p = (2 + 5i)q = (6 - i)pq = (2 + 5i) (6 - i)pq = 12 + 30i - 2i - 5i2pq = 12 + 28i + 5pq = 17 + 28iIts the same as multiplying polynomials. Just multiply all the combinations of terms, group, and simplify.
6+5i
it would go inbetwwen 12 and negative 5i hoped this helped u /\__/\( + + )| |-----------\| \( )---------( )\ \| | | | \ \___/__| /__| \--------/\|Kitty
Choose ANY number, for example, 5, -100, square root of 2, pi, or (5 + 3i). Subtract -6 minus your number, to get the other number.
The best way to factor this is to use the quadratic equation. Using an equation of the form: ax2 + bx + c = 0 You use the quadratic formula like this: x = [-b +(or -) sqrt(b2-4ac)]/2a For your problem, a = 2, b = -6, and c = 9. Therefore: x = [6 +(or -) sqrt(36 - 4(2)(9))]/4 = [6 +(or -) sqrt(36-72)]/4 = [6 +(or -) sqrt(-36)]/4 = (6 +(or -)6i)/4 = 3/2 +(or -)3i/2 Therefore the factorization is (x - 3/2 - 3i/2)(x - 3/2 + 3i/2).
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
p = (2 + 5i)q = (6 - i)pq = (2 + 5i) (6 - i)pq = 12 + 30i - 2i - 5i2pq = 12 + 28i + 5pq = 17 + 28iIts the same as multiplying polynomials. Just multiply all the combinations of terms, group, and simplify.
6+5i
it would go inbetwwen 12 and negative 5i hoped this helped u /\__/\( + + )| |-----------\| \( )---------( )\ \| | | | \ \___/__| /__| \--------/\|Kitty
Choose ANY number, for example, 5, -100, square root of 2, pi, or (5 + 3i). Subtract -6 minus your number, to get the other number.
57
The best way to factor this is to use the quadratic equation. Using an equation of the form: ax2 + bx + c = 0 You use the quadratic formula like this: x = [-b +(or -) sqrt(b2-4ac)]/2a For your problem, a = 2, b = -6, and c = 9. Therefore: x = [6 +(or -) sqrt(36 - 4(2)(9))]/4 = [6 +(or -) sqrt(36-72)]/4 = [6 +(or -) sqrt(-36)]/4 = (6 +(or -)6i)/4 = 3/2 +(or -)3i/2 Therefore the factorization is (x - 3/2 - 3i/2)(x - 3/2 + 3i/2).
-6
The idea here is to change the sign before the imaginary term. In this case, since there is a minus, you change it to a plus.
132
To multiply complex numbers you can use the same FOIL rule that you use for multiplying binomials (First, Inside, Outside, Last).(4 - 3i)(5 + 2i) = (4)(5) +(4)(2i) - (3i)(5) - (3i)(2i) = 20 + 8i-15i - 6(i)^2= 20 -7i - 6(-1) = 20 + 6 -7i = 26 -7i.
46