Question

Complex Numbers?

Answer 1

Find the real number k for which 1 + ki, (i = √-1), is a zero of the polynomial z² + kz + 5.

Answer 2

p(1+ki) = (1+ki)^2 + k(1+ki) + 5 = 0

(1 + (k^2)i^2 +2(ki) + k + k^2(i) + 5 =0

1 - k^2 + 2ki + k + (k^2)i + 5 = 0

(1+5 - k^2 + k ) + (2ki + (k^2)i ) = 0 + 0 * i

(6 - k^2 + k) + i(2k + k^2) = 0 + 0*i
6 - k^2 + k = 0

k^2 - k - 6 = 0

k^2 + 2k - 3k - 6 = 0

k(k+2) - 3(k+2) =

(k-3)(k+2) = 0

either k = 3 or k = -2

Answer 3

I don't get the first step already :(
Why are you substituting 1+ki for z? And what's p?

Answer 4

note that -2 is the only answer

Answer 5

ok @Zarkon

given p(z) = z^2 + kz + 5
and given 1 + ki is the zero of the polynomial
therefore z = 1+ki

and i wrote this as p(1+ki) p is a type of function

Answer 6

Oh, I see. I thought p was a real number. Ok thanks :)
And why is 3 not a solution?

Answer 7

3 does not satisfy the equation

i(2k + k^2) = 0*i


but -2 does