## Monday, March 22, 2010

### More complex roots

We left off with the problem to find the 4th roots of unity. This is particularly simple. As with the third roots of unity, we divide the unit circle equally into four segments with the boundaries marking the roots. (See Diagram).

We have, n = 4 with:

w_n = cos (2 pi)k/ n + isin(2 pik)/n

For k = 0, 1, 2 and 3 then:

The first root::

w0 = cos(0) + isin(0) = 1

The second root:

w1 = cos (pi/2) + isin(pi/2) = 1i = i

The third root:

w2 = cos (pi) + isin(pi) = -1

The fourth root:

w3 = cos(3 pi/2) + isin(3pi/2) = -1i = -i

And note the roots correspond to successive increases of the angle by pi/2 = 90 deg

What about complex roots not of unity?

Say: (1 + i)^1/3
Let's first examine the incorrect way to solve it, then the correct way:

The first step is to find theta using the argument function:

Theta = arctan(y/x) = arctan (1) = 45 deg

And r = (2)^1/3 = 1.25

So the (incorrect) roots are:

w0= 1.25[cos (0) + isin(0)] = 1.25

w1= 1.25[cos(2 pi/3) + isin(2 pi/3)] = 1.25(-0.5) + 1.25i(0.866) = -0.625 +1.08i

w2= 1.25[cos(4 pi/3) + isin(4pi/3)] = 1.25(-0.5) + 1.25i(-0.866) = = -0.625 - 1.08i
This isn't correct because one must first separate the root-generated angles(2 pi k/n) from the argument, theta. Then solve respectively for the independent roots.

Thus, we have:

w_n = r^1/2 [cos(theta + 2 pik/n) + isin(theta + 2 pik/n)]

We found r = [2]^1/2 = 1.25 but in fact we need: r^1/3 = [(2)^1/2]^1/3 = (2)^1/6

The roots are then defined from:

w_n = (2)^1/6 [cos(theta + 2 pik/3) + isin(theta + 2 pik/3)]

Since theta = arctan y/x = 45 deg then, theta = pi/4

So the consecutive roots are(in order):

w_0 = (2)^1/6 [cos(pi/4 + 2 pi k/3) + isin(pi/4 + 2 pi k/3)]

since k=0: w_0 = (2)^1/6 [cos(pi/4) + isin(pi/4)]

w_1 = (2)^1/6 [cos(pi/4 + 2 pi /3) + isin(pi/4 + 2 pi/3)]

Working on the interior brackets (adding fractions in pi):

(pi/4) + 2 pi/3 = (3 pi + 8pi)/ 12 pi = 11 pi/12

So:

w_1 = (2)^1/6 [cos(11 pi / 12) + isin(11 pi/12)]

and the answer can be left in this form, OR the readers can obtain cos(11pi/12) and sin(11pi/12) and work out the specific numbers, though often this isn’t done.

Finally since k =2 (for w_2):

w_2 = (2)^1/6 [cos(pi/4 + 4pi /3) + isin(pi/4 + 4 pi/3)]

Again, adding the fractions in pi in the interior brackets:

(pi/4) + (4 pi/3) = (3 pi + 16 pi)/ 12 = 19 pi/12

So:
w_2 = (2)^1/6 [cos(19 pi /12) + isin(19 pi/12)]

For those who want the exact, specific ans.

w_2 = 0.291 – 1.084i

For w_1:

w_1 = -1.084 – 0.291i

Problem:

Find all the roots of: (-2 + 2i)^1/3