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Lecture

# Homework solution 2

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McGill University

Mathematics & Statistics (Sci)

MATH 316

John Toth

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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