I have a complex quotient $ frac {(3 + i) ^ 2} {(1 + 2i) ^ 2} $

The solution in my textbook is $ -2i $, I came to different solutions and would like to know where I went wrong.

So far I've been working with the complex number i in my textbook chapter ($ sqrt {-1} $).

I understand that one can not divide by a complex number in the denominator, so I have to multiply both the numerator and the denominator with the complex conjugate.

However, I am confused in this exercise because my expression is nested in parentheses and is square. For example, if my denominator was straight $ 1 + 2i $ I know that would be the complex conjugate $ 1-2i $,

So I'm confused about what to do because the whole denominator is in parentheses and square.

Only with what I know did I try to solve the square in numerator and denominator:

$ (3 + i) ^ 2 $ = $ 3 ^ 2 + i ^ 2 $ = $ 9-1 $ = $ 8 $

For the denominator:

$ (1 + 2i) ^ 2 $ = $ 1 ^ 2 + 2 ^ 2i ^ 2 $ = $ 1 + 4 * -1 $ = $ 1-4 $ = $ -3 $

Then I would arrive $ frac {8} {- 3} $ That's not the solution.

How do I arrive? $ -2i $?