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Prove that $f(z)= (\frac{(Im(z))^3}{|z|^2}) \;\;\;\;\;\ z\neq0\\ \qquad \;\;\;\;\;\;0\qquad \quad z=0\\$

$\quad$ is continuous for all $z_0\in \mathbb{C}$.

I have done this by checking that the limit is the same over two paths, real and imaginary. I think that both paths equal 0. Is this enough to confirm continuity?

I have done the same for

Prove that $f(z)= \frac{\bar{z}}{z} \qquad z\neq0 \\ \qquad \; \; 0 \qquad z=0 \\$

is not continuous at $z_0=0$, but continuous for all other $z_0$,

and have gotten that the real and imaginary paths have different limits when tending to $0$ but i don't know how to show that it is continuous for all other $z_0$.

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  • $\begingroup$ No, you should you consider all possible paths towards zero, not just that coming through the $x$ or $y$ axis, in the first question. Remember the sequential definition of continuity, and note that nothing special is said about the sequence, it could approach the point through any direction it likes. $\endgroup$ – астон вілла олоф мэллбэрг Feb 8 '18 at 6:25
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For the first one, we have $|f(z)|=\dfrac{\left|\text{Im}(z)\right|^{3}}{|z|^{2}}\leq\dfrac{|z|^{3}}{|z|^{2}}=|z|$ for $z\ne 0$. Of course this inequality also satisfies for $z=0$. Anyway we have $\lim_{z\rightarrow 0}|f(z)|=0$ by Squeeze Theorem, then $\lim_{z\rightarrow 0}f(z)=f(0)$.

Considering only two paths to prove existence is not conclusive.

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