Complex Number 5-9

i = Imaginary Number i 2 = i 3 =i 2 i = -1*i = -i i 4 =i 2 i 2 = -1*-1 = 1 i 5 =i 4 i= 1*i= i i 6 =i 4 i 2 = 1*-1=-1 i 7 =i 4 i 3 = 1*-i=-i i 8 =i 4 i 4 = 1*1=1 Div. exp. by 4; Remainder 1  i; 2  -1; 3  –i 0  1 i 58 R: 2 so -1 7i 23 R:3 so 7(-i)=-7i 10i 13 5i 96 10i *5=50i 50i 109 R1 = 50i

Simplifying Negative Radicals i’s are only for square roots Take the i’s out first! Mult. Radicals only for pos. radicals

Adding, Subtracting, Multiply i’s a+bi Follow rules of exponents; i 2 = -1 (3 + i) – (4 + 5i)(3 + i) (4 + 5i) (3 + 4i) 2 (3 + 4i)(3 - 4i) =-1 – 4i =3+i - 4 – 5i = i + i + 5i 2 = i + i -5 =7 + 16i =9 + 12i + 12i + 16i 2 =9 + 24i - 16 = i =9-12i+12i-16i 2 =9+16 =25

Dividing: Mult. Top and bottom by i if monomial Mult. Top and bottom by conjugate if binomial. Conjugate of 4-2i is 4 + 2i. Write answer a + bi

Solving i equations Remember to use + when taking square root Solve for x in each of the following X 2 = -20 7x 2 = x 2 – 6 = 21

Other i problems 5 + 3i = m + ni Therefore 5 = m; 3 = n m + (3 + n)i = i (3 + m) + 8i = 10 + bi m = 5; 3 + n = 8 so n=5 3+m =10; 8=b m = 7