Solution set using De Moivre's theorem

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

$\displaystyle w = z^{1/7}$

So, after converting to polar form I have:

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

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?

Re: Solution set using De Moivre's theorem

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**terrorsquid** Let $\displaystyle z = \frac{128}{\sqrt2}(1 + i)$. What are all possible solutions of: $\displaystyle w = z^{1/7}$

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

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?

$\displaystyle 2^{\frac{{13}}{{14}}}(cos\left(\frac{\pi}{28} + \frac{2\pi k}7}\right)+i~sin\left(\frac{\pi}{28}+\frac{2\pi k}{7}\right))$ for $\displaystyle k=0,1,\cdots,6$

Re: Solution set using De Moivre's theorem

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

$\displaystyle w = z^{1/7}$

So, after converting to polar form I have:

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

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

Re: Solution set using De Moivre's theorem

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

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

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

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

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

What am I missing in between those steps?

Re: Solution set using De Moivre's theorem

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**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.

Re: Solution set using De Moivre's theorem

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

$\displaystyle w = z^{1/7}$

So, after converting to polar form I have:

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

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 $\displaystyle \displaystyle \frac{2\pi}{7}$. Also note that the principal solutions will be in the domain $\displaystyle \displaystyle \theta \in (-\pi, \pi]$.

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:

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

$\displaystyle w = z^{1/4}$

I convert to polar form:

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

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

I then apply the exponent 1/4 to get:

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

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

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

I can see that it makes one revolution and the answers are recorded periodically and evenly depending on the denominator of the $\displaystyle 2k\pi$ 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?

Re: Solution set using De Moivre's theorem

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**terrorsquid** I can see that it makes one revolution and the answers are recorded periodically and evenly depending on the denominator of the $\displaystyle 2k\pi$ 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 $\displaystyle z=|z|\mathbf{cis}(\theta)$

Here $\displaystyle \mathbf{cis}(\theta) $ is shorthand for $\displaystyle \cos(\theta)+\mathbf{i}\sin(\theta)$ where

$\displaystyle \theta=\text{Arg}(z)$.

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

Those n numbers are all on a circle centered at $\displaystyle (0,0)$ with radius $\displaystyle |z|^{1/n}$.

Moreover, those points are equally spaced on arcs subtended by angles of $\displaystyle \frac{2\pi}{n}$.