MAT389 Fall 2014, Midterm 1
Oct 8, 2014
• Please justify your reasoning. Answers without an explanation will not be given any credit.
• The maximum total mark in the exam is 15 points.
• Do not spend too much time on any particular problem. If you get stuck in one, go to the next.
1.1 [1pt] Write the following complex numbers in the form z = reiθ .
√
(i) [0.5pt] 3 + i,
(ii) [0.5pt] (1 + i)8 .
(i) The modulus of this complex number is
r √ 2
3 + 12 = 2.
|z| =
√
To find an argument, notice that there are two directions for which the tan θ is 1/ 3: θ = π/6
in the first quadrant, θ = 7π/6 in the third. Since both the real and imaginary part of z are
√
positive, z lies in the first quadrant. Thus, 3 + i = 2eiπ/6 .
(ii) The easy way of calculating this is to first express 1 + i in polar form. Reasoning as above
√
yields 1 + i = 2 eiπ/4 . Hence,
8 √ 8 8
√
2eiπ/4 =
2
eiπ/4 = 24 e2πi = 16.
(1 + i)8 =
1.2 [3pt] Find all solutions to each of the following equations and inequalities, and plot them.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
[0.5pt]
[0.5pt]
[0.5pt]
[0.5pt]
[0.5pt]
[0.5pt]
Re(z + 3) = Im(z − i),
Re(z 2 + 4) = 0.
(z + i)4 + 1 = 0,
1 + z + z 2 + z 3 = 0,
0 ≤ Arg z ≤ π/2, 0 < |z| < 1,
Re z 2 < 0.
(i) Expressing z = x + iy, we get
Re(z + 3) = Re(x + iy + 3) = x + 3,
Im(z − i) = Im(x + iy − i) = y − 1;
that is, the solution set of the equation in the statement is the line y = x + 4.
(ii) Writing z in terms of its real and imaginary parts, we find
Re(z 2 + 4) = Re (x + iy)2 + 4 = Re x2 − y 2 + 2xyi + 4 = x2 − y 2 + 4
The equation x2 − y 2 + 4 = 0 describes a hyperbola.
(iii) Rewriting the equation given as (z + i)4 = −1, we have that z = −i + (−1)1/4 , where (−1)1/4
denotes any fourth root of −1. We find the latter by expressing −1 in polar form:
−1 = e−iπ
⇒
(−1)1/4 = eiπ(k/2−1/4) ,
k = 0, 1, 2, 3.
Hence, the solutions of the original equation are the four points
√

2)i
−iπ/4 = 1−(1+
√

−i
+
e
k=0


√2


1+(1−
2)i
−i + eiπ/4 =
√
k=1
2 √
z=
−1+(1−
2)i

√
k=2
−i + e3iπ/4 =


2√


−i + e5iπ/4 = −1−(1+
2)i
√
k=3
2
(iv) We can write
1 + z + z2 + z3 =
1 − z4
1−z
The roots of the numerator in this fraction are the fourth roots of unity: z = ekπi/2 . The roots
of the polynomial on the left hand side are hence all of those, except for z = 1.
(v) We have 0 ≤ Arg z ≤ π/2 if and only if z belongs to the (closed) first-quadrant minus the
origin, while 0 < |z| < 1 is the punctured neighborhood of the origin of radius 1. The desired
subset is the intersection of these two.
(vi) Since Re z 2 = x2 − y 2 = (x + y)(x − y), the inequality Re z 2 < 0 is satisfied if either both
x + y > 0 and x − y < 0, or both x + y < 0 and x − y > 0. In the first case we have −y < x < y
—which can only be satisfied for positive values of y—, which yields the top wedge in the picture;
in the second, we get y < x < −y —which implies y < 0—, that is, the bottom wedge.
1.3 [1.5pt] Prove that
az + b =1
¯bz + a
¯
whenever |z| = 1.
Hint: if |z| = 1, then z = eiθ for some θ ∈ R.
If z = eiθ , we have z −1 = e−iθ = z¯. Hence,
iθ
ae + b 1 aeiθ + b −iθ aeiθ + b −iθ |aeiθ + b|
=
= e
= e = 1,
¯beiθ + a
iθ
¯ eiθ a
¯e−iθ + ¯b aeiθ + b
ae + b
where in the last step we have used the fact that the modulus of a complex number coincides with
that of its complex-conjugate.
1.4 [2pt] Find the unique M¨
obius transformation, T , that takes i 7→ 0, 1 7→ 1, ∞ 7→ i.
Hint: remember that the unique M¨
obius transformation that takes z1 7→ 0, z2 7→ 1,
z3 7→ ∞ is given by the cross-ratio
z 7→
z − z1 z 2 − z 3
z2 − z1 z − z3
Let
z−i
1−i
be the unique M¨obius transformation taking i 7→ 0, 1 7→ 1, ∞ 7→ ∞. On the other hand, the unique
M¨obius transformation taking 0 7→ 0, 1 7→ 1, i 7→ ∞ is
U1 (z) =
U2 (w) =
w−0 1−i
·
1−0 w−i
Equating U1 (z) = U2 (w) and isolating w, we find
w = T (z) = (U2−1 ◦ U1 )(z) =
z−i
1 − iz
1.5 [1.5pt] What is the image of the line Im z = −1 + Re z under the M¨
obius transformation
f (z) = 1/(z + i)?
If w = 1/(z + i), then
z=
1
−i
w
=⇒
x=
u2
u
,
+ v2
y=
−v
−1
+ v2
u2
Subsituting these values of x and y in terms of u and v in the equation y = −1 + x of the line, we find
y=
−v
u
− 1 = −1 + 2
2
+v
u + v2
u2
=⇒
−v
u
= 2
2
+v
u + v2
u2
=⇒
−v = u
In this last line, we have canceled the factor u2 + v 2 , in the understanding that the point w = 0 is
the image of the point z = ∞.
1.6 [2pt] Let S = {z ∈ C× | π/4 ≤ Arg z ≤ π/2} ⊂ C.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
[0.5pt] Classify all points in C as interior, exterior or boundary with respect to S.
[0.5pt] What are the accumulation points of S?
[0.25pt] Is S open? Is S closed?
[0.25pt] Is S bounded?
[0.25pt] Is S connected?
[0.25pt] Is S a domain? Is S a region?
Note: for (i) and (ii), you do not need to justify your answer.
(i) The points interior to S are those in the open wedge given by π/4 < Arg z < π/2. The boundary
consists of the two rays Arg z = π/4 and Arg = π/2, and the origin. The rest of the complex
plane makes up the exterior of S.
(ii) Interior points are always accumulation points. In this case, all boundary points are accumulation
points too.
(iii) S is neither open nor closed, for it includes some of its boundary points, but not all: the origin
is a boundary point of S, but it does not belong to S.
(iv) No: S contains points of arbitrarily large modulus.
(v) Yes: the straight line between any two points in S is contained in S, and so it is a path in S
connecting them.
(vi) S is not a domain, for it is not open. It is indeed a region, since the interior of S is a domain (it
is open and connected), and S can be built from its interior by adding some boundary points.
1.7 [1pt] Express f (z) = z 3 + 2z + 1 as f (z) = u(x, y) + iv(x, y).
i3
Write z = x + iy, expand the powers using the binomial theorem and recall that i2 = −1 and
= −i:
f (z) = z 3 + 2z + 1 = (x + iy)3 + 2(x + iy) + 1
= x3 + 3ix2 y − 3xy 2 − iy 3 + 2x + 2iy + 1
= (x3 − 3xy 2 + 2x + 1) + i(3x2 y − y 3 + 2y)
1.8 [3pt] For each of the functions below, determine
1. where the Cauchy-Riemann equations are satisfied, and
2. where the function is differentiable.
(i) [0.5pt+0.5pt] f (z) = ex (cos y + i sin y)
(ii) [0.5pt+0.5pt] f (z) = e−x (cos y + i sin y)
(iii) [0.5pt+0.5pt] f (z) = x2 + y 2 − x + iy
(i) We calculate
ux = ex cos y,
uy = −ex sin y,
vx = ex sin y,
vy = ex cos y.
The Cauchy–Riemann equations, ux = vy and vx = −uy , reduce to tautological statements;
hence, they are satisfied everywhere. Moreover, these partial derivatives are continuous over
the whole complex plane, and so f (z) is differentiable at every z ∈ C.
(ii) The first partial derivatives of the real and imaginary parts of f are
ux = −e−x cos y,
uy = −e−x sin y,
vx = −e−x sin y,
The Cauchy–Riemann equations, ux = vy and vx = −uy , imply
−e−x cos y = e−x cos y
=⇒
cos y = 0
vy = e−x cos y.
and
−e−x sin y = e−x sin y
=⇒
sin y = 0,
respectively. Since sine and cosine have no common zeroes, the Cauchy–Riemann equations are
not satisfied at any point of the complex plane, and so f (z) is nowhere differentiable.
(iii) We have
ux = 2x − 1, uy = 2y, vx = 0, vy = 1.
The Cauchy–Riemann equations, ux = vy and vx = −uy , say that
2x − 1 = 1
=⇒
x=1
and
0 = −2y
=⇒
y = 0,
respectively; i.e., they are satisfied only at the point z = 1. There the first partial derivatives of
u and v are continuous, from which we conclude that f (z) is differentiable only at z = 1.