# Finding real and complex roots to a 7 degree polynomial

Teacher's questions

### Finding real and complex roots to a 7 degree polynomial

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.
Guest

### Re: Finding real and complex roots to a 7 degree polynomial

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

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

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

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

There are four roots, two real and two imaginary.

$$z^3= 27$$ is a little harder. Any complex number, which includes real numbers, can be written in several different ways "a+ bi" is the same as "$$r(cos(\theta)+ isin(\theta))$$" where $$r= \sqrt{a^2+ b^2}$$ and $$theta= arctan\left(\frac{b}{a}\right)$$. 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 "$$\theta$$" for that point are the polar coordinates for the point.

At any rate, the advantage of using $$z= r(cos(\theta)+ isin(\theta))$$ is that $$z^n= r^n(cos(n\theta)+ sin(n\theta))$$ and. That is, "r" is taken to the power while $$\theta$$ 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 $$\theta= arctan(0/27)= arctan(0)= 0$$ (that is true of any real number). We can write $$27= 27(cos(0)+ isin(0))$$. First, $$27= 3^3$$ so we can immediately write $$z^3= 3^3(cos(0)+ isin(0))$$.
Taking the third root of both sides immediately gives us $$z= 3(cos(0/3)+ isin(0/3))= 3(cos(0)+ isin(0))= 3$$.

But, now, the cool part! Adding $$2\pi$$ to $$\theta$$ does not change $$cos(\theta)$$ or $$sin(\theta)$$ so we can also write $$z^3= 3^3(cos(2\pi)+ isin(2\pi))$$ (or $$z^3= 3^3e^{2\pi i}$$) but it does change when we divide that angle by 3: $$z= 3(cos(2\pi/3)+ isin(2\pi/3)= 3(-1/2+ i\sqrt{3}/2)= ]. Add [tex]2\pi$$ to $$\theta$$ again! $$z^3= 3^3(cos(4\pi)+ isin(4\pi))$$ so $$z= 3(cos(4\pi/3)+ isin(4\pi)= 3(-1/2- i\sqrt{3}/2))= -\frac{3}{2}- \frac{\sqrt{3}}{2}i$$. (Yes, those last two are "complex conjugates.)

So the 7 roots of $$(z- 2)^4(z- 27)^3= 0$$ are
$$\sqrt[4]{2}$$
$$-\sqrt[4]{2}$$
$$i\sqrt[4]{2}$$
$$-i\sqrt[4]{2}$$
$$3$$
$$-\frac{3}{2}+ \frac{\sqrt{3}}{2}i$$ and
$$-\frac{3}{2}- \frac{\sqrt{3}}{2}i$$.
Guest

### Re: Finding real and complex roots to a 7 degree polynomial

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

short life

anaallen

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