How to square imaginary numbers?
Can someone please explain to me how to do question 8b and z^-1 for 9? Thanks!
Can someone please explain to me how to do question 8b and z^-1 for 9? Thanks!
1 Answer
Explanation:
If
(a+b)^2 = a^2+2ab+b^2(a+b)2=a2+2ab+b2
So we find:
(2-i)^2 = 2^2+2(2)(-i)+(-i)^2 = 4-4i-1 = 3-4i(2−i)2=22+2(2)(−i)+(−i)2=4−4i−1=3−4i
Note that
(3+2i)^(-1) = 1/(3+2i)(3+2i)−1=13+2i
color(white)((3+2i)^(-1)) = (3-2i)/((3-2i)(3+2i))(3+2i)−1=3−2i(3−2i)(3+2i)
color(white)((3+2i)^(-1)) = (3-2i)/(3^2-(2i)^2)(3+2i)−1=3−2i32−(2i)2
color(white)((3+2i)^(-1)) = (3-2i)/(9+4)(3+2i)−1=3−2i9+4
color(white)((3+2i)^(-1)) = (3-2i)/13(3+2i)−1=3−2i13
color(white)((3+2i)^(-1)) = 3/13-2/13 i(3+2i)−1=313−213i