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How many real and non-real roots?

sweatingbear

Member
May 3, 2013
91
How many real and non-real roots does \(\displaystyle z^5 = 32\) have? \(\displaystyle z^9 = -4\)?

For \(\displaystyle z^5 = 32\): \(\displaystyle z^5 = r^5 ( \cos 5v + i \sin 5v )\) and \(\displaystyle 32 = 32 ( \cos 0 + i \sin 0 )\) yields

\(\displaystyle r = 2 \\ 5v = n \cdot 2\pi \iff v = n \cdot \dfrac {2\pi}5\)

So all roots are given by

\(\displaystyle z = 2 \left( \cos \left( n \cdot \frac {2\pi}5 \right) + i \sin \left( n \cdot \frac {2\pi}5 \right) \right)\)

where \(\displaystyle 0 \leqslant n \leqslant 4, \ n \in \mathbb{Z}\). Let us rule out all real roots by letting the imaginary part equal zero.

\(\displaystyle \sin \left( n \cdot \frac {2\pi}5 \right) = 0 \iff n \cdot \frac {2\pi}5 = k \cdot \pi \iff n = \frac {5k}2 \)

In order for \(\displaystyle n\) to be an integer, \(\displaystyle 2\) must divide \(\displaystyle k\). Thus \(\displaystyle k = 2p\) where \(\displaystyle p \in \mathbb{Z}\) and consequently \(\displaystyle n = 5p\). With \(\displaystyle 0 \leqslant n \leqslant 4\) we have

\(\displaystyle 0 \leqslant 5p \leqslant 4 \iff 0 \leqslant p \leqslant 0.8 \implies p \in \{ 0 \}\)

So there is one real root and the remaining four are non-real. Similar arguments for \(\displaystyle z^9 = -4\) yield

\(\displaystyle 9v = \pi + n \cdot 2\pi \iff v = \frac {\pi}9 + n \cdot \frac {2\pi}9\)

Equating the imaginary part to zero yields

\(\displaystyle \frac {\pi}9 + n \cdot \frac {2\pi}9 = k \cdot \pi \iff n = \frac {9k-1}2\)

\(\displaystyle 2\) must divide \(\displaystyle 9k-1\) so \(\displaystyle 9k-1 = 2p\). Using \(\displaystyle 0 \leqslant n \leqslant 8\) however yields \(\displaystyle 0 \leqslant p \leqslant 8\). The equation does not have \(\displaystyle 9\) real solutions, so what went wrong?
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
I don't know if it would solve your problem or not, but in the second problem, couldn't you let $r=-\sqrt[9]{4}$, and then use $9v=n\cdot 2\pi$, more like before? Then $n=9k/2$, so $k=2p$, and hence $n=9p$. So $0\le n\le 8$ forces $0\le 9p\le 8$, and you must have $p=0$, as before.
 

zzephod

Well-known member
Feb 3, 2013
134
How many real and non-real roots does \(\displaystyle z^5 = 32\) have? \(\displaystyle z^9 = -4\)?
Descartes rule of signs says \(\displaystyle z^5-32\) has 1 positive root and 0 negative roots, hence has exactly one real root.

For \(\displaystyle z^9+4\) Descartes tell us it has zero positive and one negative real root, so again has exactly one real root.

.
 

sweatingbear

Member
May 3, 2013
91
I don't know if it would solve your problem or not, but in the second problem, couldn't you let $r=-\sqrt[9]{4}$, and then use $9v=n\cdot 2\pi$, more like before? Then $n=9k/2$, so $k=2p$, and hence $n=9p$. So $0\le n\le 8$ forces $0\le 9p\le 8$, and you must have $p=0$, as before.
How did you end up with $9v = n \cdot 2 \pi$? $9v$ must equal the argument of the number in the right-hand side i.e. $\pi + n \cdot 2\pi$, I see no other way.

Descartes rule of signs says \(\displaystyle z^5-32\) has 1 positive root and 0 negative roots, hence has exactly one real root.

For \(\displaystyle z^9+4\) Descartes tell us it has zero positive and one negative real root, so again has exactly one real root.

.
Excellent! Thanks for that perspective, however I will not rest until I have figured out the nitty and gritty details of my approach. Thanks again.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
How did you end up with $9v = n \cdot 2 \pi$? $9v$ must equal the argument of the number in the right-hand side i.e. $\pi + n \cdot 2\pi$, I see no other way.
You can take care of the minus sign in two ways; one is through the arguments to your trig functions, and the other is in the overall multiplier. So, if you say that
$$z^{9}=- 4 \left[ \cos ( 2 \pi n )+i \sin ( 2\pi n) \right],$$
I think you'll find that equivalent to
$$z^{9}= 4 \left[ \cos ( \pi+ 2 \pi n)+i \sin ( \pi+ 2\pi n) \right].$$
The first approach is much more analogous to what you did before, and might therefore be more useful.
 

sweatingbear

Member
May 3, 2013
91
You can take care of the minus sign in two ways; one is through the arguments to your trig functions, and the other is in the overall multiplier. So, if you say that
$$z^{9}=- 4 \left[ \cos ( 2 \pi n )+i \sin ( 2\pi n) \right],$$
I think you'll find that equivalent to
$$z^{9}= 4 \left[ \cos ( \pi+ 2 \pi n)+i \sin ( \pi+ 2\pi n) \right].$$
The first approach is much more analogous to what you did before, and might therefore be more useful.
Oh right of course, the trigonometric identities! But here is a follow-up question: How would the issue be resolve if it was the case that we could not take advantage of a trigonometric identity?
 

sweatingbear

Member
May 3, 2013
91
Much appreciated you could share your thoughts, Ackbach.
 

agentmulder

Active member
Feb 9, 2012
33
How many real and non-real roots does \(\displaystyle z^5 = 32\) have? \(\displaystyle z^9 = -4\)?

If you are only interested in the NUMBER of roots and not the values of each root then there is a simple answer. Since both exponents are ODD , both equations will have only one real root. That means the first equation must have 4 complex roots , all distinct , the second equation must have 8 complex roots, all distinct. Here is why...

For real number a and positive integer n , $z^n = a$ has n dstinct roots symmetrically placed on the circle centered at the origin of the complex plane with radius $ | \sqrt[n]{a} | $. There are only 2 places on this circle where you can get real roots , $ (|\sqrt[n]{a }| , 0) $ and $ (- |\sqrt[n]{a }| , 0) $ That's it. If n is ODD then you hit exactly one of these two places , if n is even you hit both of them if a is positive or none of them if a is negative.

:)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Suppose that \(\displaystyle z^n = r \) where \(\displaystyle n\in \mathbb{Z}^+, \, r\in \mathbb{R}^+\)

Then we can rewrite as follows

\(\displaystyle \Large {z = \sqrt[n]{r}e^{\frac{2\pi k}{n}\, i }}\,\,\, 0\leq k< n\)

Then look for the solutions of the equation

\(\displaystyle \sin \left( \frac{2\pi k}{n} \right) = 0 \) to find real roots .

we know that the sin has zeros for $m \pi \,\,\, , m\in \mathbb{Z}$ so we have

\(\displaystyle 2k = m \, n \) . So we conclude that if $n$ is odd , the only real solution occurs at $k=0$ , because $k< n $, hence $z = \sqrt[n]{r}$.

Try to make a general statement for $r$ is arbitrary real.