How to calculate $\lim_{z\rightarrow \frac{1 + i \sqrt{3}}{2}} \frac{z^3 - 1}{z^2 + 2z -1}$

Question

I try to prove that a given function is continuous in the point $\frac{1 + i \sqrt{3}}{2}$, but so far I don't have been successful.

Any hints you could give me, I'm stuck.

Answer 1

With substitution the limit is $$\dfrac{-2}{-\frac12(1-3\sqrt{3}i)}=\dfrac{4}{1-3\sqrt{3}i}$$

Answer 2

If you consider $$y=\frac{z^3 - 1}{z^2 + 2z -1}$$ the denominator does not make any problem since its roots are $-1\pm\sqrt 2$. So, just plug $z=\frac{1 + i \sqrt{3}}{2}$ to get $$y=\frac{-2}{\frac{1}{2} \left(-1+3 i \sqrt{3}\right) }=-\frac{4}{-1+3 i \sqrt{3}}=\frac{1}{7} \left(1+3 i \sqrt{3}\right)$$