text: 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.