Complex Numbers in Quadratic Equations

Date: 11/09/1999 at 10:08:17
From: Takeyuki Oya
Subject: Imaginary numbers in solving quadratic equations

Hello,

I just want to know about the use of imaginary numbers in solving 
quadratic equations, particularly when the discriminant is negative.

Also if possible, could you give me a description of how solutions of 
this type of quadratic equation may be represented graphically?

Thank you.

Date: 11/09/1999 at 17:06:24
From: Doctor Rick
Subject: Re: Imaginary numbers in solving quadratic equations

Hello, welcome to Ask Dr. Math. You've asked some good questions.

>I just want to know about the use of imaginary numbers in solving 
>quadratic equations, particularly when the discriminant is negative.

Let's take an example:

     x^2 - 4x + 7 = 0

Using the quadratic equation, we get

     x = (1/2)(4 +- sqrt(16-28))
       = (1/2)(4 +- sqrt(-12))

What is the square root of -12? We can factor -12 = -1 * 3 * 4. Then

  sqrt(-1*3*4) = sqrt(-1)*sqrt(3)*sqrt(4)
               = i * sqrt(3) * 2

using the definition of the imaginary number i = sqrt(-1). Now we can 
continue simplifying x:

     x = (1/2)(4 +- 2i*sqrt(3))
       = 2 +- i*sqrt(3)

This is our final answer: the roots of the quadratic are

     x = 2 + i*sqrt(3)
and
     x = 2 - i*sqrt(3)

Whenever the discriminant is negative, its square root will be 
imaginary. The real part of the two solutions is always the same. Can 
you see why this is true? We can write out the general solution for 
the case in which the discriminant is negative:

     ax^2 + bx + c = 0

     x = (-b +- sqrt(b^2 - 4ac))/(2a)
       = (-b +- i*sqrt(4ac - b^2))/(2a)

The real part is -b/(2a), and the imaginary part is 
+-sqrt(4ac-b^2)/(2a).


>Also if possible, could you give me a description of how solutions of 
>this type of quadratic equation may be represented graphically?

In order to see the complex roots on a graph, we need to see complex 
values of x. Therefore we'll need to replace the real x-axis with a 
complex plane. So, draw a plane with two axes: Re(x), the real part of 
x, and Im(x), the imaginary part of x. Then "draw" the y-axis 
perpendicular to this plane. (You'll have to either imagine this axis 
coming out of the paper, or draw it in perspective.)

Any point in the x-plane corresponds to a complex value of x; we can 
evaluate y = ax^2 + bx + c and plot this point (Re(x),Im(x),y). We 
will end up with a surface rather than a curve: for any point in the 
x-plane, the surface will be at height y(Re(x),Im(x)) above the plane.

Wait a moment - did you notice something wrong there? If x can be any 
complex number, then y(x) may be complex itself. I was talking as if 
the height y(x) were a real number. What can we do about this?

One way to picture it is to look at the real and imaginary parts of 
y(x) separately. In other words, you can draw two surfaces, one for 
the real part of y and the other for the imaginary part of y.

We're mainly interested in the roots of the quadratic - where the 
"surface" y(x) crosses the x-plane. This will only happen when BOTH 
the real and imaginary parts of y are zero.

I won't try to describe all the work that is needed; this makes an 
interesting project. I suggest that you write the complex number x as

     x_r + i*x_i

where x_r (the real part of x) and x_i (the imaginary part of x) are 
both real. Substitute this in the quadratic:

     y = (x_r + i*x_i)^2 + b(x_r + i*x_i) + c

Separate y into real and imaginary parts. Think about what each 
surface will look like. In particular, consider: for which values of 
x_r and x_i will the imaginary part be zero? Then consider how the 
real part of y behaves for these values of x_r and x_i. You will 
discover the real parabola that you expect, plus something 
interesting!

I've probably given you more than you wanted, but this really is an 
interesting question to pursue. I hope you can try it.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/