1. ## x + iy

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

however im not sure if this is correct...

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

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

$(1+i)^3=\sum_{n=0}^{3}\binom{3}{n}1^{3-n}i^{n}=1+3i+3i^2+i^3=1+3i-3-i=-2+2i$