1. ## complex analysis

have asked this q to many people and noone is able to help would be much appreciated if anyone could solve this.

(a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
at any point (so neither function is differentiable anywhere).

(b) Consider the function f(x+iy) = p|xy|, where x, y ∈ R. Show that
f satisfies the Cauchy-Riemann equations at the origin, yet f is not
holomorphic at 0.

2. Originally Posted by edgar davids
(a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
at any point (so neither function is differentiable anywhere).
The complex function,
f(z)=z
Can also be expressed in terms of real functions,
f(x+iy)=x+iy
We note the functions x,y are differenciable (Good).
Next we check Cauchy-Riemann Equations:
The partial derivative of x along x is 1
The partail derivative of y along y is 1
So we have,
1=1 (Good)
Next we check the second equation,
The partial derivative of x along y zero,
The negative partial derivative of y along x is zero,
So we have,
0=0 (Good)
Thus, the function,
f(z)=z is homolophric on the entire complex plane. I do not see how you can say it is not.

3. Originally Posted by edgar davids
have asked this q to many people and noone is able to help would be much appreciated if anyone could solve this.

(a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
at any point (so neither function is differentiable anywhere).
Should these be Im(z) and Re(z)?

RonL

4. Originally Posted by edgar davids
have asked this q to many people and noone is able to help would be much appreciated if anyone could solve this.

(a) Verify that Im(z) and z do not satisfy the Cauchy-Riemann equations
at any point (so neither function is differentiable anywhere).
The Cauchy Riemann equations are:

for a complex valued function of a complex variable:

f(z) = u(x,y) + i v(x,y)

where z=x + i y:

du/dx = dv/dy,
du/dy = -dv/dx,

where d here denotes the curly d of partial differentiation.

Now let f(z) = Im(z) = y, then we have u(x,y)=y, v(x,y)=0,

so:

du/dy=1, dv/dx=0,

so du/dy != -dv/dx,

and the Cauchy Reimann equations are not satisfied.

A similar argument can be applied to Re(z).

RonL

5. Originally Posted by edgar davids
(b) Consider the function f(x+iy) = p|xy|, where x, y ∈ R. Show that
f satisfies the Cauchy-Riemann equations at the origin, yet f is not
holomorphic at 0.
(again d denotes the partial differential)

In this case u(x,y)=p|xy| and v(x,y)=0

now

du/dx(0,0) = lim(h->0) [p |h.0| - 0]/h = 0 =dv/dy

and similarly:

du/dy(0,0) = 0 = -dv/dx,

so f satisfies the Cauchy-Riemann equations.

Now f(z) holomorphic at 0 means that f(z) is complex differentiable
in some neighbourhood of 0. A necessary condition to be complex
differentiable at z0, is that f satisfy the Cauchy-Riemann equations
at z0. But our f(z) does not satisfy the Cauchy-Riemann equations
at any point not on the real or imaginary axes (to be shown below),
so f(z) cannot be holomorphic at 0 (because it is not complex differentiable
at some point in every neighbourhood of 0).

------------------------------------------------------------

To show that f(z) does not satisfy the Cauchy-Riemann equations
at a point not on the axes, we can restrict attention to the first

Let x>0, y>0, then:

f(x,y)=p x.y

so u(x,y)=p x.y, v(x,y)=0

Then du/dy= px, and dv/dx=0, that is:

du/dy != -dv/dx

So f(z) does not satisfy the Cauchy-Riemann equations at any point z
in C with Re(z), Im(z) >0

RonL