Solution set using De Moivre's theorem

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August 22nd 2011, 08:56 AM
terrorsquid

Solution set using De Moivre's theorem

Let $z = \frac{128}{\sqrt2}(1 + i)$ . What are all possible solutions of:

$w = z^{1/7}$

So, after converting to polar form I have:

$z^{1/7} = 2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{ \pi}{28}\right))$

I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?
August 22nd 2011, 09:35 AM
Plato

Re: Solution set using De Moivre's theorem

Quote:

Originally Posted by terrorsquid

Let $z = \frac{128}{\sqrt2}(1 + i)$ . What are all possible solutions of: $w = z^{1/7}$
$z^{1/7} =2(cos\left(\frac{\pi}28}\right)+i~sin\left(\frac{ \pi}{28}\right))$

I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?

$2^{\frac{{13}}{{14}}}(cos\left(\frac{\pi}{28} + \frac{2\pi k}7}\right)+i~sin\left(\frac{\pi}{28}+\frac{2\pi k}{7}\right))$ for $k=0,1,\cdots,6$
August 22nd 2011, 10:19 AM
Opalg

Re: Solution set using De Moivre's theorem

Quote:

Originally Posted by terrorsquid

Let $z = \frac{128}{\sqrt2}(1 + i)$ . What are all possible solutions of:

$w = z^{1/7}$

So, after converting to polar form I have:

$z^{1/7} = 2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{ \pi}{28}\right))$

I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?

The crucial step is to write $z = 2^7\bigl(\cos(\tfrac\pi4 + 2k\pi) + i\sin(\tfrac\pi4 + 2k\pi)\bigr).$
August 22nd 2011, 05:39 PM
terrorsquid

Re: Solution set using De Moivre's theorem

Quote:

Originally Posted by Opalg

The crucial step is to write $z = 2^7\bigl(\cos(\tfrac\pi4 + 2k\pi) + i\sin(\tfrac\pi4 + 2k\pi)\bigr).$

Hmm I still seem to be missing a step in my head. I don't understand how the k multiples of pi got inside the sin and cos functions. I can see that

$128(cos\left(\frac{\pi}{4}\right) + i~sin\left(\frac{\pi}{4}\right))$ can be written as $2^7(cos\left(\frac{\pi}{4}\right) + i~sin\left(\frac{\pi}{4}\right))$

but from there I thought I just applied the power $\frac{1}{7}$ to the $2^7$ and then multiplied $\theta$ by $\frac{1}{7}$ which gives me:

$2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{ \pi}{28}\right))$

What am I missing in between those steps?
August 22nd 2011, 09:09 PM
SammyS

Re: Solution set using De Moivre's theorem

Quote:

Originally Posted by terrorsquid

Hmm I still seem to be missing a step in my head. I don't understand how the k multiples of pi got inside the sin and cos functions. ...

Consider how sine & cosine are defined on the unit circle.

Adding an integer multiple of 2π to the argument of any trig function gives the same result as the original argument.
August 22nd 2011, 09:15 PM
Prove It

Re: Solution set using De Moivre's theorem

Quote:

Originally Posted by terrorsquid

Let $z = \frac{128}{\sqrt2}(1 + i)$ . What are all possible solutions of:

$w = z^{1/7}$

So, after converting to polar form I have:

$z^{1/7} = 2(cos\left(\frac{\pi}{28}\right)+i~sin\left(\frac{ \pi}{28}\right))$

I have the answer which includes 7 solutions and I'm a little confused on how to get to them. I guess this is more a trig question?

It helps if your realise that all the solutions will be evenly spaced about a circle. So they all have the same magnitude and are all separated by an angle of $\displaystyle \frac{2\pi}{7}$ . Also note that the principal solutions will be in the domain $\displaystyle \theta \in (-\pi, \pi]$ .
August 23rd 2011, 08:54 AM
terrorsquid

Re: Solution set using De Moivre's theorem

Ok, I think I get the process. My understanding is very superficial though. I only learned how to "apply the formula" and didn't cover what I am actually doing by solving one of these which is what I think my problem is.

Take a similar example:

$Let~z = 8(-1 + i\sqrt3)$ . Find all the solutions for:

$w = z^{1/4}$

I convert to polar form:

$16(-cos\left(\frac{2\pi}{3}\right) + isin\left(\frac{2\pi}{3}\right))$

I notice that $16 = 2^4$ so I figure there will be 4 solutions (not sure why).

I then apply the exponent 1/4 to get:

$2(-cos\left(\frac{\pi}{6}\right) + isin\left(\frac{\pi}{6}\right))$

Then to get my solutions I just generate 4 answers by multiplying the exponent 1/4 by $2k\pi$ and adding it to $\frac{\pi}{6}$ cycling k from 0 - 3 to get:

$2(-cos\left(\frac{\pi}{6} + \frac{2k\pi}{4}\right) + isin\left(\frac{\pi}{6}+ \frac{2k\pi}{4}\right)) = \frac{\pi}{6}, \frac{2\pi}{3}, \frac{7\pi}{6}, \frac{5\pi}{3}$

I can see that it makes one revolution and the answers are recorded periodically and evenly depending on the denominator of the $2k\pi$ increments, and I seem to be able to generate the answers for these types of problems, I just feel awkward not knowing why I'm doing each step if you understand?

Will these problems always have a coefficient that can be expressed as an integer raised to a power so you can determine how many answers there should be?
August 23rd 2011, 09:32 AM
Plato

Re: Solution set using De Moivre's theorem

Quote:

Originally Posted by terrorsquid

I can see that it makes one revolution and the answers are recorded periodically and evenly depending on the denominator of the $2k\pi$ increments, and I seem to be able to generate the answers for these types of problems, I just feel awkward not knowing why I'm doing each step if you understand?
Will these problems always have a coefficient that can be expressed as an integer raised to a power so you can determine how many answers there should be?

There are n nth roots of a complex number $z=|z|\mathbf{cis}(\theta)$
Here $\mathbf{cis}(\theta)$ is shorthand for $\cos(\theta)+\mathbf{i}\sin(\theta)$ where
$\theta=\text{Arg}(z)$ .

Those n roots are $|z|^{1/n}\mathbf{cis}\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)$ where $k=0,1,\cdots,n-1.$

Those n numbers are all on a circle centered at $(0,0)$ with radius $|z|^{1/n}$ .
Moreover, those points are equally spaced on arcs subtended by angles of $\frac{2\pi}{n}$ .