(3i+l)(3-I)
0 + 3i
The "i" in 3i means the number is imaginary.
1/(1+ 3i)
3x2 = -9 (divide both sides by 3) x2 = -3 (x would have to be the square root of -3) x = ±√-3 x = ±√3i Since you want to solve by factoring: x2 = -3 add 3 to both sides x2 + 3 = 0 x2 - 3i2 = 0 x2 - (√3i)2 = 0 Factor: (x - √3i)(x + √3i) = 0 x - √3i = 0 or x + √3i = 0 x = √3i or x = -√3i
-2 + 3iThe additive inverse: -(-2 + 3i) = 2 - 3iThe conjugate: -2 - 3i
- 2 - 3i
0 + 3i
11
The complex conjugate of 2-3i is 2+3i.
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
The "i" in 3i means the number is imaginary.
-2 - 3i
0+3i has a complex conjugate of 0-3i thus when you multiply them together (0+3i)(0-3i)= 0-9i2 i2= -1 0--9 = 0+9 =9 conjugates are used to eliminate the imaginary parts
1/(1+ 3i)
3i and or -3i
[ 2 - 3i ] is.
x2 + 9 has no real factors. Its complex factors are (x + 3i) and (x - 3i) where i is the imaginary square root of -1.