Math 249
Short Solutions to Assignment 1
[1] Let zk = r(cos φk + i sin φk ) be the polar form of zk , k = 1, 2, 3.
Suppose that z1 , z2 , z3 are vertices of an equilateral triangle. Then φ2 =
φ1 + 2π/3, φ3 = φ1 + 4π/3, and so
z1 + z2 + z3 = r(cos φ1 + i sin φ1 ) [1 + cos(2π/3) + cos(4π/3)
+i(sin(2π/3) + sin(4π/3))] = 0.
On the other hand, suppose that z1 + z2 + z3 = 0. Then
cos φ1 + cos φ2 = − cos φ3 ,
(1)
sin φ1 + sin φ2 = − sin φ3 .
(2)
Squaring and adding these equations we get
cos(φ1 − φ2 ) = −1/2,
and so
|φ1 − φ2 | = 2π/3 or 4π/3.
(3)
We can also write (1) and (2) as
cos φ1 + cos φ3 = − cos φ2 ,
sin φ1 + sin φ3 = − sin φ2 .
and in the same way we get that
|φ1 − φ3 | = 2π/3 or 4π/3.
(4)
The equations (3) and (4) imply that z2 , z2 , z3 are vertices of an equilateral
triangle (explain why).
√
[2] (a) (−1 + i 3)60 = 260 .
(b) The solutions of the equation
z 4 + 8(1 + 2i)z 2 − 33 + 56i = 0
1
are z1,2 = ±(1 − 2i), z3,4 = ±(2 − 3i).
(c) w = 1 + i.
[3] The imaginary part of x + i is in the upper half-plane. Let θ be the
argument of x + i such that 0 < θ < π. Then, by the basic trigonometry,
x = cot θ,
and
x + i = |x + i|(cos θ + i sin θ) =
(6) implies that
x−i=
and so
√
√
(5)
1 + x2 (cos θ + i sin θ).
(6)
1 + x2 (cos θ − i sin θ),
√
(x + i)n + (x − i)n = 2 1 + x2 cos nθ.
Hence, (x + i)n + (x − i)n = 0 if and only if cos nθ = 0. Since 0 < θ < π, we
have that
π
k = 0, 1 · · · , n − 1.
nθ = + kπ,
2
The relation (5) implies that the real solutions of the equation
(x + i)n + (x − i)n = 0
are
xk = cot
2(k + 1)π
,
2n
k = 0, 1, · · · , n − 1.
[4]
z − w z − w 1 z − w
1 − z̄w = 1 − z̄w z = z − w = 1.
[5] Hint: Let
z=
cos θ + i sin θ
.
sin θ
Then
1+
cos θ cos 2θ
cos nθ
+
+ ··· +
= Re(1 + z + · · · + z n )
2
sin θ
sinn θ
sin θ
n+1
z
−1
= Re
z−1
2