Originally Posted by
artvandalay11 i) $\displaystyle (2 + 3i)(2 - 3i)=2(2)+2(-3i)+3i(2)+3i(-3i)$ we just foiled
$\displaystyle =4-6i+6i-9i^2=4-9(-1)=4+9=13$ and since $\displaystyle i=\sqrt{-1}, i^2=-1$
Should there be parenthesis for the fraction in ii)? otherwise just combine the i terms
iii) $\displaystyle i^5(1 + i^5)$
$\displaystyle i^5=i^4*i=i^2*i^2*i=(-1)(-1)i=i$ so we now have
$\displaystyle i(1+i)=i+i^2=i-1=-1+i$
In general, to evaluate $\displaystyle i^n$ divide n by 4, and take the remainder as the power, for instance $\displaystyle i^{27}$ $\displaystyle \frac{27}{4}=6$ remainder 3, so $\displaystyle i^{27}=i^3=-i$
$\displaystyle i^0=1$
$\displaystyle i^1=i$
$\displaystyle i^2=-1$
$\displaystyle i^3=-i$
$\displaystyle i^4=1$ this is why we divide by 4, the cycle repeats as you see here