Question

Complex Variables:

Find the Limit using L'Hopital's Rule:

Answer 1

\[\lim_{z -> i}\frac{ z^9 + z - 2i }{ z^{15} + 1 }\]

not sure how to interpret it since it has that -2i.

Answer 2

Woahhh, Quebec friend, your problem is tough, :)

Answer 3

Did you try using DeMoivre?

Answer 4

i haven't tried

Answer 5

Because you should try evaluating it first, for \(z \sim i \)

Answer 6

\(i = \cos (90°) + i \sin (90°)\)

Answer 7

actually i figured it out but i'm not sure if this is correct. do l'hopital's once.

get:\[\lim_{z->i} \frac{ 9z^8 + 1 }{ 15z^{14} }\]

plugging in z = i gives me the textbook's answer

Answer 8

if it's a correct way to do it**