Square Roots of Complex NumbersDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/