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(2i)2=34i

(3+2i)^(-1) = 3/13-2/13 i(3+2i)1=313213i

Explanation:

If aa and bb are any numbers, then:

(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(2i)2=22+2(2)(i)+(i)2=44i1=34i

Note that z^(-1) = 1/zz1=1z and we can rationalise the denominator by multiplying both numerator and denominator by the complex conjugate, like this:

(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=32i(32i)(3+2i)

color(white)((3+2i)^(-1)) = (3-2i)/(3^2-(2i)^2)(3+2i)1=32i32(2i)2

color(white)((3+2i)^(-1)) = (3-2i)/(9+4)(3+2i)1=32i9+4

color(white)((3+2i)^(-1)) = (3-2i)/13(3+2i)1=32i13

color(white)((3+2i)^(-1)) = 3/13-2/13 i(3+2i)1=313213i