The conjugate of (84-3i) is (84+3i). This gives you a real number when multiplied.
-2 + 3iThe additive inverse: -(-2 + 3i) = 2 - 3iThe conjugate: -2 - 3i
this is a very good question. lets solve (2+3i)/(4-2i). we want to make 4-2i real by multiplying it by the conjugate, or 4+2i (4-2i)(4+2i)=16-8i+8i+4=20, now we have (2+3i)/20 0r 1/10 + 3i/20 notice that -2i times 2i = -4i^2 =-4 times -1 = 4
(3i+l)(3-I)
The "i" in 3i means the number is imaginary.
The complex conjugate of 2-3i is 2+3i.
The conjugate of 2 + 3i is 2 - 3i, and the conjugate of 2 - 5i is 2 + 5i.
11
The complex conjugate of a number in the form a + bi is simply the same number with the sign of the imaginary part changed. In this case, the number is 7 + 3i, so its complex conjugate would be 7 - 3i. This is because the complex conjugate reflects the number across the real axis on the complex plane.
[7 - 3i] To find the conjugate: the sign of the real part stays the same, and the sign of the imaginary part is reversed. So the conjugate of [7 + 3i] is [7 - 3i]
- 2 - 3i
-2 - 3i
[ 2 - 3i ] is.
The conjugate of (84-3i) is (84+3i). This gives you a real number when multiplied.
The conjugate is 7 - 3i is 7 + 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
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i