0
The easiest way to do this is treat i like a variable and multiply the binomials and combine like powers of i, then anywhere there is i², substitute it with (-1).
So (4 + 3x)(3 - 4x) = {using FOIL} 4*3 - 4*4*x + 3*3*x - 3*4*x² = 12 -7x - 12x²
So with x = i, you have 12 - 7i - 12*(i²) = 12 - 7i - -12 = 24 - 7i
One quick check to see if there is an error: The magnitude of the product of the two complex binomials will equal the product of the magnitude of each factor.
Magnitude of a + bi = sqrt(a² + b²). The magnitudes of two numbers that are multiplied are: sqrt(4² + 3²) = 5, and sqrt(3² + (-4)²) = 5. The magnitude of the answer is sqrt(24² + (-7)²) = sqrt(576 + 49) = sqrt(625) = 25, which is 5 times 5.
Wiki User
∙ 13y ago
Add your answer:
Earn +20 pts
What is the product of these complex numbers (3-4i)(1-i)?
(3-4i)(1-i) = (3x1) + (3 x -i) + (-4i x 1) + ( -4i x -i) = 3 - 3i -4i -4 = -1 - 7i
What is conjugate of 4i open bracket -2 -3i close bracket?
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
Plot the number in a complex plane -1-3i?
2
Find the absolute value of the complex number z equals 3 plus 4i?
('|x|' = Absolute value of x) |3+4i| = √(32 + 42) = √(9+16) = √25 = 5 Thus |3+4i| = 5.
When the sum of the complex numbers 3 plus 2I and 6 - 4I is graphed in which quadrant does it lie?
3+2i + 6-4i = 9-2i The real part of this number is positive, therefore it lies in Q1 or Q4. The imaginary part is negative, therefore it is in Q3 or Q4. Q4 is the common possibility, therefore 9-2i is in Q4.
Can you factor complex numbers?
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.
What is the product of these complex numbers (3-4i)(1-i)?
(3-4i)(1-i) = (3x1) + (3 x -i) + (-4i x 1) + ( -4i x -i) = 3 - 3i -4i -4 = -1 - 7i
What is conjugate of 4i open bracket -2 -3i close bracket?
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
Plot the number in a complex plane -1-3i?
2
How do you factor xsquared plus 16?
(x - 4i)(x + 4i) where i is the square root of -1
What are 10 complex numbers with the absolute value of 5?
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.
What is the product for 2 plus 4i?
A product is a binary operatoin. That is, it requires two numbers to be combined. There is only one number, 2 + 4i, in the question.
What is the equation for whose roots are 4i -3i and 3i?
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
How do you simplify 4i 3i?
Assume we have the product of these terms. Multiply these terms altogether to get 12i². Make note that i = √-1, i² = -1, i³ = -i and i⁴ = 1. Then, 12i² = 12(-1) = -12. That is the answer to the question.
What is the sum of two complex conjugate number?
Since the imaginary parts cancel, and the real parts are the same, the sum is twice the real part of any of the numbers. For example, (5 + 4i) + (5 - 4i) = 5 + 5 + 4i - 4i = 10.
What is the conjugate of the complex number 7-4i?
To get the conjugate simply reverse the sign of the complex part. Thus conj of 7-4i is 7+4i
Find the complex fourth root of 256i Express your answer in a plus bi form?
The four roots of 4√256 are {4, -4, 4i, and -4i}. Note that two of them are real numbers and the other two are pure imaginary, therefore 0 + 4i is the same as just 4i
