Let a and b be real numbers. The complex number 4 - 5i is a root of the quadratic z^2 + (a + 8i) z + (-39 + bi) = 0. What is the other root?
I tried working this problem and so far I got that (a+8i) - (4-5i)= (-39-bi)/(4-5i). Im not sure how to solve this problem.
I'd just multiply:
(4-5i)^2 + (a+8i)(4-5i) + (-39+bi) = 0+0i
Expand all that out and you have two equations in a and b. I get
a = 2
b = 18
So, that leaves us with
z^2 + (2 + 8i) z + (-39 + 18i) = 0
and the other root is thus
-6 -3i
To find the other root of the quadratic equation, we can use the fact that the sum of the roots of a quadratic equation is equal to the negation of the coefficient of the linear term divided by the coefficient of the quadratic term.
Let's set the other root as z = x + yi, where x and y are real numbers.
Since the sum of the roots is equal to (-a - 8i), we can write the equation:
(x + yi) + (4 - 5i) = -a - 8i
Now, let's equate the real and imaginary parts on both sides:
Real part: x + 4 = -a
Imaginary part: y - 5 = -8
Solving these equations, we find:
x = -a - 4
y = -3
So, the other root is z = -a - 4 - 3i.