How many real and non-real roots?

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In summary: 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 ,...In summary, z^5 = 32 has one real root and z^9 = -4 has zero real and one negative root.
  • #1
SweatingBear
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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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  • #2
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.
 
  • #3
sweatingbear said:
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.

.
 
  • #4
Ackbach said:
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.

zzephod said:
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.
 
  • #5
sweatingbear said:
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.
 
  • #6
Ackbach said:
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?
 
  • #7
Much appreciated you could share your thoughts, Ackbach.
 
  • #8
sweatingbear said:
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.

:)
 
  • #9
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.
 

FAQ: How many real and non-real roots?

What is the difference between real and non-real roots?

Real roots are values that satisfy an equation with real numbers, meaning they can be expressed as a decimal or fraction. Non-real roots, also known as imaginary roots, are values that do not satisfy the equation with real numbers and involve the use of the complex number system.

How can I determine the number of real and non-real roots of an equation?

The number of real and non-real roots of an equation can be determined by analyzing the discriminant, which is part of the quadratic formula. If the discriminant is positive, the equation will have two real roots. If it is zero, the equation will have one real root. If it is negative, the equation will have two non-real roots.

Can an equation have both real and non-real roots?

Yes, an equation can have both real and non-real roots. This occurs when the discriminant is negative, meaning there are two non-real roots, but the equation also has real coefficients, which means there are also two real roots.

What is the importance of knowing the number of real and non-real roots of an equation?

Knowing the number of real and non-real roots of an equation helps in solving the equation and understanding its behavior. It also gives insights into the graph of the equation and its intersections with the x-axis.

How does the degree of an equation affect the number of real and non-real roots?

The degree of an equation refers to the highest power of the variable in the equation. The fundamental theorem of algebra states that a polynomial equation of degree n has exactly n complex roots, which includes both real and non-real roots. Therefore, the degree of an equation can determine the maximum number of real and non-real roots it can have.

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