Interval of convergence for series with complex numbers I'm trying to find interval of convergence of this series: $$\sum_{n=1}^{\infty} \frac{7^n(z+2i)^n}{4^n+3^ni}$$ and I should draw a plot which represents the answer, this is what I've got so far: 
Using the root test
$$ \sqrt[n]{\left|\frac{7^n(z+2i)^n}{4^n+3^ni}\right|} = \left|\frac{7(z+2i)}{4}\right|<1$$
$$|(z+2i)|<\frac{4}{7}, \quad z=x+iy$$
$$|(x+i(y+2))|<\frac{4}{7}$$ 
 A: You're on the right track, but a few things need to be mentioned.
First, if the series involves complex numbers then you're actually looking for the disk of convergence, not the interval.  Second, your notation is very inappropriate.  When you apply the root test, you're applying it to $a_n$ only.  Including the summation symbol is incorrect.  Third, your first equality (as it appears as of the time I write this) is technically incorrect, even without the summation symbol.  The $n$th root of the denominator approaches 4 as $n \to +\infty$.  This is different from saying that the $n$th root of the denominator simply equals 4.  A subtle yet important detail.  Actually if you replace $\displaystyle \sum_{n=1}^{+\infty}$ with $\displaystyle \lim_{n\to+\infty}$ then that equality will be correct.
The root test can work but the ratio test may be better, and note that my second point above also applies for the ratio test.
$$
  a_n = \frac{7^n(z+2i)^n}{4^n + 3^ni},
$$
So then
$$
  a_{n+1} = \frac{7^{n+1}(z+2i)^{n+1}}{4^{n+1} + 3^{n+1}i}
          = \frac{7^n(z+2i)^n \cdot 7(z+2i)}{4^{n+1} + 3^{n+1}i},
$$
and we have
$$
\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{7^n(z+2i)^n \cdot 7(z+2i)}{4^{n+1} + 3^{n+1}i} \cdot \frac{4^n + 3^ni}{7^n(z+2i)^n}\right|
 = 7|z+2i| \cdot \frac{\left|4^n + 3^ni\right|}{\left|4^{n+1} + 3^{n+1}i\right|}
$$
Now apply the fact that $|x + iy| = \sqrt{x^2 + y^2}$ to the numerator and denominator, and take the limit as $n \to +\infty$ to end up with the same result you did.
And again, you're looking for a disk, not an interval.  And (the interior of) your disk is given by $|z + 2i| < 4/7$.  As Ivan mentioned, make sure to also check the boundary.
