7
Use the Pythagorean theorem. 5, -5, 5i, and -5i will work, as well as any combination of a real and imaginary number such that (real part) squared + (imaginary part) squared = 25, for example, 4 + 3i, 3 + 4i, 4 - 3i, etc.
The complex conjugate of 2-3i is 2+3i.
(x - 4i)(x + 4i) where i is the square root of -1
To get the conjugate simply reverse the sign of the complex part. Thus conj of 7-4i is 7+4i
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
7
2
(7 + 3i) + (8 + 9i) = (7 + 8) + (3i + 9i) = (7 + 8) + (3 + 9)i = 15 + 12i Which can also be written as: 15 + 12i = 3(5 + 4i).
The equation is:x3 - 4x2i - 9xi2+ 36i3 = 0Note: The ' i ' must be an ordinary constant, and can't be sqrt(-1).If it were sqrt(-1), then -4i would also be a root of the equation.Imaginary or complex roots always occur in conjugate pairs.If -4i is also a root of the equation and the questioner just forgot to include it,then ' i ' can be sqrt(-1), and the equation can bex4 + 25x2 + 144 = 0
(3-4i)(1-i) = (3x1) + (3 x -i) + (-4i x 1) + ( -4i x -i) = 3 - 3i -4i -4 = -1 - 7i
2ii stands for imaginary, and any negative number's square root is an i number- i, 2i, 3i, 4i, etc.
Yes. Consider, if you can factor complex numbers, then logically, you should be able to take two complex numbers, multiply them together, and get a third. That can indeed be done. For example: (4i + 7)(3i + 2) = -12 + 8i + 21i + 14 = 29i + 2 Therefore, the complex number 29i + 2 must be divisible by 4i + 7 and 3i + 2.
The square root of -1 is i, or j sometimes, if you are an engineer.Therefore it can be rewritten as 3i.
STARTING FROM TOP AND GOING COUNTER CLOCKWISE: 2I, 4E, 1I 2E 3I 1E 4I 3E e FOR EXHAUST AND I FOR INTAKE
(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
Use the Pythagorean theorem. 5, -5, 5i, and -5i will work, as well as any combination of a real and imaginary number such that (real part) squared + (imaginary part) squared = 25, for example, 4 + 3i, 3 + 4i, 4 - 3i, etc.