calculatimg $\lim_{z\to0}\left(\frac{\Re(z^{3})}{|z|^{2}}\right)$ I'm trying to solve the limit of $z$ where $z \in \mathbb{C} $

$$
\lim_{z\to0}\left(\dfrac{\Re(z^{3})}{|z|^{2}}\right)
$$
What you would use is that $|\Re z|\le|z|$ . Therefore

$$|\Re (z^3)| \le |z^3| = |z|^3 $$

So you have that:

$$\left|{\Re (z^3)\over |z|^2}\right|\le |z|$$

which will imply that $\lim_{z\to0}\Re(z^3)/|z|^2 = 0$.

You are adviced to remember the definition of limit, that for every $\epsilon&gt;0$ there exists an $\delta$ such that $|f(z)-L|&lt;\epsilon$ whenever $0&lt;|z-a|&lt;\delta$, here we have that $|\Re(z^3)/|z|^2-0|&lt;\epsilon$ whenever $0&lt;|z-0|&lt;\delta=\epsilon$.

The approach of using limits along lines is bad because limit along straigh lines doesn't imply that the limit exists as defined above. For example $f(x+iy) = x/y$ (except on $y$ axis where $f(x) = 0)$. Along straight lines it aproaches zero, but not as a limit in $\mathbb C$ (for example look at what happens if you aproach zero along the parabola $y=x^2$).
first of all we know that : $z^{n}=r^{n}cis(n\theta)$ also $|z|=r=\sqrt{x^2+y^2}$
and $\Re(z)=rcos\theta ,\Im(z)=rsin\theta$.

depending on what is above we have :

$\lim_{z\to0}(\dfrac{\Re(z^{3})}{|z|^{2}}) = \lim_{r\to0}(\dfrac{r^3cos3\theta}{r^{2}})=\lim_{r\to0}(rcos3\theta)=0
$