I have to evaluate the following limit:
$$\lim_{z\rightarrow-i}\frac{z^4}{z^3+1}$$
I found that just trying to plug in the point I get:
$$\lim_{z\rightarrow-i}\frac{z^4}{z^3+1} = \frac{(-i)^4}{(-i)^3+1} = \frac{1}{i+i} = \frac{1}{2i}.$$
Is this right? It feels like it shouldn't be this simple.
$\endgroup$ 22 Answers
$\begingroup$Looks like your $+1$ in the denominator became a $+i$. Fix that to get the correct answer of $\dfrac12 - \dfrac12i$. But otherwise, yes, it's that simple. But it won't always be that simple.
$\endgroup$ 3 $\begingroup$To highlight the previous points made:
$$\lim_{z\rightarrow-i}\frac{z^4}{z^3+1} = \frac{(-i)^4}{(-i)^3+1} = \frac{1}{i+1} =\frac{1}{i+1}(\frac{-i+1}{-i+1})= \frac{-i+1}{2}=\frac{-1}{2}i+\frac{1}{2}$$
You just needed to multiply by the conjugate and to give you the real and imaginary parts.
$\endgroup$ 3
