Algebra Honors (Periods 6 & 7)

Imaginary Numbers

x²= -16

No real number solutions

Imaginary number i us defined to be √-1

√-1   2³   ∑   ∏

I ate some pie!

All imaginary numbers involve i

3i        -5i/2   and 2i√3

i   =√-1

i²  = -1

so

x²= -16  solved over the set of imaginary numbers

x= √-16 = √16∙√-1 = 4√-1 =4i

If r > 0 then √-r = i√r

I = √-1

i² = -1

i³ = i∙i² =i(-1) = -i

i⁴ = i²∙i² = (-1)(-1) = 1