# Tag Info

18

Consider the polynomial $z-i$ . . . It's good to remember the reason the complex conjugate theorem is true in the first place: the map $z\mapsto \overline{z}$ is an automorphism of the field $\mathbb{C}$, and fixes the subfield $\mathbb{R}$ (this is a fancy way of saying $\overline{r}=r$ if $r$ is real). Thus - defining the "conjugate" $\overline{p}$ of a ...

11

HINT: Note that $$\color{Green}{x^2=-x+1}.$$ By multiplying the given expression by $x,$ we can obtained $$\color{Green}{x^3=2x-1}.$$ Again by multiplying $x$ we have $$\color{Green}{x^4=-3x+2}$$ and so on..

8

$(z-i)(z-2i)$ has two complex roots, $i$ and $2i$, but their conjugates are not roots.

6

You are dealing with the https://en.wikipedia.org/wiki/Casus_irreducibilis In the case of three real irrational roots, one may express them using cube roots of complex numbers that have both real and imaginary real part. There is no way to separate the real and imaginary parts of these cube roots in any sort of closed form. All you know is that the cube ...

5

Write the polynomial in terms of $X:=t^2$ as $X^3-10X^2+31X-30$. Its roots $\alpha$, $\beta$ and $\gamma$ can be found by the rational root test. The roots of $t^6-10t^4+31t^2-30$ are then $\pm\sqrt{\alpha}$, $\pm\sqrt{\beta}$ and $\pm\sqrt{\gamma}$.

5

This isn't true. Take for example polynomial $5x^2-6x+5$. It's easy to check it has roots $\frac{3}{5}\pm\frac{4}{5}i$, which are both on the unit circle, but neither is a root of unity. However, if you restrict your attention to monic integer polynomials, then this is indeed correct: it's a result due to Kronecker, and you can see a few proofs of this ...

5

It is known that any non-zero polynomial with real coefficients has only a finite number of roots. The constant zero polynomial has any number as its root, and is the only polynomial with infinite number of roots. Now, if $S\subset\mathbb{R}$ is an infinite set, and $\forall x \in S :p(x)=q(x)$, then $p-q$ is a polynomial that has an infinite number of ...

4

Put $$f(x) = \sum_{k=1}^n \frac{a_k}{a_k - x} - 2015$$ Then, $$f'(x) = \sum_{k=1}^n \frac{a_k}{(a_k - x)^2}$$ Notice that $f$ is then increasing over each interval of its domain. Now apply IVT $n$ times.

4

$$x^2=1-x$$ $$x^4=(1-x)^2=x^2-2x+1=1-x-2x+1=2-3x$$ $$x^6=(1-x)(2-3x)=3x^2-5x+2=3(1-x)-5x+2=5-8x$$ $$x^8=(2-3x)^2=9x^2-12x+4=9(1-x)-12x+4=13-21x$$ $$x^{10}=(1-x)(13-21x)=21x^2-34x+13=21(1-x)-34x+13=34-55x$$ Therefore, $$x^{10}+x^8+x^2+1=34-55x+13-21x+1-x+1=49-77x$$ $$x^{10}+x^6+x^4+1=34-55x+5-8x+2-3x+1=42-66x$$ Your equation ...

4

Applying polynomial long division, we have $\begin{array}{rlllllllllll} &~~1x^8-1x^7+3x^6+\dots\\ \hline x^2+x-1&|x^{10}+0x^9+x^8+0x^7+0x^6+0x^5+0x^4+0x^3+1x^2+0x+1\\ &~x^{10}+x^9-x^8\\ \hline &~~~~~-x^9+2x^8+0x^7\\ &~~~~~-x^9-x^8+x^7\\ \hline &~~~~~~~~~~~~~~~~~3x^8+\dots\\ &~~~~~~~~~~~~~~~~~~\vdots \end{array}$ Similarly, ...

4

The key lemma for the proof of Fermat's last theorem for polynomials usually is the so-called Mason-Stothers Theorem: Theorem: Let $K$ be a field and $A,B,C$ nonzero polynomials in $K[T]$ with $A+B+C = 0$ and $gcd(A,B,C) = 1$. If ${\rm deg}\, A ≥ deg \, rad \, ABC$, then $A′ = B′ = C′ = 0$. Here $rad \, f$ is the radical of $f$, i.e., if ...

4

We give a completely elementary proof simply working in the ring $\mathbb{C}[t]$ and using that it is a UFD. Suppose there are some solutions of$$a(t)^3 + b(t)^3 = c(t)^3$$in $\mathbb{C}[t]$. Choose a solution $(a(t), b(t), c(t))$ such that the maximum $m > 0$ of the degrees of $a$, $b$, $c$ is minimal possible among all solutions. Clearly, this choice ...

3

Yes, your argument is correct: it is enough to prove that $t=\sqrt[3]{2}$ has no square root in $\mathbb{Q}(t)$. To show this directly, we can calculate $(a+bt+ct^2)^2 = (a^2 + 4bc) + (2c^2 + 2ab)t + (b^2 + 2ac)t^2$. This gives us three equations to be solved in rationals: $$a^2 + 4bc = 0$$ $$2c^2 + 2ab = 1$$ $$b^2 + 2ac = 0$$ If $a=0$ or $b=0$, then the ...

3

$$n^2(n - 4)(n - 3)(n - 2)(n - 1)(n + 1)^2(3n^2 - n - 6)\\= (n - 4)(n - 3)(n - 2)(n - 1)n(n + 1) \bigl[n (n+1)(3n^2 - n - 6) \bigr]$$ Now, the product of $6$ consecutive integers is always divisible by $5$. Also note that $$3n^2 - n - 6=4n(n+1)-(n^2+5n+6)=4n(n+1)-(n+2)(n+3)$$ Thus $$n^2(n - 4)(n - 3)(n - 2)(n - 1)(n + 1)^2(3n^2 - n - 6)\\= 4(n - 4)(n ... 3 I think this is not true. For example, let$$P(x)=x^2+1, Q(x)=x^2-1$$with p=-4, q=4. Clearly |P(x)|\geq |Q(x)| for all real number x, however p\ngeq q. To prove |p|\geq |q|. Assume P(x)=ax^2+bx+c, Q(x)=dx^2+ex+f, and a>0. Case 1: If p<0. Then |P(x)|\geq|Q(x)|\Rightarrow P(x)\geq |Q(x)|\Rightarrow P(x)\geq\pm Q(x) for all real ... 3 (1) Consider the function f(x)=x^3-x^2-2x+2=(x-1)(x^2-2) we have three real roots and x+y+z=-{1\over -1}=1,{1\over x}+{1\over y}+{1\over z}=-{2\over-2}=1. (2) Consider the function f(x)=x^3-x^2-40x+20. Set the derivative equal 0 we get x=4 and x=-{10\over3} are two local extrema. When x=4, f(x)<0, when x=-{10\over 3}, f(x)>0 so there ... 3 Since it is a monic polynomial, then the only possible rational roots are integer factors of its constant term. If none of those works, then it has no linear factor, so the only possible way to factor it is in the form$$(x^2+ax+b)(x^2+cx+d)$$for some a,b,c,d. Try expanding this product, equating the coefficients, and coming up with a,b,c,d that fit the ... 3 A polynomial is usually not considered as a function, which is a key distinction, though you can use a polynomial to define a function. When we have a polynomial in a variable x, x is frequently called an indeterminate. This means that it is a symbol, not a number. The way we get a function from a polynomial is called evaluation; it is the act of ... 3 It is quite difficult to find explicitly the splitting field. The Extension is not solvable. In fact the Galois group of the splitting field of$$ f(x)=x^5-4x+2$$is isomorphic to S_5. This is because the polynomial has exactly two complex roots (Study the graph of the function to determine it), hence the complex conjugation acts on the set of roots as ... 3 A good computer-sciencey approach would be to prove generally, by structural induction: Lemma. If E is any expression built from the operators + and \times, real constants, and a single variable \mathtt X, then the function that results from evaluating E with an input value bound to \mathtt X is a polynomial. Next, if p and q are ... 3 The field is given by K=\mathbb{Q}(i,\sqrt{2}). We can construct it from \mathbb{Q} by a quadratic extension L=\mathbb{Q}(\sqrt{2}) with minimal polynomial x^2-2, and then again a quadratic extension K=L(i) with minimal polynomial x^2+1. Since clearly i\notin\mathbb{Q}(\sqrt{2}), we have$$ [\mathbb{Q}(\sqrt{2},i):\mathbb{Q}(\sqrt{2}]=2. $$Now ... 3 Hint:$$4\sqrt3+8 = (\sqrt2+\sqrt6)^2$$3 Hint. As integration and the right hand side are linear in f, the quadrature formula is exact for all polynomials of degree \le 4 iff it integrates the five functions x \mapsto x^i, i = 0,\ldots, 4 exactly. Just compute the right hand side values for those five f and you are done. 3 Although there's already an answer linking to the method known as Ruffini's Rule, I detail its computation below, because it was very popular when I was in high school. It enables us to find easily the division of a polynomial by a linear term of the form  x-x_0 , even when the remainder r(x) is not equal to 0 . In your case  x_0=3  and r(x)=0. ... 3 Let \alpha=7^{\frac13} so \alpha^3=7. We need to calculate the minimal polynomial of x=\alpha+\alpha^2. One has x^3=\alpha^3+3\alpha^2\alpha^2+3\alpha\alpha^4+\alpha^6=7+3\cdot7\alpha+3\cdot7\alpha^2+49=56+21x Thus the searched minimal polynomial is p(x)=x^3-21x-56 2 (x,y)=(2,2),(-1,-1) are the only real solutions. In the following, suppose that x\not=y. Let x+y=\alpha,xy=\beta. Then we have$$\begin{align}&(2x^3+4)+(2y^3+4)=x^2(y+3)+y^2(x+3)\\&\iff 2(x^3+y^3)+8=xy(x+y)+3(x^2+y^2)\\&\iff 2(x+y)((x+y)^2-3xy)+8=xy(x+y)+3((x+y)^2-2xy)\\&\iff 2\alpha(\alpha^2-3\beta)+8=\alpha\beta ...

2

You have $x^4-8\sqrt3x^2-16=(x^2-4\sqrt3)^2-48-16=0\Rightarrow x^2=4\sqrt3\pm8\Rightarrow x=\pm\sqrt{4\sqrt3\pm8}$ Hence $\begin {cases}x_1=\sqrt{4\sqrt3 +8}\\x_2=\sqrt{4\sqrt 3-8}\\x_3=-x_1\\x_4=-x_2\end{cases}$ Make now, if you want to simplify the irrational of degree four, $\sqrt{4\sqrt 3+8}=\sqrt a+\sqrt b$ which gives the system $a+b=8$ and ...

2

For example, let's try the equation $x^3 - 2 x^2 - x + 2 = 0$: $a= 1$, $b = -2$, $c = -1$, $d=2$, $\Delta_0 = 7$, $\Delta_1 = 20$. $$C = \sqrt[3]{10 + 9 i \sqrt{3}} = \sqrt{7} e^{i\theta} = \sqrt{7}(\cos\theta + i \sin\theta)$$ where $$\theta = \dfrac{1}{3} \arctan(9 \sqrt{3}/10)$$ Then $$\Delta_0/C = \sqrt{7} e^{-i\theta} = \sqrt{7} (\cos \theta - i ... 2 Let z be a complex root. Then its conjugate \bar{z} is also a root. so$$\sum_{i=1}^{n}\frac{a_i}{a_i-z}=2015 \tag{1}$$and$$\sum_{i=1}^{n}\frac{a_i}{a_i-\bar{z}}=2015 \tag{2}$$subtracting (2) from (1) we get$$(z-\bar{z})\sum_{i=1}^{n}\frac{a_i}{(a_i-Re(z))^2+|z|^2-(Re(z))^2}=0$$which is possible only if$$z=\bar{z} so Real roots

2

Let $|z|>1,$ and by Cauchy-Schwarz inequity \begin{align} |p(z)|&=|z|^n\left|1+\sum_{j=1}^{n}\frac{a_{n-j}}{z^j} \right|\geq |z|^n\left(1-\left|\sum_{j=1}^{n}\frac{a_{n-j}}{z^j}\right| \right)\\&\geq |z|^n\left[1-\left(\sum_{j=1}^{n}|a_{n-j}|^2\right)^{1/2}\left(\sum_{j=1}^{n}\frac{1}{|z|^{2j}} ...

Only top voted, non community-wiki answers of a minimum length are eligible