Induction: Absolute value of complex sequence

[Combinatus]

Homework Statement

Define sequence $$\left[ {z_n} \right]} _{n=1}^{\infty}$$ of points in the complex plane with $$z_1 = 1$$ and $$z_{n+1} = (4+3i)z_n - 1$$. Show that $$|z_n| \ge 4^{n-1}$$ for all $$n \ge 1$$.

Homework Equations

The Attempt at a Solution

The base case of induction is true, since $$|z_1| \ge 4^{1-1} \Leftrightarrow 1 \ge 1$$.

$$z_1 = 1$$
$$z_2 = 3+3i$$
$$z_3 = 2+21i$$
$$z_4 = -56+90i$$
$$z_5 = -2557-717i$$
$$z_6 = -8078-10539i$$
...

I figured I can find a find some $$f(n)$$, I'll perhaps be able to show the inductive step. However, I'm getting nowhere on that approach. Perhaps I am not intuitive enough, but the points seem to be all over the complex plane. Thoughts? Guidance? Ideas?!

 

[Dick]

The absolute value |4+3i|=5. Think about that.

 

[Combinatus]

I slept on it, yet no results were yielded. That usually means the problem is impossible, or retardedly simple. I'm pretty sure I know which one of those it is...

|4+3i|=5 would seem very useful if $$z_{n+1} = (4+3i)z_n$$, but unfortunately, $$z_{n+1} = (4+3i)z_n - 1$$. I don't know what to do with it, since I can't figure out how to treat $$|(4+3i)z_n - 1|$$.

 

[Dick]

It's not retardedly simple, but it's not impossible either. There's a thing called the inverse triangle inequality. |a-b|>=||a|-|b||. Can you find a way to use that as well?