How to simplify $\frac{(5+7i)}{(2+i)}*\left(\frac{(1+2i)}{(3+4i)}\right)^{-2}$

Question

I´d like to simplify the following term. But there is something I haven´t understood yet on how to deal with the negativ exponential $^{-2}$.

$$\frac{(5+7i)}{(2+i)}*\left(\frac{(1+2i)}{(3+4i)}\right)^{-2}$$

First I tried to divide both terms hence: $$\frac{(5+7i)}{(2+i)} = \frac{(17+9i)}{5}$$ and $$\frac{(1+2i)}{(3+4i)} = \frac{(11+2i)}{25}$$ therefore $$\frac{(17+9i)}{5}*\frac{1}{\left(\frac{11+2i}{25}\right)^2}$$

But this looks weird to me. First I don´t know if this is even the right approach? My next step would be to square the bracket and then simplify further but my results seem to be wrong. Did I overlook some important rules? I´m thankfull for any hints!

Answer 1

Just rationalize, compute the inverse of the square and rationalize again.

At the end you will get

$$\frac{477}{25}+\frac{61 i}{25}$$

Note

$$\left(\frac{A}{B}\right)^{-2} \equiv \frac{B^2}{A^2}$$

Answer 2

we get $$\left(\frac{11+2i}{25}\right)^2=\frac{117+44i}{625}$$ the result should be $${\frac{477}{25}}+{\frac {61\,i}{25}}$$