User:J D Bowen/sandbox2

=Homework Ch. 7 & 8=

pg 124
We are asked to show that if the points $z_0, z_1, ..., z_n \ $ are arranged in the complex plane, the area of a non-self-intersecting polygon with these points as vertices has area

$A= \frac{1}{2} \sum_{j=0}^n \mathfrak{I}(\overline{z_j}z_{j+1}) \ $

where $\mathfrak{I}(z) \ $ is the imaginary part of z. We begin by showing

$\frac{1}{2}\overline{z_j}z_{j+1} = \frac{1}{2}(\mathfrak{R}(z_j)-i\mathfrak{I}(z_j))(\mathfrak{R}(z_{j+1})+\frac{1}{2}i\mathfrak{I}(z_{j+1})) = \frac{1}{2}\left({\mathfrak{R}(z_j)\mathfrak{R}(z_{j+1})+ i\mathfrak{R}(z_j)\mathfrak{I}(z_{j+1}) - i\mathfrak{I}(z_j)\mathfrak{R}(z_{j+1}) +\mathfrak{I}(z_j)\mathfrak{I}(z_{j+1})}\right) \ $

so

$\frac{1}{2}\mathfrak{I}(\overline{z_j}z_{j+1}) = \frac{1}{2}\left({\mathfrak{R}(z_j)\mathfrak{I}(z_{j+1}) - \mathfrak{I}(z_j)\mathfrak{R}(z_{j+1})}\right)=\frac{1}{2}\begin{vmatrix} \mathfrak{R}(z_j) & \mathfrak{I}(z_j) \\\mathfrak{R}(z_{j+1}) & \mathfrak{I}(z_{j+1}) \end{vmatrix}\ $

which of course is simply the area of the triangle $0,z_j, z_{j+1} \ $.

If the points form a star-shaped region with the origin in the center, that is, a set of points which


 * 1) includes the origin in the interior of their convex hull, and
 * 2) are all visible from the origin in a line of sight which does not leave the interior of their hull,

then last formula alone is obviously sufficient to prove the theorem, as these two conditions imply the interior of the polygon is composed of a number of triangles, with empty intersections, and one vertex at the origin.

pg 126
We are asked to complete the proof of the isoperimetric inequality by demonstrating that all of the terms in the sum

$4\sum_{j=0}^{k-1} \sin \left({ \frac{\pi j}{k} }\right) \left({ \sin \left({\frac{\pi j}{k} }\right) - \tan \left({\frac{\pi}{k}}\right)\cos\left({\frac{\pi j}{k}}\right) }\right) |\hat{z}(j)|^2 \ $

are non-negative. Of course, since $|\hat{z}(j)|^2 \ $ is always non-negative and since $\sin\left({ \frac{\pi j}{k} }\right) \ $ is non-negative for $j\in\left\{{0,1,\dots,k-1}\right\} \ $, we need only show

$\sin \left({\frac{\pi j}{k} }\right) - \tan \left({\frac{\pi}{k}}\right)\cos\left({\frac{\pi j}{k}}\right)\geq 0 \ $,

at least for $j\in\left\{{1,\dots,k-1}\right\} \ $ (we need not consider the term $j=0 \ $, since for this the coefficient of this whole expression, $\sin (\pi j/k) \ $, vanishes).

For $j=1 \ $, this expression reduces 0 fairly easily.

If we make the fairly reasonable assumption that $k> 2 \ $ (reasonable since we are dealing with a polygon with k vertices), we can reduce this expression with trigonometric identities:

$\sin \left({\frac{\pi j}{k} }\right) - \tan \left({\frac{\pi}{k}}\right)\cos\left({\frac{\pi j}{k}}\right)

=\left({ \frac{ \cos (\tfrac{\pi}{k}) }{ \cos (\tfrac{\pi}{k}) } }\right)\sin \left({\frac{\pi j}{k} }\right) - \left({ \frac{ \sin (\tfrac{\pi}{k}) }{ \cos (\tfrac{\pi}{k}) } }\right)\cos\left({\frac{\pi j}{k}}\right)

=\frac{\sin \left({ (j-1)\frac{\pi }{k} }\right)}{\cos(\tfrac{\pi }{k})} \ $

As we stated before, we do not have to consider the term $j=0 \ $ any longer, which is fortunate, because then $\sin \left({ (j-1)\frac{\pi }{k} }\right) \geq 0 $, since the sine's argument will always be less than $\pi \ $. Additionally, since $k>2, \cos(\tfrac{\pi}{k})>0 \ $ and so the quotient as a whole is >0. This completes the proof.

pg. 130
We are asked to demonstrate Eulers criterion.

Since the square-roots of 1 are $1, -1 \ (\text{mod} \ p) \ $, and since $a^{p-1}= 1 \ (\text{mod} \ p), a^{(p-1)/2} \ $ is either $1, -1 \ (\text{mod} \ p) \ $, and so Eulers criterion is equivalent to stating that a is a quadratic residue modulo p if and only if $a^{(p-1)/2} =1 \ (\text{mod} \ p) \ $.

We prove each direction separately.

($\Rightarrow \ $) Assume a is a quadratic residue modulo p. We pick k such that $k^2=a \ (\text{mod} \ p) \ $. Then

$a^{(p-1)/2}=k^{p-1}=1 \ (\text{mod} \ p) \ $.

($\Leftarrow \ $) Assume $a^{(p-1)/2} =1 \ (\text{mod} \ p) \ $. Then let $y \ $ be a primitive root modulo p, so that a can be written as $y^j \ $. In particular, $y^{j(p-1)/2} =1 \ (\text{mod} \ p) \ $. By Fermat's little theorem, $p-1 | j(p-1)/2 \ $, so j must be even. Let $k=y^{j/2} \ (\text{mod} \ p) \ $. We have $k^2=y^j= a \ (\text{mod} \ p) \ $.

pg. 134
We are asked to prove lemma 3, namely:

Suppose p and q are distinct odd primes. Set

$p^{*}=(-1)^{(p-1)/2}p \ $.

Show the Gauss sum g satisfies

$g^{q-1}=\left({ \frac{p^{*}}{q} }\right) \ (\text{mod} \ q) \ $.

The Gauss sum $g=\hat{h}_p(-1) = \sum_{a=1}^{p-1} \left({\frac{a}{p}}\right)\text{exp}\left({\frac{2\pi ia}{p}}\right) \ $, and we have use of lemma 2, which tells us

$g^2=(-1)^{(p-1)/2}p \ $.

Since q is odd, we have

$g^{q-1} = (g^2)^{(q-1)/2} = (-1)^{(p-1)(q-1)/2}p^{(q-1)/2} \ $

and of course, by Eulers criterion, we have

$\left({ \frac{p^{*}}{q} }\right) = ((-1)^{(p-1)/2}p)^{(q-1)/2} =(-1)^{(p-1)(q-1)/2}p^{(q-1)/2}$

since $p^*=(-1)^{(p-1)/2}p \in \left\{{p,q-p}\right\} \ $, neither of which can divide q.

pg. 136
The use of p and q being prime in the proof of the QRL is in the application of lemma 3 (see proof, above); when we divided p out of the congruence (trivially), and when we applied lemma 4, which relies on q being prime for the division of the binomial coefficients $qCn$ by q.

Second ex, pg. 142
The Gauss sum is defined

$G_p(x) = \sum_{k\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i k^2x}{p} }\right) \ $

Show that if p does not divide x, then

$\hat{h}_p(-x)=G_p(x) \ $

We compute the transform:

$\hat{h}_p(-x) = \sum_{a=0}^{p-1} h_p(a)\text{exp}\left({\frac{2\pi i x a}{p} }\right) = \sum_{a=1}^{p-1} \left({\frac{a}{p}}\right) \text{exp}\left({\frac{2\pi i x a}{p} }\right) \ $

which is simply

$= \sum_{a=1}^{p-1} \begin{cases} \text{exp}\left({\frac{2\pi i x a}{p} }\right) \text{ if } \exists k:k^2=a \pmod{p}\\ -\text{exp}\left({\frac{2\pi i x a}{p} }\right) \text{ else } \end{cases} \ $

(From here on we simply use $m\in\mathbb{Z}_p \ $, with the understanding the sum omits $m=0 \ $. We also adopt the notation $\text{QR}=\left\{{a\in\mathbb{Z}_p:\exists k\in\mathbb{Z}_p \ \text{s.t.} \ k^2=a}\right\}$ and $\text{NQR}=\mathbb{Z}_p-\text{QR}$). Since exactly half of $\mathbb{Z}_p$ are quadratic residues and each has two possible solutions,

$G_p(x)=2\sum_{a\in\text{QR}} \text{exp}\left({ \frac{2\pi i ax}{p} }\right) \ $

and so

$G_p(x)-\hat{h}_p(-x)

=\left({ 2\sum_{a\in\text{QR}} \text{exp}\left({ \frac{2\pi i ax}{p} }\right)}\right) - \left({ \sum_{a\in\text{QR}} \text{exp}\left({ \frac{2\pi i ax}{p} }\right) + \sum_{b\in\text{NQR}} -\text{exp}\left({ \frac{2\pi i bx}{p} }\right)}\right) \ $

$=\sum_{a\in\text{QR}} \text{exp}\left({ \frac{2\pi i ax}{p} }\right)-\sum_{b\in\text{NQR}} -\text{exp}\left({ \frac{2\pi i bx}{p} }\right)

=\sum_{c\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i cx}{p} }\right)=\sum_{d\in\mathbb{Z}_p} \text{exp}\left({ \frac{2\pi i d}{p} }\right)=0 \ $

with the second to last equality possible because p is prime and the last equality because a sum of the roots of unity is 0.