If $z=(6-2i)(4-7i)$, find $z^2$

Question

If $z=(6-2i)(4-7i)$, find $z^2$

I got
$$- 5.5 - 4.5i \quad\text{for}\; n=0$$ $$\phantom{-}5.5 + 4.5i \quad\text{for}\; n=1$$

Answer required in polar and rectangular form.

Answer 1

To summarize comments:

You found correctly that $(6-2i)(4-7i)=10-50i$

Now, if we are given that $z=10-50i$ and we are interested in finding what the value of $z^2$ is, then we want to square $z$ which would give us $(10-50i)^2 = -2400-1000i$, found by just expanding the product using the usual "FOIL" method, or however you are comfortable expanding products.

On the other hand, if we are given that $z=10-50i$ and we are interested in finding what the values of $z^{1/2}$ are, that is when we would square root $z$. To do so, this is when you might invoke DeMoivre's theorem, and the results would be approximately $5.52 - 4.52 i$ and $-5.52+4.52i$. You seem to have missed a negative sign somewhere in your attempt.