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I'm stuck on the following problem:

The function $f(z)=|z|^2$ is:

  1. Differentiable only at the Origin

  2. Not differentiable anywhere

I have to determine which of the aforementioned options is true. The answer key to the problem says option 2 is true whereas I think option 1 is correct.

We see that: $$ f(z)=|z|^2 \implies u(x,y)+iv(x,y)=x^2+y^2,\;\text{where}\:z=x+iy,\:\text{say}. $$ Then: $$ u(x,y)=x^2+y^2,v(x,y)=0.\;\text{So at the Origin}\;u_x=u_y=v_x=v_y=0 $$ So, C-R equation is satisfied, and hence option 1 holds true.

Am I going in the right direction?

TheVal
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learner
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  • You're right, it is differntiable at the origin –  Aug 11 '13 at 16:13
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    How is differentiability defined? If (complex) differentiability in $z_0$ is defined by the existence of the limit $\lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}$ (as it usually is), then $\lvert z\rvert^2$ is differentiable in $0$. – Daniel Fischer Aug 11 '13 at 16:13
  • That is what I did. So,option 1 is indeed correct. Am I right,sir? – learner Aug 11 '13 at 16:16
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    Well, not if the course uses a definition of differentiability that says that $\lvert z\rvert^2$ isn't differentiable in $0$, e.g. that the function is real differentiable and satisfies the CR equations in a neighbourhood. That would be odd, but I have seen people use odd definitions. – Daniel Fischer Aug 11 '13 at 16:28
  • They might define differentiable functions on open sets , but I doubt this is the case ! – Zaid Alyafeai Aug 11 '13 at 19:44

1 Answers1

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Cauchy-Riemann equations are sufficient for differentiability only if they hold in some open subset of the complex plane.

You need to either use the definition of differentiability or alternatively check that $u$ and $v$ are real differentiable in addition to satisfying C-R. I would do the former in case you haven't covered the sufficiency of the latter in class. The latter does follow from the continuity of the partial derivatives of $u$ an $v$.

It does look like you get differentiability at the origin.

Jyrki Lahtonen
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