SOLUTIONS FOR ASSIGNMENT 4 (n = 1)
1. Problem 1
For each of the following functions, determine all its zeroes and all its isolated singular points
in the extended complex plane C (that is, the point ∞ must be investigated as well!). As long as
a point is a zero or a pole, determine the order. Explain your answer!
a) f(z) = 1−cos(z6)
(sin z)5 · e
1
π−z
b) f(z) =
√
z sin 1√
z7
(explain also why the function is holomorphic outside its isolated
singular points; the choice of Arg z for the two roots is the same).
Solution for a). First of all, the ∞ is not an isolated singularity of f(z) (and not an isolated
zero), because zeroes of the denominator z = πk accumulate to ∞. So, this is the verdict for
z = ∞: not an isolated singularity and not an isolated zero.
Potential zeroes are the points where cos(z6) = 1, that is z6 = 2πk and z = zk,l =
6
√
2πkeπil/3, k > 0, l = 0, ..., 5 or z = zk,l = 6
√
−2πkei(2πl+π)/6, k < 0, l = 0, ..., 5 (you can
use alternative representations, of course). The value k = 0 shall be considered separately. For
nonzero k, we see that the exponent and the sin are holomorphic near zk,l, and the sin does
not vanish, so zk,l are true zeroes of f(z). Let us determine their order. The easiest way is as
follows. Since the exponent and 1/(sin z)5 are holomorphic and do not vanish at zk,l (IMPORTANT!),
they do not effect the order. Hence, we just look at 1 − cos(z6). We easily find that
the first derivative 6z5 sin(z6) of the latter function vanishes at zk,l, while the second derivative
30z4 sin(z6) + 36z10 cos(z6) does not.
Hence, zk,l are all zeroes of f(z) of order 2.
Now let us deal with z = 0. The exponent is holomorphic and does not vanish at 0 (IMPORTANT!),
so it does not effect the order. Now we use the classical Taylor expansions:
1 − cos(z6
) = z12
/2 + ... = z12
(0.5 + ...), sin z = z + ... = z(1 + ...),
so
1 − cos(z6)
(sin z)5
= z7
(0.5 + ...)/(1 + ...) = 0.5z7
+ ...
(dots all the time denote a holomorphic near 0 sum of terms of higher order), so we conclude that
z = 0 is a removable singularity which is at the same tine a zero of f(z) of order 7.
Now we deal with the other isolated singularities. They corresponds to sin z = 0, z = 0, so
z = zk = πk, k = 0. The case k = 1 will be considered separately, while for k = 1 the factors
1 − cos(z6) and the exponent are holomorphic and do not vanish at zk, so they do not effect the
nature of the singularity, and we just deal with g(z) = 1
(sin z)5 . The easiest way now is to use
the property: if h(z) has at a point a a zero of order l, then 1/f(z) has at a a pole of order
l, and also the fact that the order of product is the sum of orders. From sin z = z(1 + ...) we
know that sin z has a zero of order 1 at z = 0, but sin z is (almost) periodic with period π:
sin(z + πk) = (−1)k sin z (!), so it also has a zero of order 1 at each zk. Alternatively, one can
again use derivatives.
1
2 SOLUTIONS FOR ASSIGNMENT 4 (N = 1)
So, we finally see that all zk with k = 1 are poles of f(z) of order 5.
It remains to deal with z = π. We remember that from the Taylor series of exponent,
e1/(π−z)
=
∞
j=0
1
(π − z)jj!
.
This series (which becomes the Laurent series for e1/(π−z) at z = π) has infinitely many negative
terms, and this implies that e1/(π−z) has at z = π an essential singularity. Now we shall still prove
(!) that f(z) has the same kind of singularity (counterexample: ez+1/z · e−1/z has a removable
singularity at 0, while both factors have an essential singularity!). To see this, we express the
exponent:
e1/(π−z)
= f(z)(sin z)5
/(1 − cos(z6
)).
The factor (sin z)5 is holomorphic near 0 and the factor 1/(1 − cos(z6)) has a pole at 0, so if f(z)
had a removable singularity or a pole at 0, then the product e1/(π−z) = f(z)(sin z)5/(1 − cos(z6))
would also have a removable singularity or a pole at 0, which is a contradiction.
So, z = π is an esential singularity.
Solution for a). First of all, from the classical Taylor expansion of sin z we get:
f(z) =
√
z


∞
j=0
1
√
z7
2j+1
(−1)j
(2j + 1)!

 =
∞
j=0
1
z
7j+3
(−1)j
(2j + 1)!
,
which proves that f(z) is single-valued and holomorphic in C \ {0}.
Let us start with z = ∞: setting w = 1/z, f(z) = g(w), we see that
g(w) =
∞
j=0
w7j+3 (−1)j
(2j + 1)!
,
so that g(w) is simply holomorphic at w = 0, hence z = ∞ is a removable singularity which is at
the same time a zero of order 3.
For the isolated singularity z = 0, we see from the above that the Laurent series of f(z) at
z = 0 has infinitely many negative terms, hence z = 0 is an essential singularity.
It remains to deal with the isolated zeroes. They correspond to sin 1√
z7
= 0, so we get
z = zk,l = 1
7√
π2k2
e2πil/7, k > 0, l = 0, ..., 6, or z = zk,l = 1
7√
π2k2
ei(π+2πl)/7, k < 0, l = 0, ..., 6. To
determine the order at zk,l, we first note that the factor
√
z is holomorphic and nonvanishing near
each fixed zk,l. So, it does not effect the order. Now, for the remaining function sin 1√
z7
, we
compute the derivative z−9/2 cos 1√
z7
and see that it is nonzero at zk,l. Hence, each zk,l is a
zero of f(z) of order 1.
2. Problem 2
Find Taylor/Laurent expensions of the function f(z) in all (!) possible discs and annuli with
the center z = 0:
f(z) =
z4
z(z − 1)(z − 2)
SOLUTIONS FOR ASSIGNMENT 4 (n = 1) 3
After doing so, determine the types of isolated singularity at the points z = 0 and z = ∞
(including the order, if applicable).
Solution. The function has the following singularities in C: 0, 1, 2. In view of that, there are
three principal annuli where it admits a Laurent expansion:
D1 = {0 < |z| < 1}, D2 = {1 < |z| < 2}, D3 = {2 < |z| < +∞},
and all the other annuli are contained in one of the Dj. So, it is enough to provide the Laurent
expansion in each of the Dj.
We will make use of the partial fraction decomposition:
f(z) = z3 1
(z − 1)(z − 2)
= z3 1
z − 2
−
1
z − 1
.
Let us start with D1. Since in D1 we have |z| < 1, we can use the geometric progression and get:
1
z − 1
= −
∞
j=0
zj
,
1
z − 2
= −
1/2
1 − z/2
= −
∞
j=0
1
2j+1
zj
.
Now
f(z) = z3


∞
j=0
zj
−
∞
j=0
1
2j+1
zj

 =
∞
j=0
1 −
1
2j+1
zj+3
.
We also see from here that z = 0 is a removable singularity for f(z), which is at the same time
an isolated zero of order 3.
We deal next with D2. Here |z/2| < 1, |1/z| < 1, so still using simply a geometric progression
we get:
1
z − 1
=
1/z
1 − 1/z
=
∞
j=0
1
zj+1
,
1
z − 2
= −
1/2
1 − z/2
= −
∞
j=0
1
2j+1
zj
,
f(z) = z3

−
∞
j=0
1
zj+1
−
∞
j=0
1
2j+1
zj

 = −
∞
j=0
1
zj−2
−
∞
j=0
1
2j+1
zj+3
.
It is a good example when a Laurent series actually goes from −∞ to +∞.
Finally, we deal with D3. Here |2/z| < 1, |1/z| < 1, so:
1
z − 1
=
1/z
1 − 1/z
=
∞
j=0
1
zj+1
,
1
z − 2
=
1/z
1 − 2/z
=
∞
j=0
2j
zj+1
,
f(z) =
∞
j=0
1
zj−2
2j
− 1 =
∞
j=1
1
zj−2
2j
− 1 .
Also, by setting w = 1/z, f(z) = g(w), we see that the Laurent expansion of g(w) at w = 0 has
exactly one negative term which is 1
w , and from here z = ∞ is a pole of order 1 for f(z).
3. Problem 3
Everybody succeeded, so I will not provide a solution here.