Q: What is the sum of the serious if n equals 5 and ai equals 4i plus 3?

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(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i

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.

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)

4ÃŽÂ£ (12 - 4i)i=0

The multiplicative inverse of a number a is a number b such that axb=1 If we look at (3-4i)/(5+2i), we see that we can multiply that by its reciprocal and the product is one. So (5+2i)/(3-4i) is the multiplicative inverse of (3-4i)/(5+2i)

Related questions

(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i

-6-4i.

('|x|' = Absolute value of x) |3+4i| = √(32 + 42) = √(9+16) = √25 = 5 Thus |3+4i| = 5.

Add the real and the imaginary parts separately.

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.

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.

When finding the conjugate of a binomial, you just reverse the sign. So the conjugate of 3+4i is 3-4i.

(x - 4i)(x + 4i) where i is the square root of -1

4i where i = sqrt (-1) 4i x 4 i = 16 i squared = -16

(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).

(9 + 4i)^2 = 9^2 + (2)(9)(4i) +i^2 substitute i^2 for -1; = 81 + 72i -1 = 80 + 72i

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)