0
(9 + 4i)^2
= 9^2 + (2)(9)(4i) +i^2 substitute i^2 for -1;
= 81 + 72i -1
= 80 + 72i
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How do you factor xsquared plus 16?
(x - 4i)(x + 4i) where i is the square root of -1
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
How do you simplify the expression 3i x -4i?
7
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
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.
2 plus 4i - 7 plus 4i?
(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
What number squared equals 16?
4i where i = sqrt (-1) 4i x 4 i = 16 i squared = -16
What is the conjugate of -6 plus 4i?
-6-4i.
How do you solve 3 x squared plus 51 equals 6 x?
3x2 + 51 = 6x Rewrite the equation so that it equates to zero 3x2 - 6x + 51 = 0 Simplify by dividing by 3 x2 - 2x + 17 = 0 Then use the quadratic formula x = {2 ± √[(-2)2 -(4x1x17)]} ÷ 2 = { 2 ± √-64} ÷ 2 = 1 ± 4i Then x = 1 + 4i and x = 1- 4i
What is the complex form of 4 plus 4i plus 4 plus 6i?
the problem: what is 4 + 4i + 4 + 6i what you do is add the real and imaginary parts, thus: 4+4 and 4i+6i = 8+10i answer.
What is the conjugate of 3 plus 4i?
When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.
How do you factor xsquared plus 16?
(x - 4i)(x + 4i) where i is the square root of -1
What is x squared plus 80 equals 0?
It's a second degree equation in 'x' that has no real solution. No real number in the place of 'x' can make that equation a true statement. There are only two "imaginary" numbers that 'x' can be: + 4i sqrt(5) - 4i sqrt(5)
How do you solve 7-4i-6-8i in complex number I need work answer choices are A 74-32i B -98 plus 56i C 10 plus 80i D -74-32i?
It is D.
What is 2-4i over 4 plus 2i put in the form of a plus bi?
To solve this type of problem, multiply both the numerator and denominator by the conjugate of the denominator. (2 - 4i) / (4 + 2i) = (2 - 4i)(4 - 2i) / (4 + 2i)(4 - 2i) then expand all the terms, and simplify. = (8 - 20i + 8i2) / (16 - 4i2) = (8 - 20i - 8) / (16 + 4) = -20i / 20 = -i Which in the required answer format becomes, 0 + i.
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
X2 plus x-2 equals 0?
X squared + 4 = 0 , to solve this, you would need to carry the 4 over, so X squared = -4, then move the square over so x = +4i,-4i , because when you divide the negative, it creates a imaginary -1
