Is there limitation when applying binomial theorem？

Question

Problem as title showed. $(a+b)^{-n}$. If $n$ is a positive integer. Can $a$ or $b$ be a complex number? Many thanks in advance.

Answer 1

If $\left|a/b \right|<1$, you take $b$ out and find $$ (a+b)^{-n} = b^{-n}(1+(a/b))^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} b^{-n} \left(\frac{a}{b}\right)^k = \sum_{k=0}^{\infty} \binom{-n}{k} b^{-n-k} a^k. $$ If $\left|a/b \right|>1$, swap $a$ and $b$ in the above expression.

If $\left|a/b \right|=1$, you can't normally say anything in general: consider @Alex's example, for one.

Answer 2

An example: $(1-x)^{-1} = \sum_{k=0}^{\infty} x^k$ iff $|x|<1$. You should be careful when the sum is infinite, as it may diverge. So yeah, there are restrictions on $a$ and $b$.