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Let $f:\Bbb{C}\to \Bbb{C}$ be defined as

$$f(x+iy)= \frac{x^{3}-y^{3}+i(x^{3}+y^{3})}{x^2+y^2} \text{ if} x+iy \neq 0$$

and $f(x+iy)=0$ if $x+iy=0$

Show that $f$ is continuous at $0$ and it satisfies the Cauchy-Riemann conditions but it is not differentiable

Ok, so I'm getting messed up when trying to check the continuity. Of course I have to check that

$$\lim_{x+iy \to 0}f(x+iy) = 0$$

However, my doubt is the following: I am aware of a result (I have already proved) that says that

$$\lim_{z\to z_0} f(z) = a + ib \text{ iff } \lim_{z\to z_0}\operatorname{Re}(f(z)) = a \text{ and } \lim_{z\to z_0} \operatorname{Im}(f(z)) = b$$

However, with the use of this result (or not), I'm getting mixed up with the $x+iy \to 0$. Does this imply taking $|x+iy|=x^2+y^2 < \epsilon$ for every $\epsilon >0$?

And so, how can I proceed?

Should I encounter this limit by definition or is there another way?

Another User
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Joaquin Liniado
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3 Answers3

2

The following routine calculation proves continuity at $z=0$, switching to polar coordinates at the end:

$\vert f(z)-0\vert ^{2}=\left | \frac{x^{3}-y^{3}+i(x^{3}+y^{3})}{x^2+y^2} \right |^{2}=\frac{(x^{3}-y^{3})^{2}+(x^{3}+y^{3})^{2}}{(x^{2}+y^{2})^{2}}=2\frac{x^{6}+y^{6}}{(x^{2}+y^{2})^{2}}= 2\cdot \frac{2r ^{6}(\cos^6t+\sin^6t)}{4r ^{4}}=r^2(\cos^{6}t+\sin^{6}t)\to 0\ \text{as}\ r\to 0.$

For non-differentiability at zero, we observe that along $y=0$ we have, whenever $x\neq 0$:

$\left | \frac{f(z)}{z} \right |^{2}=\frac{x^{3}(1+i)}{x^{3}}=\left ( 1+i \right )$

and along $x=0$ whenever $y\neq 0$:

$\left | \frac{f(z)}{z} \right |^{2}=\frac{y^{3}(-1+i)}{y^{3}}=\left ( -1+i \right )$

and therefore $\lim _{z\to 0}\frac{f(z)}{z}$ does not exist.

Matematleta
  • 29,139
0

The norm is continuous so $x+iy\to 0$ implies $x^2+y^2\to 0$ or you can just use the result you stated with $x+iy$ and $f$ the identity function.

karmalu
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0

You can use the fact that $$\left| \frac{x^3}{x^2+y^2} \right| = \left|x\right| \left| \frac{x^2}{x^2+y^2} \right| \leq \left|x\right| \\ \left| \frac{y^3}{x^2+y^2} \right| = \left|y\right| \left| \frac{y^2}{x^2+y^2} \right| \leq \left|y\right| \\$$

together with the triangle inequality.

N. S.
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