Question

State Cauchy-Riemann equations. Show that f(z) x*+ iy' is not analytic anywhere but the Cauchy-Riemann equations are satisfied at the origin.

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asked 2021-02-08
State Cauchy-Riemann equations. Show that f(z) x*+ iy' is not analytic anywhere but the Cauchy-Riemann equations are satisfied at the origin.

Answers (1)

2021-02-09
Step 1
Cauchy-Riemann Equations:
A necessary condition that the function f=u+iv is differentiable at a point \(\displaystyle{z}_{{0}}={x}_{{0}}+{i}{y}_{{0}}\) is that the partial derivatives \(\displaystyle{u}_{{x}},{u}_{{y}},{v}_{{x}},{v}_{{y}}\) exists and \(\displaystyle{u}_{{x}}={v}_{{y}},{u}_{{y}}=−{v}_{{x}}\) at the point \(\displaystyle{\left({x}_{{0}},{y}_{{0}}\right)}\)
Step 2
Given equation is f=x+iy
Step 3
Let, f(z)=u(x,y)+iv(x,y)
Comparing,
u(x,y)=x
v(x,y)=y
Step 4
Then,
\(\displaystyle{u}_{{x}}={1}\)
\(\displaystyle{u}_{{y}}={0}\)
\(\displaystyle{v}_{{x}}={0}\)
\(\displaystyle{v}_{{y}}={1}\)
So, \(\displaystyle{u}_{{x}}={v}_{{y}}{\quad\text{and}\quad}{u}_{{y}}=−{v}_{{x}}\) at (0,0).
So, Cauchy-Riemann equations are satisfied at the origin.
Step 5
But the Cauchy-Riemann equations are satisfied only at the point z=0.
Hence, f(z)=x+iy can not have derivative at any point \(\displaystyle{z}\ne{0}\).
So, the given function is not analytic.
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