MAT3705/103/3/2015
Tutorial Letter 103/3/2015
Complex Analysis
MAT3705
Semesters 1 & 2
Department of Mathematical Sciences
This tutorial letter contains the
memorandum for the October 2012
examination paper.
BAR CODE
Learn without limits.
university
of south africa
Memorandum MAT3705 May/June 2012
QUESTION 1
1.1
(
z−4
(z − 4)(z + 4)
) =
z+4
(z + 4)(z + 4)
(z − 4)(z + 4)
=
|z + 4|2
zz + 4z − 4z − 16
=
|z + 4|2
|z|2 − 16 4(z − z)
=
+
|z + 4|2
|z + 4|2
Since z − z is purely imaginary (take z = x + iy then z − z = 2iy) we have that
Re(
and if
|z|2 − 16
z−4
)=
z+4
|z + 4|2
|z|2 − 16
= 0 then |z| = 4.
|z + 4|2
1.2
Suppose that z = x + iy where x, y ∈ R
Then
z+4
z−4
< 1 ⇔ |z + 4| < |z − 4|
p
p
⇔
(x + 4)2 + y 2 < (x − 4)2 + y 2
⇔ (x + 4)2 + y 2 < (x − 4)2 + y 2
⇔ 8x < −8x
⇔ 16x < 0
⇔ x<0
So the region is {x + iy | y ∈ R,x < 0} which is the region left of the y=axis excluding the y=axis.
The set is open since it contains none of its boundary points.
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MAT3705/103
QUESTION 2
2.1
1 iz
(e − e−iz ) = i ⇐⇒ eiz − e−iz = −2
2i
⇔ (eiz )2 + 2(eiz ) − 1 = 0
sin z = i ⇐⇒
Substitute w = eiz to obtain the equation w2 + 2w − 1 = 0.
Then
√
−2 ± 4 + 4
w =
2√
2 2
= −1 ±
√2
= −1 ± 2
so that
eiz
= −1 ±
√
2
√
⇔ iz = ln −1 + 2 + i2nπ
or
√
iz = ln −1 − 2 + i(2n + 1)π
√
⇔ z = −i ln( 2 − 1) + 2nπ or
√
z = −i ln( 2 + 1) + (2n + 1)π
Then since
√
√
1
= 2+1
2−1
we can write the answer as
√
z = i ln( 2 + 1) + 2nπ or
√
z = −i ln( 2 + 1) + (2n + 1)π
2.2 For z = x + iy where x, y ∈ R we have
eiz = ei(x+iy) = e−y+ix = e−y eix = e−y (cos x + i sin x)
Hence
Re eiz = e−y cos x
and
e−y cos x = 0
cos x = 0 and y ∈ R
π
⇔ x = ± + 2nπ, y ∈ R
2
Hence
π
z = ( + nπ) + iy, n ∈ Z, y ∈ R
2
Re eiz
=
⇔
3
QUESTION 3
3.1
From f (z) = (2x3 + y 3 − 7y − x) + i(x3 − 2y 3 + 5y + 4x), we have
u = 2x3 + y 3 − 7y − x,
Then
ux = 6x2 − 1
uy = 3y 2 − 7
v = x3 − 2y 3 + 5y + 4x
vx = 3x2 + 4
vy == −6y 2 + 5
These first partial derivatives are all continuous everywhere, so g will therefore be differentiable
wherever the Cauchy-Riemann equations are satisfied.
(See the section on sufficient conditions for differentiability) .Now
ux = vy ⇔ 6x2 − 1 = −6y 2 + 5
⇔
x2 + y 2 = 1
Similarly
uy
= −vx ⇔ 3y 2 − 7 = 3x2 + 4
⇔ x2 + y 2 = 1
f (z) is differentiable on the circle x2 + y 2 = 1 and if (x0 , y0 ) satisfies x20 + y02 = 1 we get
f 0 (x0 + iy0 ) = ux (x0 + iy0 ) + ivx (x0 + iy0 )
= 6x20 − 1 + i(3x20 + 4)
which is also
= 6x20 − 1 + i(3(1 − y02 ) + 4)
= 6x20 − 1 + i(−3y02 + 7)
or
f 0 (x0 + iy0 ) = uy (x0 + iy0 ) + ivy (x0 + iy0 )
= 3y02 − 7 + i(−6y02 + 5 )
which is also
= 3(1 − x20 ) − 7 + i(−6y02 + 5 )
= −3x20 − 4 + i(−6y02 + 5 )
3.2
We get
∂ u
(e sin v) = eu ux sin v + eu vx cos v
∂x
and
∂2 u
(e sin v) = eu (ux )2 sin v + eu uxx sin v + eu ux vx cos v +
∂x2
eu ux vx cos v + eu vxx cos v − eu (vx )2 sin v
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MAT3705/103
Similarly
∂ u
(e sin v) = eu uy sin v + eu vy cos v
∂y
∂2 u
(e sin v) = eu (uy )2 sin v + eu uyy sin v + eu uy vy cos v +
∂y 2
eu uy vy cos v + eu vyy cos v − eu (vy )2 sin v
Hence
∂2 u
∂2 u
(e
sin
v)
+
(e sin v) = eu sin v((ux )2 − (vy )2 ) + eu sin v(uxx + uyy ) + eu cos v(vxx + vyy ) +
∂x2
∂y 2
2eu cos v(ux vx + uy vy ) + eu sin v((uy )2 − (vx )2 )
Since u + iv is analytic, it follows that u and v are harmonic and we get
uxx + uyy = 0,
vxx + vyy = 0,
ux = vy and uy = −vx
Hence also
(ux )2 = (vy )2
(uy )2 = (vx )2 and
ux vx = −uy vy
Hence
∂2 u
∂2 u
(e
sin
v)
+
(e sin v) = 0
∂x2
∂y 2
and so eu sin v is harmonic.
QUESTION 4
4.1 We have that
Hence in the circle |z| =
√
√
|z|2 = zz = ( 2)2 = 2
2 we get
z=
hence
(z)2 =
Hence
2
z
√
4
on
the
circle
C
:
|z|
=
2
z2
Z
Z
4
dz =
z 2 dz
2
z
C
C
5
4.2
We use
Z
C
f (z)
2πi (n)
f (z0 ) where C is the circle |z| = 3.
dz =
n+1
(z − z0 )
n!
Hence z0 = −1 and n = 3.
Clearly z0 = −1 lies inside the circle |z| = 3.
Also
f (z) = e2z , f 0 (z) = 2e2z , f 00 (z) = 4e2z and f 000 (z) = 8e2z
Hence f 000 (−1) = 8e−2 and so
Z
|z|=3
e2z
2πi 8
8πi
( 2) = 2
dz =
4
(z + 1)
3! e
3e
4.3
√
We have that f√(z) is analytic in the domain |z| < 1 and |f (0)| = |1 + i| = 2.
Since |f (z)| ≤ 2 = |f (0)| for |z| < 1 and 0 lies in the domain |z| < 1, it follows from the Maximum
Modulus Principle that f is constant in the domain |z| < 1.
Hence f (z) = 1 + i for |z| < 1 and so f 0 (z) = 0 for |z| < 1 and f 0 (0) = 0.
4.4
Since |g(z)| > |f (z)| ≥ 0 it follows that |g(z)| > 0 and so g(z) 6= 0 for every z ∈ C.
(z)
is also analytic on C.
Since f and g are analytic on C, fg(z)
(z)
But fg(z)
< 1 on C so by Louisville’s Theorem
QUESTION 5
Partial fractions:
f (z)
g(z)
is constant on C.
A
B
1
=
+
(z − 1)(z − 2)
z−1 z−2
i.e.
1 = A(z − 2) + B(z − 1)
If z = 1 then A = −1; if z = 2 then B = 1.
Hence
−1
1
1
=
+
(z − 1)(z − 2)
z−1 z−2
Since 1 < |z| < 2 we get z1 < 1 and z2 < 1.
Hence
∞
∞
X
−1
1
1
1X 1 n
(
=−
=
−
)
=
−
valid for |z| > 1
1
n+1
z−1
z n=0 z
z
z(1 − z )
n=0
Also
∞
1
1
1X z n
( )
=
=
−
z−2
−2(1 − z2 )
2 n=0 2
∞
X
zn
= −
valid for |z| < 2
2n+1
n=0
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MAT3705/103
Hence
∞
∞
X 1
X zn
1
= −
−
valid for 1 < |z| < 2
(z − 1)(z − 2)
z n+1 n=0 2n+1
n=0
= ... −
z
1
1
1
1 1
z2
−
−
−
−
−
−
− ...
z4 z3 z2 z
2 22 23
QUESTION 6
6.1
z2 − 1
z2 − 1
=
z 2 − 5iz − 4
(z − i)(z − 4i)
Hence we have isolated singularities at z = i and z = 4i.
Let p(z) = z 2 − 1 and q(z) = z 2 − 5iz − 4.
Since p(i) 6= 0, p(4i) 6= 0 , q 0 (i) = 2i − 5i 6= 0, q 0 (4i) = 8i − 5i 6= 0, both of these singularities are
simple poles.
Hence
p(z)
p(i)
i2 − 1
−2
−2i
Resz=i
= 0
=
=
=
and
q(z)
q (i)
2i − 5i
−3i
3
Resz=4i
p(4i)
(4i)2 − 1
−17
17i
p(z)
= 0
=
=
=
q(z)
q (4i)
2(4i) − 5i
3i
3
and
6.2.1
Both i and 4i lie inside the circle |z| = 6.
Hence
Z
z2 − 1
17i 2i
15i
dz = 2πi(
− ) = 2πi(
) = −10π
2
3
3
3
|z|=6 z − 5iz − 4
6.2.2
Since |4i − i| = 3 > 1 and |i − i| = 0 < 1 only the singularity at z = i lies inside the circle |z − i| = 1.
Hence
Z
2i
4π
z2 − 1
dz = 2πi(− ) =
2
3
3
|z−i|=1 z − 5iz − 4
QUESTION 7.
7.1 We use the parametric representation z = eiθ (0 ≤ θ ≤ 2π) to describe C .
Furthermore dz
= ieiθ = iz i.e. dθ = dz
dθ
iz
and from cos θ = 21 (eiθ + e−iθ ),we obtain cos θ = 12 (z + z −1 )
Z
0
2π
dθ
=
3 − 2 cos θ
Z
|z|=1
Z
=
|z|=1
Z
=
|z|=1
Z
=
|z|=1
1
1
dz
3 − 2 (z + z −1 ) iz
dz
iz(3 − z − z1 )
dz
i(3z − z 2 − 1)
idz
2
z − 3z + 1
1
2
7
The integrand has singularities at
z=
Note that
Then
√
3+ 5
2
3±
√
√
9−4
3± 5
=
2
2
> 1 so this singularity lies outside |z| = 1)
Res
√
3− 5
2
z2
i
i
i
√
==
= −√
− 3z + 1
5
2( 3−2 5 ) − 3
Therefore by the residue theorem we get
Z 2π
dθ
=
3 − 2 cos θ
0
Z
|z|=1
z2
idz
− 3z + 1
i
= 2πi (− √ )
5
2π
= √
5
7.2 Question: Show that f (z) − z 4 has exactly four solutions inside the unit circle under the condition
that |f (z)| < 1 for |z| = 1.
Let
g(z) = −z 4
Then g(z) is analytic inside and on |z| = 1 and has 4 zeros (counting multiplicities) inside |z| = 1.
Also
|g(z)| = −z 4 = 1 > |f (z)| on |z| = 1.
Thus by Rouche’s theorem g(z) and f (z) + g(z) = f (z) − z 4 have the same number of zeros inside
|z| = 1
i.e. f (z) − z 4 has exactly four solutions inside the unit circle.
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