MHB How many real and non-real roots?

[SweatingBear]

How many real and non-real roots does $$z^5 = 32$$ have? $$z^9 = -4$$?

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

$$r = 2 \\ 5v = n \cdot 2\pi \iff v = n \cdot \dfrac {2\pi}5$$

So all roots are given by

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

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

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

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

$$0 \leqslant 5p \leqslant 4 \iff 0 \leqslant p \leqslant 0.8 \implies p \in \{ 0 \}$$

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

$$9v = \pi + n \cdot 2\pi \iff v = \frac {\pi}9 + n \cdot \frac {2\pi}9$$

Equating the imaginary part to zero yields

$$\frac {\pi}9 + n \cdot \frac {2\pi}9 = k \cdot \pi \iff n = \frac {9k-1}2$$

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

 

[Ackbach]

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]

How many real and non-real roots does $$z^5 = 32$$ have? $$z^9 = -4$$?

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

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

.

 

[SweatingBear]

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 $$z^5-32$$ has 1 positive root and 0 negative roots, hence has exactly one real root.

For $$z^9+4$$ 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]

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]

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]

Much appreciated you could share your thoughts, Ackbach.

 

[agentredlum]

How many real and non-real roots does $$z^5 = 32$$ have? $$z^9 = -4$$?

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.

:)

 

[alyafey22]

Suppose that $$z^n = r $$ where $$n\in \mathbb{Z}^+, \, r\in \mathbb{R}^+$$

Then we can rewrite as follows

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

Then look for the solutions of the equation

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

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

$$2k = m \, n $$ . 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.