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Lösningar till tentamen.
partial derivatives are identically zero, then u must
be constant.
Analytic Functions I,
2014-10-25
3. We first note the expansion
3
1 − ez = 1 − (1 + z 3 + O(z 6 )) = −z 3 + O(z 6 ).
It follows that
1. The multivalued inverse sine-function is given by
sin−1 z = −i log iz + (1 − z 2 )1/2 .
z
z
=
3 =
3
z
−z + O(z 6 )
1−e
1
1
1
= − 2 + O(z).
z 2 1 + O(z 3 )
z
The branch which coincides with arcsin on ] − 1, 1[ We see that the singularity is a pole of order 2.
is the one obtained by using the principle branch of Since there is no 1/z-term in the expansion, the
the logarithm to define both the log and the power Residue is 0.
1/2 in the above formula, i.e.
2
1
Arcsin z = −iLog iz + e 2 Log(1−z )
4. a) We put g(z) = 11z 3 . The clearly |g(z)| = 11
We obtain
2
1
Arcsin(±i) = −iLog ∓1 + e 2 Log(1−(±i) ) =
−iLog ∓i + e
1
2 Log 2
= −iLog(∓1 +
√
−iLog( 2 ∓ 1),
√
2) =
which can in fact also be written as
√
±iLog( 2 + 1).
−
when |z| = 1. We also note that note |f (z)−g(z)| =
|z 5 − 2z 4 + 6z 2 + 1| ≤ |z 5 | + | − 2z 4 | + |6z 2 | + |1| =
1 + 2 + 6 + 1 = 10 when |z| = 1.
It follows from Rouché’s theorem that f (z) and
g(z) have the same number of zeros inside the circle
|z| = 1. Since g(z) has a triple zero at the origin it
follows that f (z) has 3 zeros inside |z| = 1.
b) Here we use g(z) = 10z 3 which clearly satisfies
|g(z)| = 10 when |z| = 1. However, in this case the
simple estimate |f (z)−g(z)| = |z 5 −2z 4 +6z 2 +1| ≤
|z 5 | + | − 2z 4 | + |6z 2 | + |1| = 1 + 2 + 6 + 1 = 10 won’t
do, because we only get |f (z)−g(z)| ≤ |g(z), but we
need strict inequality to apply Rouché’s theorem.
2. a) If v is a harmonic conjugate of u, then
f = u + iv and hence also ef = eu+iv is analytic. The solution to the problem consists in observing
5
4
Since eu cos v = Re(eu+iv ), it follows that eu cos v that we actually have a strict inequality |z − 2z +
2
5
4
2
6z +1| < |z |+|−2z |+|6z |+|1|. This is because
is harmonic.
for any (non-zero) complex numbers z1 , z2 , . . . , zn ,
b) We compute the Laplacian of eu :
we only have equality in |z1 + z2 + . . . + zn | ≤ |z1 | +
|z2 | + . . . + |zn | if all the numbers have the same
∂ u
∂2 u
0 u
00 u
0 2 u
argument. But −2z 4 , 6z 2 and 1 can not have the
e = ux e ,
e = uxx e + (ux ) e ,
2
∂x
∂x
same argument for any z ∈ C. In fact, 6z 2 and 1
can only have the same argument if z is real. But
∂2 u
∂ u
00 u
0 2 u
then the arguments of −2z 4 and 1 differ by π.
e = u0y eu ,
e
=
u
e
+
(u
)
e
,
yy
y
∂y
∂y 2
⇒ ∆(eu ) = (∆u)eu + ((u0x )2 + (u0y )2 )eu
If now ∆(eu ) = ∆u = 0, then it follows that
((u0x )2 + (u0y )2 )eu = 0 ⇒ (u0x )2 + (u0y )2 = 0 ⇒
u0x = u0y = 0 at all points in the plane. But if the
5. Using cos x = Re eix , we can rewrite the integral:
Z ∞
Z ∞
eix dx
cos x dx
=
Re
.
2 2
2 2
−∞ (1 + x )
−∞ (1 + x )
The second integral can be calulated by Residue
calculus in the upper half plane:
where the principal branch of the logarithm gives
the square root which maps to the target region.
Hence the problem is solved by the composition of
If we to the interval IR = [−R, R] add the semi- these maps, i.e.
circle γR in the upper half-plane with center at
the origin and radius R, we get a closed contour
f (z) = z 2 + (z 4 − 1)1/2 .
ΓR (which we consider to be oriented in the positive direction) over which the integral of f (z) =
eiz /(4 + z 4 ) can be computed. Since the integral of An alternative solution consists in observing that
f (z) over γR tends to zero by standard theory (the the Möbius map
degree of the denominator is 4 and the numerator
1+z
,
g(z) =
is bounded), and since f (z) has one pole of order
1−z
two in the upper half-plane at the point i, we get
which maps the upper half plane onto itself, maps
Z
Z ∞
eiz dz
eix dx
= lim
= 2πi (Res(i)) . the first quadrant onto the given target region. In
2 2
R→∞ Γ (1 + z 2 )2
fact, the positive imaginary axis is mapped onto
−∞ (1 + x )
R
the upper half of the unit circle.
The residue is computed as
d
eiz
2
(z − i)
Res(i) = lim
=
7. By the Cauchy’s estimates in a circle of radius
z→i dz
(1 + z 2 )2
R around the origin we get
iz
eiz
d
e (iz − 3)
1
lim
= lim
=
(C + DRN )n!
z→i dz
z→i
(z + i)2
(z + i)3
2ei
→ 0 if n > N.
|f (n) (0)| ≤
Rn
which gives that the original integral is equal to
(the real part of)
Since all derivatives of order larger than N vanish
at the origin, f (z) must be a polynomial. If remains
1
π
to show that a periodic polynomial must be a con2πi ·
= .
2ei
e
stant. Suppose that f (z) = w for some z and w.
Then also f (z + ka) = w for all integers k. Since a
polynomial which is not constant can have only a
3
6. a) The map z 7→ z maps the first quadrant finite number of roots, it follows that f (z) = w for
onto the three first quadrants. To get the requi- all z which proves that f (z) is constant. Q.E.D.
red map, we just have to turn the complex plane
counterclockwise by 90 degrees, which is achieved /Martin Tamm, 141025/
by multiplication by i. Thus, the problem is solved
by the conformal map
f (z) = iz 3 .
b) In this case, we can start by mapping the first
quadrant onto the upper half plane by the map z 7→
z 2 . We also know that the map
1
1
z=
w+
2
w
takes the gives target region onto the upper half
plane. Solving for w, we get
w = z + (z 2 − 1)1/2 ,