How to square imaginary numbers?

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

1 Answer
Feb 28, 2018

(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