Use the Cauchy-Riemann equations to show that f(z)=bar(z) is not analytic.

Step 1
A function is said to be analytic when Cauchy-Riemann equations are satisfied.
The Cauchy-Riemann equations are satisfied when ${\displaystyle u_x=v_y\text{and}{v}_x=-u_y}$.
The given function is ${\displaystyle f\left(z\right)=\bar{z}}$
Rewrite the given function as follows.
${\displaystyle f\left(z\right)=\bar{z}}$
=x-iy
Here u(x,y)=x and v(x,y)=-y.
Step 2
Evaluate ${\displaystyle u_x}$ as follows.
${\displaystyle u_x=\frac{\partial }{\partial x}\left(x\right)}$
=1
Thus, ${\displaystyle u_x=1}$.
Evaluate ${\displaystyle u_y}$ as follows.
${\displaystyle u_y=\frac{\partial }{\partial y}\left(x\right)}$
=0
Thus, ${\displaystyle u_y=0}$.
Step 3
Evaluate ${\displaystyle v_x}$ as follows.
${\displaystyle v_x=\frac{\partial }{\partial x}(-y)}$
=0
Thus, ${\displaystyle v_x=0}$.
Evaluate ${\displaystyle v_y}$ as follows.
${\displaystyle v_y=\frac{\partial }{\partial y}(-y)}$
=-1
Thus, ${\displaystyle v_y=-1}$.
Clearly, ${\displaystyle u_x\ne v_y}$. So, the function did not satisfy Cauchy-Riemann equations.
Therefore, the given function is not analytic.