Complex Number: $\frac{(3+i)^2}{(1+2i)^2}$ - cannot arrive at textbook solution I have a complex quotient $\frac{(3+i)^2}{(1+2i)^2}$
The solution provided in my textbook is $-2i$. I arrived at different solutions and I'd like to know where I went wrong.
Till now in my textbook chapter I have been working with the complex number i ($\sqrt{-1}$).
I understand that one cannot divide by a complex number in the denominator so I must multiply both the numerator and denominator by the complex conjugate.
However, for this exercise I'm confused since my expression is nested inside parenthesis and is squared. So, if for example my denominator was just $1+2i$ I know that the complex conjugate would be $1-2i$.
So I'm confused about what do do since the whole denominator is within parenthesis and squared.
Just using what I know I tried solving for the squared term in both 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 at $\frac{8}{-3}$ which is not the solution.
How can I arrive at $-2i$?
 A: You accidentally fell into the trap of using The Freshman's Dream.
Exponentiation does not distribute over sums.  In most every context $(a+b)^n$ is not equal to $a^n + b^n$ and in particular $(a+b)^2$ is not simply $a^2+b^2$ but is instead $a^2+2ab+b^2$.  For larger exponents, see the binomial theorem.
In your attempt you tried to calculate the numerator as $(3+i)^2 = 3^2+i^2$ which is incorrect.  Rather, you should remember to use the FOIL method correctly or learn the correct outcome.  Instead, it should have been $(3+i)^2 = 3^2\color{red}{+2\cdot 3\cdot i}+i^2 = 9+6i-1=8+6i$.
You made a similar mistake for the denominator.  Once those mistakes are corrected, you should be able to continue on your own using the method you allude to in your post of multiplying both top and bottom by the complex conjugate of the denominator and simplifying after.
A: Expanding we get $$\frac{9-1+6i}{1-4+4i}=\frac{8+6i}{-3+4i}$$ and now multiply numerator and denominator by $$-3-4i$$
A: It might help to note that this expression is
$$\left(\frac{3+i}{1+2i}\right)^2=\left(1-i\right)^2=-2i$$
assuming that you know how to divide $3+i$ by $1+2i$.
