COMPLEX ANALYSIS: SOLUTIONS 5
1. Find the poles and residues of the following functions
1
,
z 4 + 5z 2 + 6
1
,
(z 2 − 1)2
π cot(πz)
,
z2
1
(m, n ∈ Z>0 )
z m (1 − z)n
Solution: Throughout we use the following formula for calculating residues: If f (z)
has a pole of order k at z = z0 then
dk−1 1
k
(z
−
z
)
f
(z)
.
res(f, z0 ) =
0
k−1
(k − 1)! dz
z=z0
In particular, if f (z) has a simple pole at z0 then the residue is given by simply
evaluating the non-polar part: (z − z0 )f (z), at z = z0 (or by taking a limit if we have
an indeterminate form).
Let
1
1
1
√
√
√
√ .
= 2
=
2
2
+ 5z + 6
(z + 2)(z + 3)
(z + i 2)(z − i 2)(z + i 3)(z − i 3)
√
√
This has simple poles at z = ±i 2, ±i 3 with residues
f (z) :=
z4
√
√
res(f, i 2) = (z − i 2)
√
1
1
1
√
√
=
=
,
√
√
z 4 + 5z 2 + 6 z=i 2 (z + i 2)(z 2 + 3) z=i 2 2i 2
√
res(f, −i 2) = (z + i 2)
1
1
√ = − √ ,
4
2
z + 5z + 6 z=−i 2
2i 2
√
1
res(f, i 3) = − √ ,
2i 3
√
res(f, −i 3) =
1
1
√ .
2i 3
COMPLEX ANALYSIS: SOLUTIONS 5
2
For the second one let
f (z) =
(z 2
1
1
=
.
2
2
− 1)
(z + 1) (z − 1)2
This has double poles at ±1. From the formula we get
1
d
res(f, 1) =
= −1/4,
dz (z + 1)2 z=1
1
d
res(f, −1) =
= 1/4.
2
dz (z − 1) z=−1
For the third let
π cot(πz)
.
z2
Now, cot(πz) has poles wherever sin(πz) = 0, so at z = n ∈ Z. About these points
we have
(z − n)3
(z − n)2
− π 3 cos(πn)
+ ···
sin(πz) = sin(πn) + π cos(πn)(z − n) − π 2 sin(πn)
2!
3!
2
n
2 (z − n)
=(−1) π(z − n) 1 − π
+ O((z − n)4 )
3!
and
(z − n)3
(z − n)2
+ π 3 sin(πn)
+ ···
cos(πz) = cos(πn) − π sin(πn)(z − n) − π 2 cos(πn)
2!
3!
(z − n)2
=(−1)n 1 − π 2
+ O((z − n)4 ) .
2!
Hence, for z close to n ∈ Z, we have
f (z) =
1 − π 2 (z − n)2 /2 + O((z − n)4 )
π(z − n) 1 − π 2 (z − n)2 /6 + O((z − n)4 )
1 − π 2 (z − n)2 /2 + O((z − n)4 ) 1 + π 2 (z − n)2 /6 + O((z − n)4 )
=
π(z − n)
1
π
=
− (z − n) + O((z − n)3 )
π(z − n) 3
cot(πz) =
Therefore, f (z) = π cot(πz)/z 2 has simple poles at z = n 6= 0 and a triple pole at
z = 0. For the simple poles we have
π cot(πz) 1
res(f, n) = (z − n)
= 2.
2
z
n
z=n
COMPLEX ANALYSIS: SOLUTIONS 5
3
For the triple pole at at z = 0 we have
f (z) =
π2 1
1
+ O(z)
−
z3
3 z
so the residue is −π 2 /3.
Finally, the function
1
− z)n
has a pole of order m at z = 0 and a pole of order n at z = 1. From the formula for
residues we have
1
dm−1
1
res(f, 0) =
m−1
n
(m − 1)! dz
(1 − z) f (z) =
z m (1
z=0
n(n + 1) · · · (n + m − 2)
=
(m − 1)!
(n + m − 2)!
=
(n − 1)!(m − 1)!
and
1
dn−1 (−1)n res(f, 1) =
(n − 1)! dz n−1
zm
z=1
−m(m + 1) · · · (m + n − 2)
=
(n − 1)!
(m + n − 2)!
=−
(m − 1)!(n − 1)!
= − res(f, 0).
2. Use the substitution eiθ = z along with the residue theorem to show that
Z 2π
2π
dθ
=√ .
2 + cos θ
3
0
Solution: As suggested we let eiθ = z so that dθ = dz/(iz) and the integral becomes
Z
Z
1
dz
2
dz
=
.
−1
2
i |z|=1 z(2 + (z + z )/2)
i |z|=1 z + 4z + 1
√
Now z 2 + 4z + 1 has zeros of order 1 at z = z± = −2 ± 3 and so the integrand has
simple poles at z+ and z− . Only z+ lies in the unit disk and therefore by the residue
COMPLEX ANALYSIS: SOLUTIONS 5
4
theorem
2
i
Z
|z|=1
dz
2
1
=
×
2πi
×
res
,
z
+
z 2 + 4z + 1 i
z 2 + 4z + 1
1
=4π(z − z+ ) 2
z + 4z + 1 z=z+
1
=4π(z − z+ )
(z − z+ )(z − z− ) z=z+
1
=4π
z+ − z−
1
=4π √
2 3
2π
=√ .
3
3. Evaluate the following integrals via residues. Show all estimates.
(i)
Z ∞
x2
dx
x4 + 5x2 + 6
0
(ii)
Z ∞
x sin x
dx, a real
x 2 + a2
0
(iii)
Z ∞
log x
dx
1 + x2
0
Solution: (i) Since the integrand is an even function the integral in question is equal
to I/2 where
Z ∞
x2
I=
dx.
4
2
−∞ x + 5x + 6
√
√
As a function of a complex variable, the integrand has simple poles at ±i 2, ±i 3.
We will be considering a semicircular
√ √ contour in the upper half plane so we only need
calculate the residues at z = i 2, i 3. A slight modification of the first calculation
in question 1 gives
√
√
√
z2
(i 2)2
i 2
√ =
res 4
,i 2 =
z + 5z 2 + 6
2
2i 2
COMPLEX ANALYSIS: SOLUTIONS 5
5
and
√
√
√
(i 3)2
i 3
z2
,i 3 = − √ = −
.
res 4
z + 5z 2 + 6
2
2i 3
Now, Consider the semicircular contour ΓR , which starts at R, traces a semicircle
in the upper half plane to −R and then travels back to R along the real axis. Then,
on taking R large enough, by the residue theorem
√
√
Z
X
√ √
i 2 i 3
z2
dz
=
2πi×
residues
inside
Γ
=
2πi(
−
)
=
π(
3− 2).
R
4
2
2
2
ΓR z + 5z + 6
On the other hand
Z
lim
R→∞
ΓR
z2
dz = I + lim
R→∞
z 4 + 5z 2 + 6
Z
γR
z2
dz
z 4 + 5z 2 + 6
where γR is the semicircle in the upper half plane. But by the Estimation Lemma
Z
z2
z2
1
dz 6 πR max 4
→0
4
2
2
z∈γ
R
R z + 5z + 6
γR z + 5z + 6
√
√
as R → ∞. Hence, I = π( 3 − 2) and so
Z ∞
√
x2
π √
dx
=
(
3
−
2).
x4 + 5x2 + 6
2
0
The last estimate in the inequality for the integral over γR should be clear since the
dominant term is the z 4 term in the denominator, but for completeness:
π
z2
1
πR max 4
= max 2
−2
−4
z∈γR z + 5z + 6
R z∈γR 1 + 5z + 6z
π
1
6 max
−2
R z∈γR 1 − 5|z| − 6|z|−4
π
1
=
−2
R 1 − 5R − 6R−4
where we have used |z + w| > |z| − |w| in the second line. On taking R > 5, say, this
is 6 2π/R and the constant 2π is absorbed into the symbol.
(ii). We have
Z ∞
0
1
x sin x
dx
=
x 2 + a2
2
Z
∞
−∞
Z ∞
x sin x
1
xeix
dx = =
dx .
2
2
x 2 + a2
2
−∞ x + a
Denote this last integral by J. Again, we will consider J as the horizontal section of
the contour ΓR from part (i).
COMPLEX ANALYSIS: SOLUTIONS 5
6
In the upper half plane the integrand has a simple pole at z = ia with residue
zeiz
e−a
zeiz iae−a
=
.
,
ia
=
(z
−
ia)
=
res 2
z + a2
z 2 + a2 z=ia
2ia
2
Hence, by the residue theorem
Z
−a
πie = lim
R→∞
ΓR
zeiz
dz = J + lim
R→∞
z 2 + a2
Z
γR
zeiz
dz.
z 2 + a2
Thus it remains to show that this last integral vanishes in the limit. This is similar
to question 7 (ii) of Problems 3; a trivial estimate of the integrand is 1/R which is
not enough for the Estimation Lemma. Instead we apply integration by parts which
is probably the quickest way (see Problems 3). Integrating eiz and differentiating the
rest gives
−R
Z
Z zeiz
1
zeiz 1
2z 2
−
dz =
− 2
eiz dz.
2 + a2
2 + a2 ) 2 + a2
2 )2
z
i(z
i
z
(z
+
a
γR
γR
R
The first term on the right is −2R cos R/i(R2 + a2 ) 1/R. For the integrals we use
the Estimation Lemma to give
Z
eiz eiz
1
1
6 πR
dz 6 πR max 2
,
2
2
2
2
2
z∈γR z + a
R −a
R
γR z + a
Z
γR
z 2 eiz z 2 eiz
1
1
6 πR max 6 πR3
dz
,
z∈γR (z 2 + a2 )2
(z 2 + a2 )2
(R2 − a2 )2
R
as R → ∞. Hence, J = πie−a and so
Z ∞
1
π
x sin x
dx = =(J) = e−a .
2
2
x +a
2
2
0
(iii) This is quite hard and, as I discovered recently, the solution is in Conway’s
book anyway (pg. 117–118). My bad. It’s still instructive to attempt this before
reading Conway though.
We see that, as a function of a complex variable, the integrand has a branch cut
and simple poles at z = i, −i. Taking the branch of the log with −π < arg(z) < π,
we would like to choose a contour which lies just above and below the cut and that
also picks up the residues at i, −i. A natural choice is the contour we used in the
lectures. This consisted of a small circle about z = 0, horizontal lines just above and
below the negative real axis, and a large circle completing the contour.
COMPLEX ANALYSIS: SOLUTIONS 5
7
Unfortunately, with this choice the integrals over the horizontal lines L1 , L2 are
given by
Z
Z
log z
log z
dz +
dz
2
2
L2 1 + z
L1 1 + z
Z 0
Z −∞
log(x + iδ)
log(x − iδ)
=
dx +
dx
2
1 + (x − iδ)2
−∞ 1 + (x + iδ)
0
Z −∞
Z 0
log |x| + πi
log |x| − πi
dx +
dx
→
2
1+x
1 + x2
−∞
0
Z 0
1
=2πi
dx
2
−∞ 1 + x
and the integral we’re after has disappeared.
This motivates the choice of a new contour: we want something with a horizontal
line over the whole real axis, since then the integral over this line is given by
Z ∞
Z ∞
log |x|
log x
dx = 2
dx
2
1 + x2
−∞ 1 + x
0
and we avoid the problem of cancellation. Consequently, we choose a branch of log z
with a branch cut along the negative imaginary axis. To avoid z = 0 and the branch
cut we indent our contour with a small circle in the upper half plane. We complete
the contour with a large circle.
Thus, let Γr,R be the contour consisting of a line from r to R, a semicircle in the
upper half plane traced from R to −R, a line from −R to −r, and a semicircle in
the upper half plane traced from −r to r. Follwing the details in Conway we find
Z ∞
log x
dx = 0.
1 + x2
0
4. Let ΓN be the square which crosses the real axis at ±(N + 1/2) with N ∈ N.
(i) Show that cot(πz) is bounded on ΓN and hence show that
Z
π cot(πz)
lim
dz = 0.
N →∞ Γ
z2
N
(ii) For a given N compute the above integral via residues. Conclude something
interesting.
Solution: (i) We have
eπiz + e−πiz
e2πiz + 1
e2πix e−2πy + 1
cot(πz) = i πiz
= i 2πiz
= i 2πix −2πy
.
e − e−πiz
e
−1
e e
−1
COMPLEX ANALYSIS: SOLUTIONS 5
8
Thus, if y > 1
| cot(πz)| 6
1 + e−2π
1 + e−2πy
6
.
1 − e−2πy
1 − e−2π
Since cot(πz) = cot(πz) we obtain the same bound for y 6 −1. It remains to show
that cot(πz) is bounded for z = ±(N + 1/2) + iy with |y| < 1. But in this case
±2πi(N +1/2) −2πy
e
| − e−2πy + 1|
e
+
1
e2π − 1
=
| cot(πz)| = ±2πi(N +1/2) −2πy
<
=: C.
e
e
− 1 | − e−2πy − 1|
e−2π + 1
Since C is bigger than the bound for |y| > 1, we may say that | cot(πz)| < C on all
of ΓN .
(ii) By the estimation lemma
Z
cot(πz) π cot(πz) C
8πC
< 8π(N +1/2)
dz 6 8π(N +1/2) max =
2
2
2
z∈ΓN
z
z
(N + 1/2)
N + 1/2
ΓN
and this tends to zero as N → ∞.
On the other hand, for a fixed N this integral is given by 2πi × the sum of residues
inside ΓN . Using the results from question 1 then gives
Z
X 1
π cot(πz)
π2 dz
=
2πi
−
.
2
z2
n
3
ΓN
−N 6n6N
n6=0
Therefore, on letting N → ∞ we have
∞
X
1
π2
1
π2
0=
−
=
2
−
n2
3
n2
3
−∞6n6∞
n=1
X
n6=0
and so
∞
X
1
π2
ζ(2) =
=
,
n2
6
n=1
which is interesting.
5. (i) Let f (z) = z 6 + cos z. Find the change in argument of f (z) as z travels once
around the circle of radius 2, center zero, in the positive direction.
(ii) How many solutions does 3ez − z = 0 have in the disk |z| 6 1?
(iii) Use Rouche’s Theorem to prove that a polynomial of degree n has n roots in C.
COMPLEX ANALYSIS: SOLUTIONS 5
9
Solution: (i) By the argument principle, the change in argument of f (z) as z travels
around the circle is equal to (2π ×) the number of zeros minus poles of f (z) inside
the circle, so just the number of zeros then.
To find the number of zeros of f (z) inside the circle we compare f (z) with its
dominant term z 6 and apply Rouche’s Theorem. So let g(z) = z 6 . On the circle we
have
|g(z)| = |z|6 = 26 = 64
and
|f (z) − g(z)| = | cos(z)| 6
|eiz | + |e−iz |
6 e|z| = e2 < 64.
2
Thus, Rouche’s Theorem applies and f (z) has the same number of zeros inside the
circle as z 6 , which is 6. Hence, the change in argument is 2π × 6 = 12π.
(ii). Rephrasing the question we ask for the number of zeros of f (z) = 3ez − z in
the closed unit disk. We apply Rouche’s Theorem again. This time the dominant
term is g(z) = 3ez . On the unit circle we have
|g(z)| = 3e<(z) > 3e−1 > 1
and
|f (z) − g(z)| = |z| = 1.
By Rouche’s Theorem 3ez − z has the same number of zeros in the unit disk as 3ez ,
which is none. So the answer is no solutions.
(iii) In the lectures we showed that the polynomial z 6 + z + 2 has 6 zeros in the
disk |z| 6 2 by comparing it with z 6 and applying Rouche’s Theorem. We generalise
this by comparing a general polynomial
f (z) = an z n + an−1 z n−1 + · · · + a0 ,
an 6= 0
with its leading term
g(z) = an z n
on arbitrarily large circles. On the circle |z| = R > 1 we have
|g(z)| = |an |Rn
COMPLEX ANALYSIS: SOLUTIONS 5
10
and
|f (z) − g(z)| =|an−1 z n−1 + an−2 z n−2 + · · · + a0 |
6|an−1 |Rn−1 + |an−2 |Rn−2 + · · · + |a0 |
<n max |aj |Rn−1
06j6n−1
n maxn−1 |aj | j=0
|an |Rn
=
|an |R
Therefore, on a circle of radius R > n maxn−1
j=0 |aj |/|an | we have |f (z) − g(z)| < |g(z)|.
By Rouche’s Theorem f (z) has the same number of zeros inside the circle as g(z),
which is n.
6. Prove that the function
f (z) =
X
n∈Z
1
(n + z)2
is meromorphic on C.
Solution: Clearly, f (z) has poles only at the integers so we need to show that f (z) is
analytic on C\Z. By the results given in the lectures, this will follow if we can show
that the series is uniformly convergent on compact subsets of said region. Since
X
1
1
f (z) = 2 +
z
(n + z)2
n6=0
it suffices to show the uniform convergence of this last sum.
So let |z| 6 R, let z be fixed away from integers: |z + n| > ρ >P
0 ∀n ∈ Z, and let
−M be the nearest integer to z. Note that the series looks like 2 n6=0 n−2 which is
convergent. So we first pull out the convergent part by writing
X
X 1
1
.
(n + z)2 =
2
2
n
|1
+
z/n|
6=0
n6=0
Then if |n| 6 |M |
|z + n| > |z + m| =⇒ |1 + z/n| > (|M |/|n|)|1 + z/m| > |1 + z/M | > ρ/|M |.
If |n| > |M | + 1
|1 + z/n| > |1 + <(z)/n| > 1 −
|<(z)|
|M | + 1
1
>1−
=
.
|n|
|M | + 2
|M | + 2
COMPLEX ANALYSIS: SOLUTIONS 5
Therefore,
X
|M |2
1
(n + z)2 6 ρ2
n6=0
X
06=|n|6|M |
1
+ (|M | + 2)2
2
n
X
|n|>|M |+1
11
1
1
+
2
n
|z ± (M ± 1)|2
2
62ζ(2)
(R + 1)
4
2
+
2ζ(2)(R
+
3)
+
ρ2
ρ2
<∞.
Hence, by the Weierstrass M test with Mn = n−2 max((R + 3)2 , (R + 1)2 /ρ2 ) the
series is uniformly convergent on compact subsets of C\Z.
7. Let ϕ : R → R be a smooth function whose derivatives ϕ(k) (t), k > 0, are of rapid
decay at ∞ i.e.
lim tA ϕ(k) (t) = 0
t→∞
for all A ∈ R and all k > 0. The Mellin transform of ϕ is defined as
Z ∞
ϕ̃(z) =
ϕ(t)tz−1 dt.
0
(i) Prove that ϕ̃(z) is analytic in the region <(z) > 0.
(ii) Use integration by parts to show that ϕ̃(z) can be analytically continued to
C\Z60 , with possible simple poles at z = −n.
(iii) Find the residues at these poles and compute the formal sum
∞
X
res(ϕ̃(z)t−z , −n).
n=0
What does this look like?
(iv) Given part (iii), suggest an argument that would prove
Z
1
ϕ̃(z)t−z dz = ϕ(t)
2πi <(z)=c
where the integral is over the vertical line from c − i∞ to c + i∞ with c > 0.
Solution: (i). Since ϕ is smooth on R it is bounded in a neighbourhood of t = 0:
|ϕ(t)| 6 C for t ∈ [0, t0 ], say. Then
Z t0
Z t0
<(z)
z−1
|ϕ(t)t |dt 6 C
t<(z)−1 dt = Ct0 /<(z) < ∞.
0
0
COMPLEX ANALYSIS: SOLUTIONS 5
12
Also, since ϕ is of rapid decay, for any z there exists a t1 = t1 (<(z)) such that
|ϕ(t)tz−1 | 6 Dt−2 for all t > t1 . Then
Z ∞
|ϕ(t)tz−1 |dt 6 D/t1 < ∞.
t1
Rt
Since the remaining integral t01 |ϕ(t)tz−1 |dt is finite the integral is absolutely convergent in the region <(z) > 0.
Let γ be a closed curve in the region <(z) > 0. By absolute convergence and
Fubini’s Theorem
Z
Z Z ∞
Z ∞
Z
z−1
−1
ϕ̃(z)dz =
ϕ(t)t dtdz =
ϕ(t)t
tz dzdt.
γ
γ
0
0
γ
But since tz is analytic in the region <(z) > 0 for all t ∈ [0, ∞) this last inner integral
is zero by Cauchy’s Theorem. Hence,
Z
ϕ̃(z)dz = 0.
γ
The continuity of ϕ̃(z) follows exactly as it does for the Gamma function, hence it
is analytic by Morera’s Theorem.
(ii). Integrating by parts once gives
∞
Z
Z
ϕ(t)tz 1 ∞ 0
1 ∞ 0
z
ϕ̃(z) =
−
ϕ (t)t dt = −
ϕ (t)tz dt.
z t=0 z 0
z 0
R∞ 0
By the same reasoning used in part (i), the integral 0 ϕ (t)tz dt is analytic in the
region <(z) > −1. Hence, the above expression provides a meromorphic continuation
of ϕ̃(z) to said region. At z = 0 we have a residue of
Z
Z ∞
1 ∞ 0
z z·−
ϕ (t)t dt
=−
ϕ0 (t)dt = ϕ(0).
z 0
0
z=0
In particular, if ϕ(0) = 0 then there is no pole at z = 0.
More generally, integrating by parts n times gives
Z ∞
(−1)n
(1)
ϕ̃(z) =
ϕ(n) (t)tz+n−1 dt.
z(z + 1) · · · (z + n − 1) 0
Again, this expression is seen to be analytic in the region <(z) > −n, with the
exception of possible simple poles at z = 0, −1, . . . , −n + 1. Since n is arbitrary,
we’re done.
COMPLEX ANALYSIS: SOLUTIONS 5
13
(iii). From equation (1) we see that
Z ∞
(−1)n+1
(n+1)
z+n ϕ
(t)t dt
res(ϕ̃(z), z = −n) =(z + n)
z(z + 1) · · · (z + n) 0
z=−n
Z ∞
n+1
(−1)
=
ϕ(n+1) (t)dt
(−n)(−n + 1) · · · (−1) 0
Z
1 ∞ (n+1)
=−
ϕ
(t)dt
n! 0
1
= ϕ(n) (0).
n!
In particular, if ϕ(t) is of rapid decay at zero as well as at infinity, then ϕ̃(z) is entire.
Since all the poles are simple, an extra factor of t−z in the above residue calculations
gives a factor of t−z |z=−n = tn and hence
∞
X
∞
X
1 (n)
res(ϕ̃(z)t , −n) =
ϕ (0)tn .
n!
n=0
n=0
−z
This, of course, looks like ϕ(t) in the form of a Taylor expansion about t = 0.
(iv). Assuming that the above series converges and equals ϕ(t) (so ϕ is analytic
in the real analysis sense) then we would like a contour integral which captures all
these residues. As suggested, we should look at
Z
1
ϕ̃(z)t−z dz.
2πi <(z)=c
We would like to truncate this integral at heights z = c ± iT and then consider
the truncated integral as part of a rectangular contour which encloses some of the
residues. To estimate these integrals we need bounds on ϕ̃(z).
From equation (1) we see that for fixed <(z), ϕ̃(z) → 0 as |=(z)| → ∞. More
precisely, as =(z) → ∞ equation (1) gives
Z ∞
C<(z),n
1
|ϕ̃(z)| 6
|ϕ(n) (t)|t<(z)+n−1 dt 6
,
|z(z + 1) · · · (z + n − 1)| 0
|=(z)|n
for some constant C<(z),n . Consequently,
Z
Z c+iT
Z
Z −T i
1
1
1h ∞
−z
−z
ϕ̃(z)t dz =
ϕ̃(z)t dz +
+
ϕ̃(c + iy)t−c−iy dy
2πi <(z)=c
2πi c−iT
2π T
−∞
Z c+iT
1
=
ϕ̃(z)t−z dz + O(T −n+1 )
2πi c−iT
COMPLEX ANALYSIS: SOLUTIONS 5
14
We now consider this last integral as part of a rectangular contour Γ whose left edge
crosses the real axis at −N − 1/2. Then,
Z c+iT
Z
Z
1
1
1
−z
−z
ϕ̃(z)t dz =
ϕ̃(z)t dz −
ϕ̃(z)t−z dz
2πi c−iT
2πi Γ
2πi other edges of Γ
Z
N
X
1
1 (n)
n
ϕ (0)t −
=
ϕ̃(z)t−z dz
n!
2πi other edges of Γ
n=0
by the residue theorem. Using our bound on |ϕ̃(z)| we can estimate the remaining
integrals. The largest contribution comes from the integral over the left edge of Γ,
and this is seen to be O(tM +1/2 ). The other integrals are small in size. On letting
T → ∞ we get
Z
N
X
1 (n)
1
−z
ϕ̃(z)t dz =
ϕ (0)tn + O(tN +1/2 ) = ϕ(t).
2πi <(z)=c
n!
n=0