What can be said about $\lim_{n\to\infty} z^n$, when $z$ is complex? Can it be expressed in terms of $a,b$ where $z = a + bi$? Is the formula $z^n = r^n(cos(n\theta) + i\sin(n\theta))$ helpful here -- if so, I'm not seeing it yet.
Asked
Active
Viewed 916 times
2
-
Ah, OK, that makes sense. I know it's trivial, but would you like to make your comment an answer so that I can accept it? – bosmacs Oct 15 '11 at 20:21
1 Answers
8
If $|z|<1$, then the limit is $0$. If $z=1$, then the limit is $1$. Otherwise it doesn't exist.
hmakholm left over Monica
- 286,031