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### MC III HW4 help

MC III HW4 help

I have difficulty doing HW4 problems. Could you please provide detailed hints for problem 2.26-2.35 except 2.27? thanks

Re: MC III HW4 help

I will give some hints here.

2.26: The divisor $x^4+x^3+x^2+x+1$ is related to the 5th roots of unity.  Let $\epsilon$ be any of the imaginary 5th roots of unity (there are 4 of them, any of them will do), then $\epsilon^5=1$, and

$$x^4+x^3+x^2+x+1=(x-\epsilon)(x-\epsilon^2)(x-\epsilon^3)(x-\epsilon^4).$$

Now write

$$x^{1234} + x^{2341} + x^{3412} + x^{4123} = (x^4+x^3+x^2+x+1)Q(x) + R(x)$$

where $R(x)$ has degree at most 3, and try to plug in $\epsilon$ and $\epsilon^2$ to figure out $R(x)$.

2.28: Consider $\omega$ the cube root of unity: $\omega=-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$.  Plug in to the equation.

2.29: This can be done by expanding everything and clearing the denominators. Show some more steps and ask further if you need more help.

2.30: This is an AMC 12 problem.  The polynomial looks like $(z+i)^4$ (not quite exactly), so go from there.

2.31: Consider this as coordinate geometry.

Re: MC III HW4 help

2.32: This is similar to Example question 2.15.  In order to find the sums separately, it is easier to combine them into a complex number.

Let $A = \sum_{k=1}^{n-1}k\cos\frac{2k\pi}{n}, \quad B = \sum_{k=1}^{n-1}k\sin\frac{2k\pi}{n},$ and also define $\epsilon = \cos\dfrac{2\pi}{n} + i\sin\dfrac{2\pi}{n}$, then $A+Bi = \epsilon + 2\epsilon^2 + 3\epsilon^3 + \cdots + (n-1)\epsilon^{n-1}.$

After calculating the sum, the real part is $A$ and the imaginary part is $B$.

2.33: This is similar to Example question 2.14.  Consider binomial expansion of $(1+x)^n$, then plug in the cube root of unity, $\omega$ and $\omega^2$, using the fact that $\omega^3=1$ and $1+\omega+\omega^2=0$. To get the required terms, you need to multiply $\omega$ or $\omega^2$ on to the binomial expansions.

2.34 is similar to 2.33.

Re: MC III HW4 help

2.35(a): I will give a full solution here.  2.35(b) is similar.

The question is: Find the product: $\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}.$ Solution: Let $\displaystyle\epsilon=\text{cis}\frac{2\pi}{2n}=\text{cis}\frac{\pi}{n}$ be the imaginary root of unity of degree $2n$. Then, $1, \epsilon, \epsilon^2, \ldots, \epsilon^{n-1}, -1, \epsilon^{n+1}, \ldots,\epsilon^{2n-1}$ are all the $2n$ roots of unity of degree $2n$, and $z^{2n}-1 = (z-1)(z-\epsilon)(z-\epsilon^2)\cdots(z-\epsilon^{n-1})(z+1) \cdots(z-\epsilon^{2n-1}).$ For the root $\epsilon^k$ ($1\leq k\leq n-1$), match it with its conjugate, which is $\epsilon^{2n-k}$, and we get $\begin{array}{rcl} (z-\epsilon^k)(z-\epsilon^{2n-k}) &=&\displaystyle \left({z-\cos\frac{k\pi}{n}-i\sin\frac{k\pi}{n}}\right) \left({z-\cos\frac{k\pi}{n}+i\sin\frac{k\pi}{n}}\right)\\ \\ &=& \displaystyle z^2 - 2z\cos\frac{k\pi}{n} + 1. \end{array}$ Hence in the following factorization, $\begin{array}{cl} & z^{2n}-1 \\ =& (z-1)(z-\epsilon)(z-\epsilon^2)\cdots(z-\epsilon^{n-1})(z+1) \cdots(z-\epsilon^{2n-1}) \\ \\ =&\displaystyle (z-1)(z+1)\left({(z-\epsilon)(z-\epsilon^{2n-1})}\right) \left({(z-\epsilon^2)(z-\epsilon^{2n-2})}\right)\cdots \left({(z-\epsilon^{n-1})(z-\epsilon^{n+1})}\right) \\ \\ =& \displaystyle(z^2-1)\left({z^2 - 2z\cos\frac{\pi}{n} + 1}\right) \left({z^2 - 2z\cos\frac{2\pi}{n} + 1}\right) \cdots\left({z^2 - 2z\cos\frac{(n-1)\pi}{n} + 1}\right) \\ \\ =& \displaystyle (z^2-1)\prod_{k=1}^{n-1}\left({z^2 - 2z\cos\frac{k\pi}{n} + 1}\right). \end{array}$ Therefore, $\frac{z^{2n}-1}{z^2-1} = \prod_{k=1}^{n-1}\left({z^2 - 2z\cos\frac{k\pi}{n} + 1}\right),$ so $z^{2n-2} + z^{2n-4} + \cdots + z^4 + z^2 + 1 = \prod_{k=1}^{n-1}\left({z^2 - 2z\cos\frac{k\pi}{n} + 1}\right).$ Let $z=1$ in the above identity, $n = \prod_{k=1}^{n-1}\left({2-2\cos\frac{k\pi}{n}}\right) = \prod_{k=1}^{n-1} 4\sin^2\frac{k\pi}{2n} = 2^{2n-2}\prod_{k=1}^{n-1}\sin^2\frac{k\pi}{2n}.$ Therefore $\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n} = \frac{\sqrt{n}}{2^{n-1}}.$