# Thread: Solution set using De Moivre's theorem

1. ## 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?

2. ## Re: Solution set using De Moivre's theorem

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 $k=0,1,\cdots,6$

3. ## Re: Solution set using De Moivre's theorem

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 $z = 2^7\bigl(\cos(\tfrac\pi4 + 2k\pi) + i\sin(\tfrac\pi4 + 2k\pi)\bigr).$

4. ## Re: Solution set using De Moivre's theorem

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?

5. ## Re: Solution set using De Moivre's theorem

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.

6. ## Re: Solution set using De Moivre's theorem

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 $\displaystyle \theta \in (-\pi, \pi]$.

7. ## 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?

8. ## Re: Solution set using De Moivre's theorem

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 $\frac{2\pi}{n}$.