# Thread: show that f'(c) exists

1. ## 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

2. ## Re: show that f'(c) exists

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:

$\displaystyle \frac{f(z)-f(c)}{z-c} = \frac{u+iv}{x+iy} = \frac{(u+iv)(x-iy)}{x^2+y^2}.$

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

3. ## Re: show that f'(c) exists

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

4. ## Re: show that f'(c) exists

Originally Posted by Opalg
To find the real and imaginary components, rationalise the denominator:

$\displaystyle \frac{f(z)-f(c)}{z-c} = \frac{u+iv}{x+iy} = \frac{(u+iv)(x-iy)}{x^2+y^2}.$

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

$\displaystyle \frac{(x-a)[u(x,y)-u(a,b)]+(y-b)[v(x,y)-v(a,b)]+i[(x-a)[v(x,y)-v(a,b)]-(y-b)[u(x,y)+u(a,b)]}{(x-a)^2-(y-b)^2}.$

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

5. ## Re: show that f'(c) exists

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.

6. ## Re: show that f'(c) exists

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.

7. ## Re: show that f'(c) exists

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

$\displaystyle [(f(z)-f(c)][(x-a)+i(y-b)]$

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

8. ## Re: show that f'(c) exists

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.