Show a limit doesn't exists, complex case

[moxy]

Homework Statement

$f(z) = |z|$
I'm looking to show that $f'(z)$ does not exist for any $z \in ℂ$.

Homework Equations

$f'(z) = \lim_{z_0 → 0}{\frac{f(z) - f(z_0)}{z - z_0}}$

$z = x + iy = Re(z) + i Im(z)$

$|z| = \sqrt{x^2 + y^2}$

The Attempt at a Solution

Clearly I just have to show that $\lim_{z_0 → 0}{\frac{f(z) - f(z_0)}{z - z_0}}$ does not exist. However, I'm confused about how to do this. I'm unsure of how to show that a limit doesn't exist in complex analysis.

I tried to take the limit in two cases, when $Re(z) = Re(z_0)$ and then again when $Im(z) = Im(z_0)$. Is this the correct approach?

 

[Dick]

Have you done the Cauchy-Riemann equations?

 

[moxy]

I was only trying to prove it with limits because it's an exercise in my complex analysis book that comes before the Cauchy-Riemann equations are introduced.

$f(z) = U(x,y) + iV(x,y)$

$f(z) = |z| = |x + iy| = \sqrt{x^2 + y^2}$

$U(x,y) = \sqrt{x^2 + y^2}$
$V(x,y) = 0$

$U_x = 0.5(x^2 + y^2)^{-1/2}(2x) = x(x^2 + y^2)^{-1/2}$
$U_y = y(x^2 + y^2)^{-1/2}$
$V_x = 0$
$V_y = 0$

$U_x ≠ V_y$
$U_y ≠ -V_x$However, does the derivative exist when

$x(x^2 + y^2)^{-1/2} = 0$
$y(x^2 + y^2)^{-1/2} = 0$

i.e. when z = 0?

 

[Dick]

I was only trying to prove it with limits because it's an exercise in my complex analysis book that comes before the Cauchy-Riemann equations are introduced.

$f(z) = U(x,y) + iV(x,y)$

$f(z) = |z| = |x + iy| = \sqrt{x^2 + y^2}$

$U(x,y) = \sqrt{x^2 + y^2}$
$V(x,y) = 0$

$U_x = 0.5(x^2 + y^2)(2x) = x(x^2 + y^2)$
$U_y = y(x^2 + y^2)$
$V_x = 0$
$V_y = 0$

$U_x ≠ V_y$
$U_y ≠ -V_x$


However, does the derivative exist when

$x(x^2 + y^2) = 0$
$y(x^2 + y^2) = 0$

i.e. when z = 0?

U_x=x/sqrt(x^2+y^2). That's not what you wrote. And the derivatives don't exist at (x,y)=(0,0) either, for the same reason the derivative of |x| with respect to x doesn't exist at x=0.

 

[moxy]

Yeah, I forgot the (-1/2) exponents when I typed it out.

Okay, so z = 0 is a "potential" solution after I check the C-R equations, but then when I check it case by case, I see that the derivative doesn't exist when z = 0. Is this a correct?

 

[Dick]

Yeah, I forgot the (-1/2) exponents when I typed it out.

Okay, so z = 0 is a "potential" solution after I check the C-R equations, but then when I check it case by case, I see that the derivative doesn't exist when z = 0. Is this a correct?

Right. U_x and U_y at z=0 have the form 0/0. That doesn't tell you much. But if you look at the details, they don't exist at z=0 either.

 

[moxy]

Thank you! I'm still confused about limits in the complex plane and my original approach to this problem, but I'll save that for another time and another problem :) I really appreciate your help.