Finding Roots of Complex NumbersDate: 09/01/2005 at 00:10:05 From: Kira Subject: How do you find the nth roots of a complex number a + bi How do you find the nth roots of a complex number a + bi? Date: 09/01/2005 at 08:58:47 From: Doctor Jerry Subject: Re: How do you find the nth roots of a complex number a + bi Hello Kira, I'll answer by showing how to find the fifth roots of 2 + 3i. We convert 2 + 3i to polar form and look for complex numbers in the polar form r*e^{i*t} = r*(cos(t) + i*sin(t)). Since 2 + 3i = sqrt(13)*e^{i*arctan(3/2)}, we want ( r*e^{i*t} )^5 = sqrt(13)*e^{i*arctan(3/2)} This is r^5*e^{i*5t} = sqrt(13)*e^{i*arctan(3/2)} We see that r^5 = sqrt(13) and so r = [13^(1/2)]^(1/5)] or 13^(1/10) and also that 5t - arctan(3/2) = 2*pi*n, n = 0,1,2,3,4 So, t = [2*pi*n + arctan(3/2)]/5, n = 0,1,2,3,4. Let's look at one of these "fifth roots." Take n = 3. r = 13^{1/10} = 1.29239222078083 t = [2*pi*3 + arctan(3/2)]/5 = 3.96646992895723 r*e^{i*t} = -0.87707824256 - i*0.949216207595 The fifth power of this complex number is 2 + 3*i. Please write back if my comments are not clear or if you have further questions on this process. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/