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Finding Roots of Complex Numbers

Date: 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


  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 
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers

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