MATH 440/508 - Midterm
Friday, Oct 23
This exam has 5 pages. Be sure to show your work.
1. Define what it means for a function f : C → C to be meromorphic.
Use your definition to determine if tan(z) = sin(z)/ cos(z) is meromorphic.
SOLUTION: A function is meromorphic on C if it is holomorphic (differentiable) at all points in C, except at most at a countable set of singularities,
which are poles.
Note for students: This definition says nothing about the behaviour at
infinity. The set of singularities could be finite or infinite, but the fact the
poles are isolated implies automatically that the set is countable (i.e. a finite
or infinite sequence). So you don’t have to say that. It also is automatic
that the singularities do not have a limit point in C, because if they did, that
limit point would be a singularity that is not a pole. However, it does not
hurt to mention these facts in the definition.
tan(z) has singularities only at the zeros of cos(z), which occur at the
points z = (n + 1/2)π, n ∈ Z. The zeros of cos(z) are order one, so the
singularities of tan(z) are simple poles.
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2. Compute these three integrals, with a (brief) justification of your answers.
Z ∞
Z
Z
cos(x) − 1
1
ez
dz
dx
dz
2
2
2
3
x
−∞
|z|=2 (z − 1) (z + 1)
|z|=2 z − 1
SOLUTION: For the first, integrate (eiz − 1)/z 2 on a semicircular region
in the upper half plane. The integral along the upper circle of radius R decay
like 2πRe−y /R2 , so it vanishes as R → ∞. Thus
Z ∞ iz
eiz − 1
e −1
dx
=
πiRes
= πi(i) = −π.
0
z2
z2
−∞
The integral with the cosine is the real part of this, since cos(x) = Real(eix ),
so we are are done.
Note to students: It does not work to integrate (cos(z) − 1)/z 2 on the
semicircular region, because cos(iy) = cosh(y) blows up exponentially.
By the residue theorem,
Z
ez
ez
ez
dz
=
2πi(Res
+
Res
)
z=−1
z=+1
2
(z − 1)(z + 1)
(z − 1)(z + 1)
|z|=2 z − 1
= 2πi(e/2 − e−1 /2) = πi(e − 1/e).
This is just like the assignment question. The hard way to do it is to
compute the two residues at z = ±1, using Cauchy integral formula for
derivatives, and so on. The easy way is to notice, by Cauchy theorem, that
we can replace the circle |z| = 2 with any circle |z| = R > 2 and not change
the answer. Then the integral grows like 2πR/R5 , which goes to zero as R
tends to infinity. So the original must be
Z
1
dz = 0.
2
3
|z|=2 (z − 1) (z + 1)
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3. Determine how many zeros there are in the disk of radius one, for the
function
f (z) = ez + 4z n ,
where n ≥ 0 is an integer.
SOLUTION: Use Roche’s theorem, with F (z) = 4z n and G(z) = ez . On
the unit circle, we have |F (z)| = 4|z|n = 4 and |G(z)| = ex ≤ e1 = 2.7....
Thus on the circle, |F (z)| > |G(z), so by Rouche’s theorem F + G has the
same number of zeros as F , which has exactly n zeros.
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4. Verify that the real-valued function u(x, y) = x3 − 3xy 2 is harmonic on
the whole plane. Then find its harmonic conjugate.
SOLUTION: The fast way is to notice that u(x, y) = Real(z 3 ) thus it is
harmonic, with a conjugate v(x, y) = Imag(z 3 ) = 3x2 y − y 3 .
Only a little slower is to note that u is twice continuously differentiable
(since it is a polynomial), and a quick computation shows uxx = 6x, uyy =
−6x. Thus we have uxx + uyy = 0 on the plane, so the function is harmonic.
Use the Cuachy Rieman equations to find
vx = −uy = −6xy, vy = ux = 3x2 − 3y 2 ,
then integrate to solve for v.
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5. Suppose that f (z) is an entire function solving the equation
f (z)2 = A + Bz + Cz 2 + Dz 3 ,
where A, B, C, D are complex constants. Show the only possible solutions
are of the form f (z) = a + bz.
SOLUTION: For large z we have |f (z)|2 ≤ M |z|3 and thus |f (z)| ≤
M |z|3/2 . By the Cauchy inequalities, we can conclude that for large R, we
have the second derivative bounded
R3/2
= M R−1/2 ,
R2
which tends to zero as R tends to infinity. Thus the second derivative is zero
everywhere, so f 0 is constant and f is linear, az + b.
|f (2) (z)| ≤ M
Note to students: Some tried to do this by expanding the function f as a
power series, computing the coefficients of f 2 and then suggesting that since
all the higher terms of f 2 must be zero, that f must be a polynomial. That
argument does not really work, because it is always possible to solve this
equation locally:
√
f (z) = A + Bz + Cz 2 + Dz 3 ,
away from any zeros of that cubic. So you can get nice analytic solutions,
locally, and solve for coefficients. Somehow you need to invoke the fact that
the required solution is entire, in order to get the desired result.
For example, take A, C = 1, B, D = 0 and try√to solve f (z)2 = 1 + z 2 .
Recall from calculus that the real function h(x) = 1 + x is nice and smooth
near x = 0 so it has a Taylor series expansion
1
1 x2 3 x3
+
+ ···.
h(x) = 1 + x −
2
4 2!
8 3!
Recall, we compute the coefficients of the expansion by taking derivatives
of (1 + x)1/2 and evaluating at x = 0, so it’s easy enough to figure out the
coefficients.
This is a perfectly good Taylor series, and its radius of convergence is
one. Now define
1
1 z4 3 z6
f (z) = h(z 2 ) = 1 + z 2 −
+
+ ···.
2
4 2! 8 3!
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This defines a holomorphic function on the unit disk, again since the radius
of convergence is one. But now
√
(f (z))2 = ( 1 + z 2 )2 = 1 + z 2 .
So in fact, I CAN find a non-trivial series expansion that solves f (z)2 = 1+z 2 .
But notice, this f is NOT entire, since its radius of convergence is only 1.
And the singularities at z = ±i cannot be removed.
Also, if you prefer algebra, notice if we take the first few terms in the
series and square it,
1 z4 2
1
1
1 z4
1
) = 1+2 z 2 +( z 2 )2 −2
+ higher order = 1+z 2 +0z 4 + higher order
(1+ z 2 −
2
4 2!
2
2
4 2!
so the z 4 term cancels out. If we compute higher order terms, we will see
they cancel too.
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