complex analysis: differentiability of a complex function?

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February 16th 2011, 01:12 AM
chocaholic

complex analysis: differentiability of a complex function?

Hi I'm having some trouble with showing a function is analytic at a point. If someone could help me with these questions I'd appreciate it.

(a) Find all points where the function f(z)=(x^3+y^3+3y) + i(y^3-x^3+3y) is differentiable and compute the derivative at those points.
my thoughts:
using the cauchy riemann formulae i found that this funtion is differentiable at all points z=x+iy where x^2=y^2+1 or y^2=x^2-1 which, by taking the square root on both sides, is undefined at x=0 and hence the derivative of f at a point z=x+i(sqrt(x^2-1)) is f'(x+i(sqrt(x^2-1)))=3x^2-i3x^2. Is this correct? If not, then where am I going wrong.

I'm having more difficulty with this next question:
(b)Is the function in (a) analytic at any point? Justify your answer clearly.
my thoughts:
by the definition:A function f(z) is said to be analytic at a point z0 if there exists some ε-spherical neighborhood of z0 at all points of which f '(z) exists. I'm having a hard time using this definition to show that it is analytic at all points z=x+iy where x^2=y^2+1 or y^2=x^2-1 or am I wrong in thinking that it's analytic at these points? I just don't know how to approach this and reason it out.

Thanks in advance for any help.
February 16th 2011, 01:47 AM
Prove It

Remember that the Cauchy Riemann equations are

$\displaystyle \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ AND $\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ .

I think you'll find that you have evaluated the second of these equations incorrectly...
February 16th 2011, 02:55 AM
chocaholic

I've done it over a couple of times and I still come up with the same answer
{partial derivative u}/{partial derivative y} = 3y^2+3
-{partial derivative v}/{partial derivative x} = -(-3x^2)= 3x^2

hence 3y^2+3=3x^2 and so y^2+1=x^2

where am I going wrong?
February 16th 2011, 03:25 AM
DrSteve

It doesn't look like you're doing anything wrong. The function is differentiable on the hyperbola $x^2-y^2=1$ .
February 16th 2011, 03:28 AM
Prove It

Quote:

Originally Posted by chocaholic

I've done it over a couple of times and I still come up with the same answer
{partial derivative u}/{partial derivative y} = 3y^2+3
-{partial derivative v}/{partial derivative x} = -(-3x^2)= 3x^2

hence 3y^2+3=3x^2 and so y^2+1=x^2

where am I going wrong?

Sorry, you are right, I misread what you wrote as you having found two separate equations that could satisfy the CR equations...
February 16th 2011, 03:34 AM
chocaholic

O ok sorry I should have stated that I was simplifying. So the differentiability part is correct but what about the point at which it is analytic?
February 16th 2011, 08:33 AM
FernandoRevilla

Quote:

Originally Posted by chocaholic

by the definition:A function f(z) is said to be analytic at a point z0 if there exists some ε-spherical neighborhood of z0 at all points of which f '(z) exists.

As DrSteve said, there exists $f'(z)$ iff $z=x+iy$ satisfies $H\equiv x^2-y^2=1$ so, there is no $\epsilon$ -spherical neighborhood of $z_0\in H$ contained in $H$ . Hence, $f$ is nowhere analytic.

Fernando Revilla