Question

How to square imaginary numbers?

Can someone please explain to me how to do question 8b and z^-1 for 9? Thanks!

Answer 1

`(2-i)^2 = 3-4i`

`(3+2i)^(-1) = 3/13-2/13 i`

Explanation:

If `a` and `b` are any numbers, then:

`(a+b)^2 = a^2+2ab+b^2`

So we find:

`(2-i)^2 = 2^2+2(2)(-i)+(-i)^2 = 4-4i-1 = 3-4i`

Note that `z^(-1) = 1/z` and we can rationalise the denominator by multiplying both numerator and denominator by the complex conjugate, like this:

`(3+2i)^(-1) = 1/(3+2i)`

`color(white)((3+2i)^(-1)) = (3-2i)/((3-2i)(3+2i))`

`color(white)((3+2i)^(-1)) = (3-2i)/(3^2-(2i)^2)`

`color(white)((3+2i)^(-1)) = (3-2i)/(9+4)`

`color(white)((3+2i)^(-1)) = (3-2i)/13`

`color(white)((3+2i)^(-1)) = 3/13-2/13 i`