H. Quotient Groups Isomorphic to the Circle Group Every complex number a + bi ma
ID: 3115727 • Letter: H
Question
H. Quotient Groups Isomorphic to the Circle Group Every complex number a + bi may be represented as a point in the complex plane. Imaginary axis cos x +isin x Real axis The unit circle in the complex plane consists of all the complex numbers whose distance from the origin is 1; thus, clearly, the unit circle consists of all the complex numbers which can be written in the form for some real number x. # 1 For each x R, it is conventional to write cis x cos x 1 sin x. Prove that eis (x +y)-(cis x)(cis y) 2 Let T designate the set {cis x: x ER), that is, the set of all the complex numbers lying on the unit circle, with the operation of multiplication. Use part 1 to prove that Tis a group. (T' is called the circle group.) 3 Prove that f(x) -cis x is a homomorphism fromR onto T. 4 Prove that kerf= {2nT. n z} = 27 5 Use the FHT to conclude that Te R/(2) 6 Prove that g(x) = cis 2TX is a homomorphism from R onto T, with kernel Z. 7 Conclude that TE R/Z.Explanation / Answer
1. cis (x + y) = cos (x+y ) + i sin (x+y)
cos (x +y) = cos x cos y - sin x sin y
sin (x + y) = sin x cos y + cos x sin y
LHS = cis (x + y) = cos x cos y - sin x sin y + i ( sin x cos y + cos x sin y)
= cos y ( cos x + i sin x) + i sin y ( cos x + i sin x)
= cos y cis x + i sin y cis x
= cis x ( cos y + i sin y)
= cis x cis y
= RHS
6. Let g: R -> T by g(x) = cis(2 x).
g is subjective since ever element of T is of the from cis(a) = cis(2 (a/2 ) = g(a/2 ) for some a R. The kernel of g is the set of x R such that cis(2 x) = 1.
This equation only holds true if and only if 2 x = 2 k for some k Z. Divide by 2 then you get ker(g) = Z. Hence, g is a homomorphism with a kernel of Z
7. By the FHT g: R -> T is a homomorphism and R is subjective to T. Since Z is the kernel of g then H is isomorphic to R/Z .
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.