# Thread: Simple Complex Analysis Question

1. ## Simple Complex Analysis Question

Hi I would just like for someone to check my proof/point out anything I am missing. Thanks!

Prove that the function $f(z)= \sqrt{|xy|}$ is not diff at origin.

Proof:
If it was differentiable at the origin, then $f(z)$ should have the same derivative no matter how you approach the origin. We consider two paths approaching the origin from the x axis (positive side) and the $x=y$ line from the 1st quadrant.

Case 1 (x-axis):
$\lim_{(x,0)\to(0,0)}\frac{f(x,0)-f(0,0)}{x} = \lim_{(x,0)\to(0,0)}\frac{0}{x}$

Case 2 (x=y line):
$\lim_{(x,x)\to(0,0)}\frac{f(x,x)-f(0,0)}{(x,x)} = \lim_{(x,x)\to(0,0)}\frac{x-0}{x+ix}= \lim_{(x,x)\to(0,0)}(1/(1+i))$

Since these two limits are different. The function is not differentiable at the origin.

2. Originally Posted by arsenicbear
Hi I would just like for someone to check my proof/point out anything I am missing. Thanks!

Prove that the function $f(z)= \sqrt{|xy|}$ is not diff at origin.

Proof:
If it was differentiable at the origin, then $f(z)$ should have the same derivative no matter how you approach the origin. We consider two paths approaching the origin from the x axis (positive side) and the $x=y$ line from the 1st quadrant.

Case 1 (x-axis):
$\lim_{(x,0)\to(0,0)}\frac{f(x,0)-f(0,0)}{x} = \lim_{(x,0)\to(0,0)}\frac{0}{x}$

Case 2 (x=y line):
$\lim_{(x,x)\to(0,0)}\frac{f(x,x)-f(0,0)}{(x,x)} = \lim_{(x,x)\to(0,0)}\frac{x-0}{x+ix}= \lim_{(x,x)\to(0,0)}(\sqrt{2}/2-\sqrt{2}/2*i)$

Since these two limits are different. The function is not differentiable at the origin.
Look at that last limit again, but yeah your method works.

3. Originally Posted by arsenicbear
Hi I would just like for someone to check my proof/point out anything I am missing. Thanks!

Prove that the function $f(z)= \sqrt{|xy|}$ is not diff at origin.

Proof:
If it was differentiable at the origin, then $f(z)$ should have the same derivative no matter how you approach the origin. We consider two paths approaching the origin from the x axis (positive side) and the $x=y$ line from the 1st quadrant.

Case 1 (x-axis):
$\lim_{(x,0)\to(0,0)}\frac{f(x,0)-f(0,0)}{x} = \lim_{(x,0)\to(0,0)}\frac{0}{x}$

Case 2 (x=y line):
$\lim_{(x,x)\to(0,0)}\frac{f(x,x)-f(0,0)}{(x,x)} = \lim_{(x,x)\to(0,0)}\frac{x-0}{x+ix}= \lim_{(x,x)\to(0,0)}(1/(1+i))$

Since these two limits are different. The function is not differentiable at the origin.
To make it easier on yourself and assume it's diff. and use the chain rule to show it's composition with another function (which will simplify things) is diff. but it won't be.

Spoiler:
$z^2$