So I have:
\frac{z^2+i}{z^4-1}=\frac{z^2+i}{(z^2-1)(z^2+1)}=\frac{z^2+i}{(z-1)(z+1)(z-i)(z+i)}
Am I missing something in the numerator?
EDIT: Would multiplying by the numerators conjugate be beneficial?
Wow, can't believe I didn't realize that. It helped me solve a), which I ended up getting to be 6i, but b) cannot be factored (I don't think?).
If it were z^4 + 1 in the denominator then I could, but I'm pretty sure I cannot factor anything in that problem?
Homework Statement
a) \lim_{z\to 3i}\frac{z^2 + 9}{z - 3i}
b) \lim_{z\to i}\frac{z^2 + i}{z^4 - 1}
Homework Equations
?
The Attempt at a Solution
I'm assuming both of these are very, very similar, but I'm not quite sure how to solve them. I would like a method other than using ε...