just a quick question on expressing $\displaystyle (1 + i) ^3$ in the form x + iy
the answer i came to was:
$\displaystyle 2 + 4i + i^2 + i^2
= 2 + 4i -1 -1
= 0 + 4i$

just a quick question on expressing $\displaystyle (1 + i) ^3$ in the form x + iy
the answer i came to was:
$\displaystyle 2 + 4i + i^2 + i^2
= 2 + 4i -1 -1
= 0 + 4i$

however im not sure if this is correct...

I am not sure how you expanded the above but it can be treated like a polynomial in i.

Using the binomial theorem or by the distributive law you get