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Square Roots of Complex Numbers


Date: 03/28/99 at 15:40:20
From: Jaimee
Subject: Roots of Imaginary Numbers

I am currently working on a math problem that asks me to devise at 
least two methods for finding the square root of (a+bi).  

Here is what I have tried. First I thought I could solve it using 
the formula for finding the root of any complex number, but when 
I was finding the arg to put it into mod arg form I got it as being 
tan^-1 (b/a). I don't think that this helps. So then I let square root 
of a + ib = x + iy and squared both sides. Next I equated the 
imaginary and real sections, but now I am stuck.

Please help


Date: 03/29/99 at 07:03:23
From: Doctor Floor
Subject: Re: Roots of Imaginary Numbers


Hi,

Thanks for your question!

Your first idea was good.

Write a+bi in the form  

   r*e^(i*t)=r*(cos(t)+i*sin(t))

The square root becomes 

   sqrt(r)*(cos(t)+i*sin(t))^(1/2)

By DeMoivre's theorem this becomes 

   sqrt(r)*(cos(t/2)+i*sin(t/2))

We are used to being able to find a second square root, the opposite 
of the first. This one can also be found by taking t'=t+2*pi for t.

So that is one way to compute the square root of a complex number.

To learn more about DeMoivre and his theorem, visit the following 
answers in the Dr. Math archives:

   De Moivre's Theorem
   http://mathforum.org/library/drmath/view/54047.html   

   Usefulness of De Moivre's Theorem 
   http://mathforum.org/library/drmath/view/53975.html   

   DeMoivre's Formula
   http://mathforum.org/library/drmath/view/53854.html   

   History of DeMoivre's Theorem 
   http://mathforum.org/library/drmath/view/52533.html   

The second way you started was good too. Suppose that x+yi is the 
square root of a+bi. Then 

    (x+yi)(x+yi) = a+bi
  x^2-y^2 + 2xyi = a+bi

  So           a = x^2-y^2 [1].

  And          b = 2xy, or y = b/(2x) [2].

Substituting [2] in [1] gives:

               b^2
   a = x^2 -  ----
              4x^2

   4ax^2 = 4x^4 - b^2

   4x^4 - 4ax^2 - b^2 = 0

   x^4 - ax^2 - b^2/4 = 0

This can be solved for x^2 using the quadratic formula:

        a +/- sqrt(a^2 + b^2)
  x^2 = ---------------------
                2

Since |a| < sqrt(a^2+b^2) only one of the roots for x^2 is positive, 
which is necessary for x to be a real number.

We find two values for x, and using [2] this gives two solutions for 
x+yi.

So that is the second way to compute the square root of a complex 
number.

I hope this helped!

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Imaginary/Complex Numbers
High School Trigonometry

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