0 + 3i
(3i+l)(3-I)
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
It depends what level of mathematics you are talking about. There is no real number whose square is -9. Once we introduce complex numbers, there are two possibilities, 3i and -3i. Whether you allow complex numbers depends on what you are trying to do. They are unlikely to show up in mathematics before college level, but they are useful in engineering and other areas.
0 + 3i
The complex conjugate of 2-3i is 2+3i.
(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i
-2 - 3i
11
(3i+l)(3-I)
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
3i and or -3i
[ 2 - 3i ] is.
1/(1+ 3i)