1. ## Complex Number

Have a question:

Let f be the complex polynomial
f(x)=(3+i)x^2+(-2-6i)x+12
Calculate the following values f(i) and f(2+3i)
I calculated these values as such:
f(i)=(3+i)i^2+(-2-6i)i+12
Doing the math f(i)=15-3i
f(2+3i)=(3+i)(2+3i)(2+3i)+(-2-6i)(2+3i)+12
Doing the math f(2+3i)=-1+13i

Is this right?

Also, it then asks find a complex number z such that the equation f(x)=z has a unique solution (use the formula for solving a quadratic equation). How do you do this?

2. Hello, JaysFan31!

Let f be the complex polynomial: f(x) .= .(3 + i)x² + (-2 - 6i)x + 12
Calculate the following values: f(i) and f(2 + 3i)

I calculated these values as such:

f(i) .= .(3 + i)i^2 + (-2 - 6i)i + 12
. . Doing the math: .f(i) .= .15 - 3i

f(2 + 3i) .= .(3 + i)(2 + 3i)(2 + 3i) + (-2 - 6i)(2 + 3i) + 12
. . Doing the math: .f(2 + 3i) .= .-1 + 13i

Is this right? . . . . They look good to me!

Also, it then asks find a complex number z
such that the equation f(x) = z has a unique solution.
(Use the Quadratic Formula. .How do you do this?

We have: .(3 + i)x² + (-2 - 6i)x + 12 .= .z

Or: . (3 + i)x² + (-2 -6i)x + 12 = z .= .0

We have a quadratic equation with: a = (3 + i), b = (-2 - 6i), c = (12 - z)
. . . . . . . . . . . . . . . . . . . . . . . . . _____________________
. . . . . . . . . . . . . . . . -(-2 - 6i) ± √(-2 - 6i)² - 4(3 + i)(12 - z)
Then we have: . x .= . ----------------------------------------------
. . . . . . . . . . . . . . . . . . . . . . . . . . 2(3 + i)

A quadratic will have one solution if its discriminant equals zero.
That is: . (-2 - 6i)² - 4(3 + i)(12 - z) .= .0

Now solve for z . . .

3. Thanks a lot. You're very helpful. Where did you teach?

4. Hello again, JaysFan31!

I taught at a tiny two-year state college,
. . Massachusetts Bay Community College, in Wellesley, MA.

I enjoyed it so much, I taught there for 36 years.