Is the angle on on Cartesian coordinate system between dots of all complex roots of polynomial with real coefficients the same?

Question

So far I realized that any polynomial with complex roots has the even number of complex roots.Because for every $(x-(a-b*i))$ there is $(x-(a+b*i)) $ in order for coefficiants to be real.That means that odd degree polynomials have at least one real root.

Because of symmetry of complex numbers roots e.g. $a-b*i$ and $a+b*i$ the angle between them is the same regarding $ x$ axis.

But is it the same for other roots. Here is example of polynomial: $x^4 + 3x +21$

Here is graph,but I am not sure are all angles the same,at least I dont see a geometric reason for that. complex roots on coordinate system for x^4 + 3x +21

And why are all complex roots on circle with semidiameter 1 on the same distance? x^10 -1 complex roots on coordinate system

Answer 1

Let $Z$ be a complex number

I'll start from the basics. $Z$ can be expressed in the form of $x+iy$. $$Z=x+iy$$ This can be plotted in 2 dimensional Argand plane,in which y axis is imaginary while x axis is real.

Here $x=a,y=b$. Notice that we can also say $$a=\sqrt{a^2+b^2}\cosθ$$ and $$b=\sqrt{a^2+b^2}\sinθ$$. Hence $$Z=\sqrt{a^2+b^2}(\cosθ+i\sinθ)$$ The above formula is called "Euler's formula", named after Leonhard Euler.

$\sqrt{a^2+b^2}$ is known as modulus of $Z$ and also written in the form $|Z|$. $θ$=Amplitude, the clockwise angle from x axis to $a+bi$

Using Taylor series, it can be proven that $$(\cosθ+i\sinθ)=e^{iθ}$$ I will link the proof below.

So $$Z=\sqrt{a^2+b^2}e^{iθ}=|Z|e^{iθ}$$ Also if you take $θ=π$, you get $$e^{iπ}=\cos(π)+i\sin(π)=-1$$ This is a famous equation regarding Euler's identity

For your first doubt, regarding $$x^4+3x+21=0$$ You might notice that the coefficient of $x^3=0$.

Formula for sum of roots in a polynomial= $\frac{-b}{a}$. Here $a$=coefficient of $x^4$,$b$=coefficient of $x^3$. Here sum of roots $=0$. Let them the roots be $α_1,α_2,α_3,α_4$. $$α_1+α_2+α_3+α_4=0$$ You have stated yourself that complex roots occur in pairs. Let's assume $α_1,α_2$ and $α_3,α_4$ are pairs. Hence $α_1+α_2=0$ and $α_3+α_4=0$. So $α_2$ is mirror image of $α_1$ w.r.t $x$ axis. Same is the case for $α_3,α_4$. Hence it appears symmetric in the Argand plane.

For your second doubt,let $$Z^n=1 $$ Thus this polynomial must have n number of roots of unity.(A principle of algebra)

$$Z= (1)^\frac{1}{n}$$

Here $|Z|=1$. This is why, it creates an unit circle. ( Eqn for circle is $x^2 + y^2 = 1$)

But $(1)^\frac{1}{n}$ can be rewritten as ${((-1)^{2k})}^{\frac{1}{n}}$ because $(-1)^{2k} , k=0,1,2,...,k∈${$0,R^+$}$ $ is always 1.

Now we substitute the value of -1 from earlier, to get

$$Z={((e^{iπ})^{2k})}^{\frac{1}{n}}$$ Or $$Z=e^{\frac{2kπi}{n}}$$

Hence,as you learnt earlier, $\frac{2kπ}{n}$ will be the amplitude of our root.

But $k$ can have multiple values, starting from $0$. But we know, there must be $n$ roots. Thus, k will have values from $0$ to $(n-1)$. So the roots will be:-

$$e^0,e^{\frac{2π}{n}},e^{\frac{4π}{n}},...e^{\frac{2(n-1)π}{n}}$$

These represents all the possible roots of $n$. For $n=10$, which is given in the Wikipedia example, there exists 10 such roots with all these different angles, which lie on the unit circle. If connected, these points form a regular polygon of n sides.

Proof of $(\sinθ+i\cosθ)=e^{iθ}$

Answer 2

I see different angles in your figure.

Here is a case where the difference is more obvious: the roots of $x^4 - 2 x^3 + 3 x^2 - 2 x + 2$ plotted on the complex plane.

If you only meant is the angle the same within each conjugate pair of roots, but can be a different angle for another pair of roots, then of course the answer is yes for the reason you have already given.

If you want to know how I came by this example, I decided what I wanted my four roots to be (remembering that I needed the ones in the lower plane to be mirror images of the ones in the upper plane), then I wrote out the polynomial in the form $(x-r_1)(x-r_2)(x-r_3)(x-r_4)$ and multiplied it out. (Technically, I took the lazy way out and had Wolfram Alpha multiply it for me, as well as generate the graph of the roots.)