Evaluating $\lim_{n\to\infty}\left(\frac{1-i}{4}\right)^n$

Question

It's been a while since I have taken Calculus II so my experiences on sequences and series has gone down the drain.

I'm trying to find the limit of the sequence $\displaystyle\left(\frac{1-i}{4}\right)^n$ as $n\to \infty$.

I originally was going to try to split everything up, then I got a little worried that I was doing one of those $(x+y)^2 = x^2 + y^2$ phenomena which we all know is not correct. This is where I get stuck.

Thank you.

Answer 1

$$\left(\frac{1-i}{4} \right)^1=\frac{1}{4}-\frac{i}{4}$$

$$ \left(\frac{1-i}{4} \right)^2=-\frac{i}{8}$$

$$\left(\frac{1-i}{4} \right)^3=-\frac{1}{32}-\frac{i}{32}$$

$$\left(\frac{1-i}{4} \right)^4=-\frac{1}{64}$$

$$...$$

$$\lim_{n \to \infty} = 0$$

Answer 2

HINT:

Assuming $i=\sqrt{-1}$

$$\left|\frac{1-i}4\right|=\frac{\sqrt{1^2+(-1)^2}}4=\frac{\sqrt2}4\text{ which is }<1$$

$$\lim_{n\to\infty}a^n=?\text{ for }|a|<1$$

Again, $$(x+y)^2=x^2+y^2\iff 2xy=0\implies$$ at least one of $x,y$ is zero