Class Notes (810,422)
MATH 316 (5)
John Toth (5)
Lecture

# Homework solution 2

5 Pages
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School
McGill University
Department
Mathematics & Statistics (Sci)
Course
MATH 316
Professor
John Toth
Semester
Fall

Description
Homework 1 Solutions Problem 1 ▯ 4 ▯ 4i 7 ▯ 4 + 4i▯7 + 2 + 2i 2 ▯ 2i ▯ ▯ 4▯4i 7 The second term is the conjugate of the ▯rst thus it is equal to2+2i ▯ 4 ▯ 4i▯7 ▯ 1 ▯ i▯7 27 = 27 = ((1 ▯ i) ) = (▯2i) = (▯1) 2 i = 128i 2 + 2i 1 + i 27 ▯ ▯7 Therefore we have the equation equal to 2Re2+2i = 2Re(128i) = 0 Problem 2 This is clear sinc▯z = ▯z▯▯ =▯zz▯ Problem 3 We have z 1 ▯ + i▯;z 1 ▯ ▯ i▯, so jz1+ z 2 = 2j▯j = a. Try ▯ = a=2 p p 1 p 1 1 jz1j + 2z j = ▯ + ▯ + ▯ + ▯ = 2 ▯ + ▯ = so ▯ + ▯ =2 a a 4a2 q 1 1 2 substituting back ▯ we get ▯ = 2 a ▯ a which gives us r r a i 1 2 a i 1 2 z1= + 2▯ a and z2= ▯ 2▯ a 2 2 a 2 2 a Problem 4 p i▯ p ▯1 1 i(▯▯2k▯) ▯ 3 + i = 2e 6 so we have (▯ 3 + i) 5 = p52e 6 5 for k = 0;1;2;3;4 equivalently p ▯ ▯ ▯ ▯ ▯▯ (▯ 3 + i) 5= p1 cos 5▯ + 12k▯ ▯ isin 5▯ + 12k▯ for k = 0;1;2;3;4 5 2 30 30 Problem 5 a) We can extract the term (z ▯ 1)(z + zn▯1+ ▯▯▯ + z + 1) = z(z + z n▯1+ ▯▯▯ + z + 1) ▯ 1(z + z▯1 + ▯▯▯ + z + 1) = zz + (zz n▯1 + zzn▯1+ z) ▯ (z + zn▯1 + ▯▯▯ + z) ▯ 1 = z n+1 ▯ 1 b) Use the previous result 5 4 3 2 z ▯ 1 z + z + z + z + 1 = for z 6= 1 z ▯ 1 So the roots of above is roots of z ▯ 1 = 0 so the answer is z = ci) for k = 1;2;3;4. 5 Notice there is one root missing, it is because z 6= 1 so k = 0 is not allowed. Problem 6 a) Lets compute both expressions ▯ ▯ ▯ ▯ ▯ ▯ 1 n n▯ 2▯k n n▯ 2▯k (z )m = r cis + = r cis cis for k = 0;1;▯▯▯ ;m ▯ 1 m m m m ▯ ▯ ▯ ▯ ▯ ▯ 1 n n▯ 2▯nk n n▯ 2▯nk (z ) = r cis + = rm cis cis for k = 0;1;▯▯▯ ;m ▯ 1 m m m m ▯2▯1▯ Both set o▯ val▯es are identical only when for each mis we can ▯nd k2such that it is equal to cis▯nk2 for k ;k 2 f0;1;▯▯▯ ;m▯1g.This is true since gcd(m;n) = 1 so there m 1 2 exist a;b 2 Z such that am + bn = 1, thus bn = 1 mod (m) where we can replace b by2k since b = Nm+k f2r some N 2 Z. Therefore we have nk =21 mod (m) so by multiplying each side with 1 we are able to get 2k = 1 mod (m). Therefore for any k1we can ▯nd ▯2▯k1▯ ▯2▯nk2 nk2such that they are equal up to mod (m) which implies cis m = cis m as we needed. b) ▯ ▯ ▯▯ 2 12 2▯k (1 ) = cis = ▯1 for k = 0;1;2;3 4 ▯ ▯ ▯▯ 21 1 2▯k (1 ) = 1 = cis = ▯1;▯i for k = 0;1;2;3 4 Problem 7 1 It is a circle of radius half center2d at ( ;0), since 1 1 x + y = x x ▯ x + y = 0 (x ▯ ) + y = ( ) 2
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