Exam 558

T: for any positive integer k,

Iz^kI = IzI^k, arg(z^k) = karg(z), prove using induction

T: for any complex number z and w,

IzwI = IzI IwI and arg(zw) = arg(z) + arg(w)

D: two integers men are relatively prime

iff gcd(m,n) = 1

T P: for n > 1, the sum of nth roots of unity

is 0

D: a polynomial equation anxn + an-1xn-1 +...+a1x + ao = 0

is solvable by radicals iff all solutions can be written in terms of integers, and the coefficients using addition, subtraction, multiplication, and division, and any roots (rational roots)

D: the smallest order of a root of unity zeta

is the smallest positive integer m such that zeta^m = 1. write o(zeta) = n

D: for int a,b and positive integer n, a & b are multiplicative inverses

mod n iff a*b =_ 1modn

D: the modulus of a complex number z=a+bi is

r = sqrt a^2 + b^2. IzI = I a+bi I = sqrt a^2 + b^2

T: if abc element of complex numbers, a =/o, then the quadratic equation

say equation has two complex (not necessarily distinct) solutions

T: if n >_ 3 x^n -1 = 0 is

solvable by radicals then the regular n-sided polygon is constructible using a straight edge and compass

D: Primitive nth root of unity is a complex number zeta

such that zeta^n = 1 and ord(zeta) = n. so cis (2pi/n) is a primitive nth root of unity

T: let m= ord(zeta) where zeta is any nth root of unity

then n is a multiple of m (m div n) mIn

T: if abc are real numbers, a=/ o, and b^2 -4ac >0,

then the quadratic equation (formula) has two (not necessarily distinct) real solutions. the real number r is a solution iff r = opposite plus minus etc.

T: for a,b element of integers, not both 0. if g=gcd (a,b)

then there exists integers x and y such that ax+by = g

T P: If z is a complex number with IzI = 1

then z bar = 1/z

T: Gen demoiv form: If z is a nonzero complex number with argument theta and r is any rational number,

then z^r = {IzI^r cis (r(Thta + 2pik)} k is element of integers

T P: if zeta is a complex root of unity of order m and n is anyinteger

then zeta^n = 1 iff mIn

A T: If zeta is a complex root of unity of order m and n is any integer,

then zeta^n = 1 iff n is a multiple of m.

Division Algorithm: Given any integer n and any positive integer m,

there exists unique integers q and r such that 0 <_ r <_ m-1 and n=qm+r

D: Let n be a positive integer, 2 integers a and b are congruent module n

written a =_ b (modn) iff nI(a-b)

T: let n be a positive integer and z a nonzero complex number,

z= IzI cis theta. Write zeta = cis 2pi/n. Then nsqrtz = {( nsqrt IzI cis theta/n)(zeta^k); o <- k <- n-1

T: for any n, let zeta equal cis 2pi/n

zeta^k = cis 2pik/n. Then nsqrt 1 = {2, zeta, zeta^2, ... zeta^n-1}. Where zeta = cis 2pi/n