-y) \\ =0 \\ \frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(-y) \\ =-1 \end{gathered}$$

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

Therefore, the given function is not analytic.