Finding the Square Root of a Quadratic Function

Date: 10/30/2002 at 15:09:12
From: Ryan MacRae
Subject: Finding the square root of a Quadratic Function.

I was receintly assinged the problem 

   Find the square root of 3+4i

I am in advanced college math and am totally stumped by this question. 
Please help me...

Thank you for your time. 
Ryan

Date: 10/30/2002 at 17:46:55
From: Doctor Paul
Subject: Re: Finding the square root of a Quadratic Function.

If sqrt(3+4*i) = a + b*i where a and b are real numbers, then squaring 
both sides gives:

   (a + b*i) * (a + b*i) = 3 + 4*i

     a^2 + 2*a*b*i - b^2 = 3 + 4*i

equating coefficients gives:

               a^2 - b^2 = 3
      and
                   2*a*b = 4

Solve for b in the second equation and substitute into the first 
equation:

      b = 2/a

so

      a^2 - (2/a)^2 = 3

        a^2 - 4/a^2 = 3

multiply both sides by a^2:

    a^4 - 3*a^2 - 4 = 0

This is quadratic in a^2 so we make a substitution:

Now let x = a^2

So we have:

   x^2 - 3*x - 4 = 0

     (x-4)*(x+1) = 0

   So x = 4 or x = -1

This gives

   a = 2, -2, i, -i

We said above that a had to be real, so it must be the case that 
a = 2 or a = -2.

Now, if a = 2 then b = 1

and  if a = -2 then b = -1

So we have:

   sqrt(3 + 4*i) = (2 + i) or -(2 + i)

I hope this helps.  Please write back if you'd like to talk about 
this some more.

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