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MA74 PRACTICE WITH COMPLEX NUMBERS1. Groups within the Complex NumbersDefinition: We define the set of complex numbers:C := {a + ib | a, b ∈ R, i2= −1}We define two operations on C, addition + and multiplication ·, in terms of thesame operations on R. Given z1, z2∈ C, write z1= a1+ ib1and z2= a2+ ib2.z1+ z2:= (a1+ a2) + i(b1+ b2)z1· z2:= (a1· a2− b1· b2) + i(a1· b2+ b1· a2)Notice that the expression for multiplication is “forced” by the relation i2= −1,along with the usual rules for multiplying binomials (distributive property, etc). Ifz = a + bi then we refer to a as the real part and b as the imaginary part of z.Two complex numbers are equal if they have the same real and imaginary parts,by definition. If the real or imaginary part of a complex number is zero, then wedo not write that term, as in 2 = 2 + i0 and i√3/2 = 0 + i√3/2.Definition: The modulus r of a complex number z = a + ib is its distance fromthe origin when plotted as (a, b) ∈ R2. Accordingly, we have r =√a2+ b2, and wewrite |z| = r.Homework due 4/20: Problems (3), (4), (6), (7). (You should work all of thefollowing, but hand in these four. Problem (1) was a homework problem last week.)(1) Show that C − {0} is a group under multiplication.(2) Show that C is a group under addition.(3) Show that for z = a + ib ∈ C, we have |z| = 1 if and only if there exists θ ∈ Rsuch that a = cos(θ) and b = sin(θ).(4) Look up the Taylor series for each of the real valued functions ex, sin(x), andcos(x). Then “show” by computation that for all θ ∈ R we have:eiθ= c os(θ) + i sin(θ)Deduce a famous formula relating e, i, π, and −1.(5) Show that for z ∈ C, there exists a unique r ∈ [0, ∞) and a unique θ ∈ [0, 2π)such that z = r · eiθ. (Hint: z = |z| ·z|z|, and |z|z|| = 1.)12(6) Show that if z1= r1· eiθ1and z2= r2· eiθ2, then we havez1· z2= (r1· eiθ1) · (r2· eiθ2) = (r1· r2)ei(θ1+θ2)(The LHS is defined according to the multiplication in C, and the RHS accordingto the formula in (4), which is known as Euler’s identity. You will have to use trigidentities as well as the formula for r in terms of the real and imaginary parts of z.Euler’s identity is an easy way to remember trig identities.)(7) Show that U := {z ∈ C, |z| = 1} is a group under multiplication.(8) Let m ∈ N≥2. Show thatG := {z ∈ C | ∃k ∈ {0, 1, ..., m − 1} such that z = ei2πkm}is a group under multiplication. Sketch this group in R2.(9) Show that G = {z ∈ C | zm− 1 = 0}. (You may assume the fact that apolynomial of degree m has at most m roots.) Deduce thatzm− 1 =m−1Yk=0(z − ei2πkm)Finally, equate coefficients on both sides to get:m−1Xk=0ei2πkm= 0m−1Yk=0ei2πkm=


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Berkeley MATH 74 - PRACTICE WITH COMPLEX NUMBERS

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