Thread: show that f'(c) exists

let f(z) = u + iv, z = x + iy
let u and v be defined as follows:
u(x,y) = (x^3 - y^3) / (x^2 + y^2) if (x,y)≠(0,0), u(0,0)=0
v(x,y) = (x^3 + y^3) / (x^2 + y^2) if (x,y)≠(0,0), v(0,0)=0

solve lim [f(z) - f(c)] / (z-c) as z approaches c. that is, show f'(c) exists

our instructor told us a hint: solve it component-wise

the thing is, I do not know how to apply that in such a complex function please help me.. teach me some sorcery

edit: btw, this limit is defined to be the derivative

Originally Posted by StillInUse

let f(z) = u + iv, z = x + iy
let u and v be defined as follows:
u(x,y) = (x^3 - y^3) / (x^2 + y^2) if (x,y)≠(0,0), u(0,0)=0
v(x,y) = (x^3 + y^3) / (x^2 + y^2) if (x,y)≠(0,0), v(0,0)=0

solve lim [f(z) - f(c)] / (z-c) as z approaches c. that is, show f'(c) exists

our instructor told us a hint: solve it component-wise

the thing is, I do not know how to apply that in such a complex function please help me.. teach me some sorcery

edit: btw, this limit is defined to be the derivative

To find the real and imaginary components, rationalise the denominator:



Now substitute the formulas for u and v into the fraction ... .

uhm could you kindly explain what happened to f(c)?

Originally Posted by Opalg

To find the real and imaginary components, rationalise the denominator:



Now substitute the formulas for u and v into the fraction ... .

Hello Sir Opalg i take it that what you gave me is a hint..

after multiplying a conjugate of the denominator and then some further algebraic manipulations I ended up with these bunch of letters



is this, by any chance, going towards the right direction? i still do not know how to take the limit of this function though

Originally Posted by StillInUse

uhm could you kindly explain what happened to f(c)?

Good point! I also left the c out of the denominator. I must have been subconsciously assuming that c=0. In fact, I think that c probably should be 0, because actually f(z) is not differentiable at any other point.

Originally Posted by Opalg

Good point! I also left the c out of the denominator. I must have been subconsciously assuming that c=0. In fact, I think that c probably should be 0, because actually f(z) is not differentiable at any other point.

f'(c) exists if the Cauchy-Riemann conditions are satisfied at c. Look them up. You have to compute some partial derivatives on u and v.

Originally Posted by Hartlw

f'(c) exists if the Cauchy-Riemann conditions are satisfied at c. Look them up. You have to compute some partial derivatives on u and v.

exactly. now, the dilemma here is how can i prove that f'(c) exist? i mean, does rationalizing the denominator then separating the real and imaginary parts of



enough of a reason to say that the limit as z approaches to c do exist?

Originally Posted by Opalg

Good point! I also left the c out of the denominator. I must have been subconsciously assuming that c=0. In fact, I think that c probably should be 0, because actually f(z) is not differentiable at any other point.

By C-R f'(z) doesn't exist anywhere except possibly at z=0.

If you let x^2 + y^2 = r^2 and note |x|<r and |y|<r, you can show limit of u and v and their derivatives approach 0 as r approaches 0. If you then define u and v and their drivatives = 0 at z=0 they are continuous at z=0 and C-R is satisfied at z=0 so that f'(0)=0.