Homework 1 Solutions.

1. Calculate

(a) 5−i3+i,

Solution: 5−i3+i

= (5−i)(3−i)(3+i)(3−i) = 14−8i

10= 7−4i

5

(b) (1− i)100.Solution: 1− i =

√2(cos(−π

4) + i sin(−π

4))

(1− i)100 = 250(cos(−100π4

) + i sin(−100π4

) = −250

2. Let a ∈ C. Mapsl1(z) = az,

l2(z) = az̄

are called complex linear and complex antilinear correspondingly. Thinkof C as a vector space over R with a basis (0, 1) and (1, 0). Writematrices for l1 and l2. Show that each linear map L : R2 → R2 is asum of a complex linear and complex antilinear maps.

Solution: Let a = a1 + ia2, then the matrices for l1 has the form(a1 −a2a2 a1

)and the matrix for l2 has the form

(a1 a2a2 −a1

).

The matrix

(α βγ δ

)=

(α+δ2−γ−β

2γ−β2

α+δ2

)+

(α−δ2

γ+β2

γ+β2−α−δ

2

)3. Prove that if complex numbers z1 and z2 are thought of as vectors in

R2 their dot product equals Rez1z̄2.

Solution: z1 = x1 + iy1, z2 = x2 + iy2, z̄2 = x2 − iy2Rez1z̄2 = x1y1 + x2y2

4. Let S2 = {x21 + x22 + x23 = 1}, N = (0, 0, 1). Let p : S2 → C be astereographic projection. Check that p is given by the formula

p(x1, x2, x3) = x1+ix21−x3 .

Solution: The line passing through (0, 0, 1) and (x1, x2, x3) has theform (tx1, tx2, tx3 + 1− t).The stereographic projection is the intersection of this line with theplane, third coordinate is 0:

1

tx3 + 1− t = 0, t = 11−x3 .

Hence, p(x1, x2, x3) = x1+ix21−x3

5. Find all eighth roots of 1. Draw them. Check that they form a groupunder multiplication. Describe the group.

Solution: 1 = cos 0 + i sin 0, z8 = 1, zk = cos(kπ8

) + i sin(kπ8

)), k =0, . . . , 7. The picture is a regular octagon.

z0 = 1 is the identity in the group, zkzn = zm, where m is the residueof k + n mod 8, z8−k is the inverse of zk. The group is a cyclic groupwith the generator z1.

6. Let z = r(cos θ + i sin θ). Recall De Moivre’s Formula:

zn = rn(cosnθ + i sinnθ).

Use De Moivre’s Formula to express cos 3θ and sin 3θ as polynomialsin cos θ and sin θ.

Solution: cos 3θ+ i sin 3θ = (cos θ+ i sin θ)3 = cos3 θ+ 3i cos2 θ sin θ−3 cos θ sin2 θ − i sin3 θ

Hence, cos 3θ = cos3 θ − 3 cos θ sin2 θ, sin 3θ = 3 cos2 θ sin θ − sin3 θ

2