# Assignment 3 - Complex Analysis MATH 440/508 – M.P. Lamoureux Solutions

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```Assignment 3 - Complex Analysis
MATH 440/508 – M.P. Lamoureux
Solutions
Note: These are sketches of the main ideas for solutions.
1. Page 106, # 16. Continuity of zeros.
Proof: We use Roche’s theorem to show that f (z) = f (z)+g(z) has only
one zero for small enough. Let M be the max of |g(z)| on the unit circle,
m the minimum of |f (z) on the same circle. Both are finite and non-zero by
compactness. (We assume g is not identically zero, otherwise the problem is
trivial.) Then for || smaller than m/M we have that
|f (z)| &gt; |g(z)|
on the unit circle. So by Roche’s, the two functions f and f + g have the
same number of zeros in the disk, which is 1.
To show continuity of the zero, the only challenge here is to be careful in
deciding what continuity means. It means sufficiently small changes in will
give small changes in the zero, z . After that, it is just Roche’s again.
We will show continuity at 0 . So f0 has a single zero at z0 in the unit
disk. Fix any r &gt; 0. We will find a δ so that if changes by less than δ, then
the zero moves no further than r.
First, if r is too big, reduce it to something smaller than the distance
from z0 to the unit circle. On the circle of radius r centred at f0 , let M be
the max of |g(z)| on that small circle, and m the minimum of f0 . Again,
both are non-zero, so by the argument above, the function f = f0 + ( − 0 )g
has exactly one zero inside this small disk, if | − 0 | is smaller than m/M .
End of proof. But notice we can also see this directly by the integral
formula
Z
zf0 (z)
dz
z =
C f (z)
1
where we integrate over the unit circle. For small enough, the denominator does not vanish, so the integrand is uniformly continuous and thus the
integral is a continuous function.
2. Page 106, # 19. Max modulus principle for harmonic functions.
(a). If u is harmonic and non-constant, at an interior maximum (x0 , y0 )
we can find a harmonic conjugate v in a small disk around (x0 , y0 ). The
holomorphic function f (z) = u(z) + iv(z) maps that disk into the vertical
strip Real(z) ≤ M ax(u), and touches the boundary. Which contradicts the
open mapping property of holomorphic functions. Another way to see this
is that | exp(f (z))| = exp(u(z)) takes its maximum at the same place that u
does, since exp is strictly increasing on the real line. Ad that would contradict
the maximum modulus principle for holomorphic functions.
Looking at −u, we also conclude a harmonic function can’t have an interior minimum.
(b). When Ω is compact and u continuous on it, it does achieve its max
by the maximization property of continuous functions on compact sets. By
part (a), that max must be on the boundary. So we get that supΩ u ≤ supΩ u.
(The book actually asks to prove this for the absolute values. I think that is
a typo, but if you really must, break up into two cases, where 1) the absolute
maximum occurs at a large positive value for u, or 2) the abs max occurs at
a large negative value for u.
3. Page 109, # 3. Laurent series expansions.
Following the hint, we have
f (z) =
1 Z f (w)
1 Z f (w)
dw −
dw.
2πi C2 w − z
2πi C1 w − z
In the first integral, we we have |z − z0 | &lt; |w − z0 |, since w is on the big
circle containing z, so we expand the factor 1/(w − z) as a geometric series
1
(w − z0 )(1 −
z−z0
)
w−z0
=
∞
1 X
z − z0 n
(
) .
w − z0 n=0 w − z0
By uniform convergence, we can take the sum outside the integral and get
Z
∞
∞
X
X
1 Z f (w)
1 Z
(z − z0 )n
f (w)
n 1
dw =
f (z)
dw =
(z−z0 )
dz.
n+1
2πi C2 w − z
(w − z0 )
2πi C2 (w − z)n+1
n=0 2πi C2
n=0
So that gives us the positive powers in the Laurent expansion.
2
We get the negative powers by examining the second integral in the hint,
and expanding 1/(w − z) in a geometric series in (w − z0 )/(z − z0 ), since this
ratio is lest than one on the second, smaller circle.
M h(x)
dx.
4. Page 110, # 5. The function g(z) = −M
x−z
a). This h(x)/(x − z) is a parameterized integrand (real parameter is
x) which is jointly continuous in x, z and holomorphic in z, except where
z = x, so we have a theorem that the integral of such things is holomorphic.
Thus, holomorphic on the complex plane, except the real interval [−M, M ]
(which is where the denominator vanishes in the parameter set of x.) An
easy estimate shows |g(z)| is bounded by max|h|/(|z| − M ) for z large, which
clearly goes to zero as z tends to infinity.
R
!
1
1
1 ZM
h(x)
−
dx
g(xo + i) − g(xo − i) =
2πi −M
x − xo − i) x − x0 + i)
Z
1ZM
=
h(x)
dx = P (xo − x)h(x)dx,
π −M
(x − x0 )2 + 2
where P (x) is the Poison kernel for the upper half plane. So the limit as
→ 0 will just recover h.
b). Only asks to prove uniqueness. So if G is another function with the
same properties as g, we let f = G − g. Then the functions f (x + i) and
f (x − i) converge uniformly to functions f+ (x), f− (x) on the interval, whose
difference is zero. Thus f is continuous on [−M, M ] and also holomorphic on
the rest of C, so in particular it is continuous along the real axis. Since it is
also holomorphic in the lower and upper half plane, by Morera’s theorem it
is also homolmorphic on the the real line. So f is entire, vanishes at infinity,
so it must be constant. And the constant is zero. Therefore g = G.
5. Page 132, # 2.
We did this in detail in class.
5. Page 133, # 3.
2
a). Consider the function f (z) = F (z)e−z . It is also bounded, and goes
2
2
2
to zero as z → ∞ along the strip, since |e−z | = e−x +y . On the lower
boundary line, |f (z)| ≤ 1 while on the upper boundary, |f (z)| ≤ e . By the
max modulus principle, since we can ignore the zero limits out at the end (
e.g. use a rectangle to limit the domain), we have |f (z)| ≤ e in the strip.
3
2
Thus F (z) = f (z)e+z is no bigger than e2 in the strip. Letting → 0 gives
the bound of 1 that we want.
b). This is an easy re-scaling of the problem. Given F in the strip, let
f (z) = F (z)M0−iz−1 M1iz .
Then f is holomorphic, and on the lower strip |f (x)| = |F (x)|M0−1 which is
bounded by one, while on the upper strip |f (x + i)| = |F (x + i)|M
1 −1 which
is also bounded by one. Thus f is bounded by one in the strip, and so
|F (z)| ≤ |M0+iz+1 M1−iz | = M01−y M1y , where z = x + iy.
And so we can take sups over x.
c) Again, a rescaling. Define S(y) = supx |F (x + iy)|. Given any y0 &lt; y1
in the interval [0, 1], define F1 (z) = F ((y1 − y0 )z + iy0 ). Apply part b) to
this function to obtain
S((y1 − y0 )t + y0 ) ≤ S(y0 )1−t S(y1 )t , all 0 ≤ t ≤ 1.
Taking logs gives
log S((1 − t)y0 + ty1 ) ≤ (1 − t) log S(y0 ) + t &middot; log S(y1 ),
which shows convexity.
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