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Patrick Corn Math 185, 7/26/ Solution Set 9
Problem 1: Let f (z) =
∏n i=1(z^ −^ ai). Then^ f^ is a polynomial, and so it’s holomorphic every- where. Note that |f (0)| = 1, so since the maximum of |f | over the unit disk is attained on the boundary, there must be a point p with |p| = 1 such that |f (p)| ≥ 1. And this is what we wanted.
Problem 2: Let R be a large enough real number, i.e. large enough that all the roots of q are enclosed by the circle CR. Note that ∣ ∣∣ ∣
CR
p(z) q(z)
dz
∣ ≤^2 πRMR
where MR is the maximum value of
∣ p q((zz))
∣ on^ CR.^ Now let^ n^ = deg(q)^ −^ deg(p), and note that
limz→∞ z
n (^) p(z) q(z) is a finite number.^ So lim supR→∞^ R
nMR is finite. Since n > 1, it follows that
limR→∞ RMR = 0, so that
lim R→∞
CR
p(z) q(z)
dz = 0
But (^) ∫
CR
p(z) q(z)
dz =
C
p(z) q(z)
dz
for large enough R, by Cauchy’s theorem for an annulus. So then
C
p(z) q(z) dz^ = 0. Problem 3: (a) The polynomial in the denominator has four distinct roots, equal to e^2 πik/^5 for 1 ≤ k ≤ 4. These are all simple poles. Also, it is not hard to see that ∞ is a simple pole:
f
z
z
z^4 + z^3 + z^2 + z + 1
and the latter fraction is holomorphic near 0. (b) It is not hard to see that the zeroes of sin z are precisely the integer multiples of π. They are all simple zeroes, because the derivative of sin z doesn’t vanish at any of them. Therefore z = kπ is a pole of order 2 (a double pole) for all k ∈ Z. The case of ∞ is more interesting. Be careful–in fact ∞ is not an isolated singularity at all! There is no punctured disk around ∞ contained in the region G on which (^) sin^12 z is holomorphic, because kπ /∈ G for all k ∈ Z. (c) First of all, ∞ is a removable singularity, because sin z is holomorphic at 0. The only other possible problem is at z = 0. We can write the Laurent series for sin (^1) z in powers of (^1) z :
sin
z
∑^ ∞
n=
(−1)nz−^2 n−^1 (2n + 1)!
so 0 must be an essential singularity.
Problem 4: For the finite set of points F = {z 1 ,... , zn}, we can find a nonnegative integer pk, 1 ≤ k ≤ n, such that (z − zk )pk^ f (z) can be extended to a holomorphic function in a neighborhood of zk (because each zk is either removable, so pk = 0, or a pole of order pk). Then consider
g(z) = f (z)
∏^ n
k=
(z − zk)pk
1
Then g(z) is an entire function, and by assumption it has a non-essential singularity at ∞. As we have seen in class, this implies that g(z) is a polynomial. (It is equal to its Taylor series centered at 0, which can only have finitely many nonzero terms.) Thus f is a rational function.
Problem 5: We can write h(z) = (z − z 0 )h 1 (z) with h 1 holomorphic in a neighborhood of z 0 , and h 1 (z 0 ) 6 = 0. Then g(z) h(z)
= (z − z 0 )−^1
g(z) h 1 (z) and g/h 1 is holomorphic in a neighborhood of z 0 , hence representable by a Taylor series
n=0 an(z− z 0 )n^ there. Then the coefficient of (z − z 0 )−^1 in the Laurent series for g/h is equal to a 0 = (^) hg 1 ((zz^00 ) ).
But, as we have seen many times already, h 1 (z 0 ) = h′(z 0 ).
Problem 6: (a) There are q singularities of the form ζk = e^2 πik/q, 0 ≤ k ≤ q − 1. By the previous exercise, the residue at ζk is equal to z
p −qzq−^1 evaluated at^ ζk. This equals ζkp −qζ kq−^1
ζ kp+ q
e^2 πik(p+1)/q q (b) There are double poles at ±1, so we cannot use the previous exercise. Consider the function f 1 (z) = z^5 /(z + 1)^2. It has a Taylor series
n=0 an(z^ −^ 1)
n, and the Laurent series of our function
around 1 is equal to this Taylor series divided by (z − 1)^2. Then the coefficient of (z − 1)−^1 in that Laurent series will be equal to a 1. This equals f 1 ′(1), and now we must do a little computation to get f 1 ′(1) = 1, so the residue at z = 1 is 1. Similarly let f− 1 (z) = z^5 /(z − 1)^2. The residue at z = −1 will equal f (^) −′ 1 (−1) = 1. So both residues are 1. (c) There are two simple poles, ζ and ζ^2 , where ζ = e^2 πi/^3. We can use the previous exercise to find that the residue at ζk^ is the function 2 cosz+1^ z evaluated at ζk. Let’s just leave it like that:
resz=ζk
cos z 1 + z + z^2
cos ζk 2 ζk^ + 1
, k = 1, 2.
(d) There are simple poles at z = kπ, k ∈ Z, so we can use the above exercise to see that
resz=kπ
sin z
cos kπ
= (−1)k.
Problem 7: Let f (z) = (z − z 0 )mf 1 (z), where f 1 (z) is holomorphic in a neighborhood of z 0 and f 1 (z 0 ) 6 = 0. Then
f ′(z) f (z)
m(z − z 0 )m−^1 f 1 (z) + (z − z 0 )mf 1 ′(z) (z − z 0 )mf 1 (z)
m z − z 0
f 1 ′(z) f 1 (z)
and the latter fraction is a holomorphic function in a neighborhood of z 0 , so the principal part of the Laurent series of f ′/f is exactly (^) z−mz 0. So the residue is m, as desired.
Problem 8: First, if g(z) is identically zero, there’s nothing to prove. Otherwise, the zeroes of g(z) are isolated, as we’ve seen, so h(z) = f (z)/g(z) is holomorphic everywhere except for isolated singularities. Let z 0 be one such singularity; we see immediately that
lim sup z→z 0
|h(z)| < ∞
since |h(z)| ≤ 1 on a punctured neighborhood of z 0. Thus z 0 is a removable singularity. So all the (non-∞) singularities of h(z) are removable. So h(z) is actually an entire function. But it’s bounded! So by Liouville’s theorem it is constant.
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