Mylo O'Moore

2021-02-08

State Cauchy-Riemann equations. Show that f(z) x*+ iy

mhalmantus

Skilled2021-02-09Added 105 answers

Step 1

Cauchy-Riemann Equations:

A necessary condition that the function f=u+iv is differentiable at a point$z}_{0}={x}_{0}+i{y}_{0$ is that the partial derivatives $u}_{x},{u}_{y},{v}_{x},{v}_{y$ exists and $u}_{x}={v}_{y},{u}_{y}=-{v}_{x$ at the point $({x}_{0},{y}_{0})$

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,

${u}_{x}=1$

${u}_{y}=0$

${v}_{x}=0$

${v}_{y}=1$

So,$u}_{x}={v}_{y}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{u}_{y}=-{v}_{x$ 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$z\ne 0$ .

So, the given function is not analytic.

Cauchy-Riemann Equations:

A necessary condition that the function f=u+iv is differentiable at a point

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,

So,

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

So, the given function is not analytic.