Can we apply the infinite sum formula if the common ratio is a complex number$?$ 
Suppose there is a series, an infinite series $$a+bi,(a+bi)^2,(a+bi)^3,\cdots$$

Can we apply the formula of $G.P$ to calculate its sum $\frac{a_1}{1-r}?$
Here the common ratio is a complex number. To my knowledge, the above formula can only be applied if $|r|<1$.
I was doing a problem in which we have to calculate limit $n$ tending to $\infty$ of a similar series. There, I mistakenly applied the above formula and got the same answer as given in the solution. But why is this so$?$ The complex number there was $$\frac{1+i\sqrt{5}}{6}$$
Any help is greatly appreciated.
 A: Yes, you can.  The justification is basic algebra.
Letting $(z)$, rather than $(x)$ denote a complex variable, you have that
$$(1 + z + z^2 + \cdots + z^n)(1 - z) = 1 - z^{(n+1)}.$$
So, just as in Real Analysis, the issue is whether
the limit, as $n \to \infty$ of $z^n$ is $(0)$.  Putting this another way, just as in Real Analysis, when you estimate
$$(1 + z + z^2 + \cdots + z^n)$$
by
$$\frac{1}{1 - z},$$
the error in the estimation is
$$\frac{-z^{(n+1)}}{1-z}.$$
Since $z$ is presumably a fixed value, not equal to $[1 + i(0)]$, then the error in the estimation goes to $(0)$ if and only if
$$\lim_{n\to\infty} z^n = 0. \tag1 $$
The one wrinkle is that determining whether (1) above is true is a little tricky.  One way is that (1) is true if and only if
$$\lim_{n\to\infty} |z|^n = 0 \iff |z| < 1. \tag2 $$
The alternative approach is to express $z^n$ as $u(n) + iv(n)$, where $u$ and $v$ are real valued functions whose domain is the positive integers.
Then, you have that (1) above is true if and only if both of the following are true:
$$\lim_{n\to\infty} u(n) = 0, ~~~ \lim_{n\to\infty} v(n) = 0.$$
