Expert: Paul Klarreich - 10/12/2008

Question
Find all fifth roots of 32i and express in rcis.  Pick one of them and express in a + bi form.  I am studying complex numbers in polar form.  I do not even know where to start with this question.

Answer
Questioner:   Melissa
Category:  Advanced Math
Private:  No
 
Subject:  fifth roots
Question:  Find all fifth roots of 32i and express in rcis.  Pick one of them and express in a + bi form.  I am studying complex numbers in polar form.  I do not even know where to start with this question.
........................................
Hi, Melissa,

There are three parts:

1. Convert 32i  INTO r-cis form.  Write the FIVE alternative forms.
2. Apply deMoivre's theorem to each, using the z^(1/5) rule.
3. Convert them (well, only one) BACK to  a + bi form.

You may, alas, have to use all that fine trigonometry you learned last year.

1. Draw the diagram:

        |
    32i +
        |
        |
        |
        |
---------+-----------
        |
        |

Your distance is 32 = r and your polar angle is  pi/2.  (yes it's 90 degrees, but degrees are for children)

You will do:

32i = 32 cis (pi/2 + 0 pi)
32i = 32 cis (pi/2 + 2 pi)
32i = 32 cis (pi/2 + 4 pi)
32i = 32 cis (pi/2 + 6 pi)
32i = 32 cis (pi/2 + 8 pi)

32i = 32 cis ( pi/2)
32i = 32 cis (5pi/2)
32i = 32 cis (9pi/2)
32i = 32 cis (13pi/2)
32i = 32 cis (17pi/2)


2. Now you will 'root' each of those:  The 5th root of 32 is 2, so that will be your new r.

You will take 1/5 of each polar angle;

fifth root of 32i = 2 cis (pi/10)
fifth root of 32i = 2 cis (5pi/10)
fifth root of 32i = 2 cis (9pi/10)
fifth root of 32i = 2 cis (13pi/10)
fifth root of 32i = 2 cis (17pi/10)

3. You will now convert them all (not really all) back to rectangular (a + bi) form. Since you only have to do one,  I will pick the second:

fifth root of 32i = 2 cis (5pi/10)
fifth root of 32i = 2 cis (pi/2)
= 2 (cos pi/2 + i sin pi/2)

= 2 (0     + i (1))

= 0 + 2i

You can easily check that  (2i)^5 = 32i.