Finding real and complex roots to a 7 degree polynomial

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Finding real and complex roots to a 7 degree polynomial

Postby Guest » Thu Feb 28, 2019 7:49 am

Find all complex solutions of the equation z^7 + 27(z^4) − 2(z^3) − 54 = 0, justifying your answer. How many distinct complex solutions are there? How many distinct real solutions are there? How many distinct rational solutions are there?
Hint: the polynomial has a factorisation of the form ((z^n) + a)((z^m) + b) for a, b ∈ Z.

I've managed to simplify it down to ((z^4)-2)((z^3)+27) so essentially I've arrived at n=4, a=-2, m=3, b=27, but I don't know where to go from there.
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Re: Finding real and complex roots to a 7 degree polynomial

Postby Guest » Wed May 22, 2019 8:05 am

You've done the hard work factoring it as [tex](z^4- 2)(z^3- 27)[/tex]! If that is equal to 0 then either [tex]z^4- 2= 0[/tex], which means that [tex]z^4= 2[/tex], or [tex]z^3- 27= 0[/tex], which means that [tex]z^3= 27[/tex].

From [tex]z^4= 2[/tex] take the square roots of both sides to get [tex]z^2= \sqrt{2}[/tex] and [tex]z^2= -\sqrt{2}[/tex].

Taking square roots of both sides of [tex]z^2= \sqrt{2}[/tex] we get [tex]z= \sqrt[4]{2}[/tex] and [tex]z= -\sqrt[4]{2}[/tex]. Those can also be written as [tex]z= 2^{1/4}[/tex] and [tex]z= -2^{1/4}[/tex].

Taking square roots of both sides of [tex]z^2= -\sqrt{2}[/tex] gives the same thing except that we also have the "square root of -1" which is "i". The two roots are [tex]z= i\sqrt[4]{2}[/tex] and [tex]z= -i\sqrt[4]{2}[/tex]. Those can also be written as [tex]z= 2^{1/4}i[/tex] and [tex]z= -2^{1/4}i[/tex].

There are four roots, two real and two imaginary.

[tex]z^3= 27[/tex] is a little harder. Any complex number, which includes real numbers, can be written in several different ways "a+ bi" is the same as "[tex]r(cos(\theta)+ isin(\theta))[/tex]" where [tex]r= \sqrt{a^2+ b^2}[/tex] and [tex]theta= arctan\left(\frac{b}{a}\right)[/tex]. The idea is that since "a+ bi" involves two real numbers, a and b, we can associate the number with the point (a, b) in a Cartesian graph. The "r" and "[tex]\theta[/tex]" for that point are the polar coordinates for the point.

At any rate, the advantage of using [tex]z= r(cos(\theta)+ isin(\theta))[/tex] is that [tex]z^n= r^n(cos(n\theta)+ sin(n\theta))[/tex] and. That is, "r" is taken to the power while [tex]\theta[/tex] is multiplied by the power.

And that works for fractional powers as well! The third root is the 1/3 root so we can apply those formulas with n= 1/3.

27, a real number, is of the form 27+ 0i so [tex]\theta= arctan(0/27)= arctan(0)= 0[/tex] (that is true of any real number). We can write [tex]27= 27(cos(0)+ isin(0))[/tex]. First, [tex]27= 3^3[/tex] so we can immediately write [tex]z^3= 3^3(cos(0)+ isin(0))[/tex].
Taking the third root of both sides immediately gives us [tex]z= 3(cos(0/3)+ isin(0/3))= 3(cos(0)+ isin(0))= 3[/tex].

But, now, the cool part! Adding [tex]2\pi[/tex] to [tex]\theta[/tex] does not change [tex]cos(\theta)[/tex] or [tex]sin(\theta)[/tex] so we can also write [tex]z^3= 3^3(cos(2\pi)+ isin(2\pi))[/tex] (or [tex]z^3= 3^3e^{2\pi i}[/tex]) but it does change when we divide that angle by 3: [tex]z= 3(cos(2\pi/3)+ isin(2\pi/3)= 3(-1/2+ i\sqrt{3}/2)= ].

Add [tex]2\pi[/tex] to [tex]\theta[/tex] again! [tex]z^3= 3^3(cos(4\pi)+ isin(4\pi))[/tex] so [tex]z= 3(cos(4\pi/3)+ isin(4\pi)= 3(-1/2- i\sqrt{3}/2))= -\frac{3}{2}- \frac{\sqrt{3}}{2}i[/tex]. (Yes, those last two are "complex conjugates.)

So the 7 roots of [tex](z- 2)^4(z- 27)^3= 0[/tex] are
[tex]\sqrt[4]{2}[/tex]
[tex]-\sqrt[4]{2}[/tex]
[tex]i\sqrt[4]{2}[/tex]
[tex]-i\sqrt[4]{2}[/tex]
[tex]3[/tex]
[tex]-\frac{3}{2}+ \frac{\sqrt{3}}{2}i[/tex] and
[tex]-\frac{3}{2}- \frac{\sqrt{3}}{2}i[/tex].
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Re: Finding real and complex roots to a 7 degree polynomial

Postby anaallen » Fri Oct 25, 2019 11:57 pm

:) wow thanks a lot...
You are really great..

short life

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