Question

Use the Cauchy-Riemann equations to show that \(\displaystyle{f{{\left({z}\right)}}}=\overline{{{z}}}\) is not analytic.

Answer 1

Step 1
A function is said to be analytic when Cauchy-Riemann equations are satisfied.
The Cauchy-Riemann equations are satisfied when \(\displaystyle{u}_{{x}}={v}_{{y}}{\quad\text{and}\quad}{v}_{{x}}=-{u}_{{y}}\).
The given function is \(\displaystyle{f{{\left({z}\right)}}}=\overline{{{z}}}\)
Rewrite the given function as follows.
\(\displaystyle{f{{\left({z}\right)}}}=\overline{{{z}}}\)
=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.
\(\displaystyle{v}_{{x}}=\frac{\partial}{{\partial{x}}}{\left(-{y}\right)}\)
=0
Thus, \(\displaystyle{v}_{{x}}={0}\).
Evaluate \(\displaystyle{v}_{{y}}\) as follows.
\(\displaystyle{v}_{{y}}=\frac{\partial}{{\partial{y}}}{\left(-{y}\right)}\)
=-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.