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Cauchy-Reimann from first principles

  • #1

Homework Statement



Prove from first principles that f(z) = [itex]\overline{z}^2[/itex] is not differentiable at any point z ≠ 0

Homework Equations





The Attempt at a Solution



So i guess i Have to show [itex]\stackrel{lim}{h\rightarrow0} \frac{f(z+h)-f(z)}{h}[/itex] is equal to zero right?

[itex] \frac{\overline{z+h}^2-\overline{z}^2}{h}[/itex] but not sure where to go from here.
Do i use z=(x+iy) then [itex]\overline{z} = x-iy [/itex] and so [itex]\overline{z}^2[/itex] = [itex]x^2-y^2-i2xy [/itex]
 

Answers and Replies

  • #2
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i Have to show [itex]\stackrel{lim}{h\rightarrow0} \frac{f(z+h)-f(z)}{h}[/itex] is equal to zero right?
No, you need to prove that it does not exist. Start by formulating "limit does not exist" in the epsilon-delta language.
 
  • #3
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You could assume the limit exists (and hence f is differentiable), and calculate it twice using two sequences which converge to zero, and find that you get two different limits. In this case, your assumption becomes false, so f isn't differentiable.

You do know that the definition of differentiability you have is equivalent to having the limit
[itex]\lim_{n\rightarrow\infty}\frac{f(z+x_n)-f(z)}{x_n}[/itex]
exist for every (complex!) sequence [itex]x_n\rightarrow 0[/itex], right?

In this case, I recommend you work with the definition you have, and see what happens when you approach zero along these two sequences:
[itex] x_n=1/n[/itex]
[itex]y_n=i/n[/itex].

In general, such an approach is usually the best way to show that something -isn't- differentiable. Unless of course you have the Cauchy-Riemann equations :)
 
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