M337 2011
M337 2011
M337 2011
2011 SOLUTIONS
Page 1
Question 1
(a)
1 i
1
1 i
Since i(1- i) = 1 + i then
= 3 = i .
=
i
1+ i
i (1 i )
[[ I have included a couple of other methods. ]]
1
1+
2 exp( / 4 )
= ( exp( / 2 ) ) 3 = exp( 3 / 2) = i .
2 exp( / 4 )
3
i
=
i
3
3
(1 i ) 2
1 i
2i
3
=
= i = i.
1+ i
2
(1 + i )(1 i )
3
(b)
(c)
e
2
3+ i
[[ Parts c and d are very similar to those on the 2007 paper. In both cases the
value in the bracket can be written as exp(i). ]]
1+ i 3
3
2
Alternatively using A2, Sect 5, Para. 1
1+ i 3
1+ i 3
+ iArg 1 + i 3 = log e 1 + i = i
= log e
Log
2
3
3
2
3 i
(d)
1+ i 3
1+ i 3
= exp ( 3 i ) i = exp i + = exp using result from part (c)
3
3
2011 SOLUTIONS
Page 2
Question 2
(a)
B-A
(b)
A
B-A
(i) open
A3, Sect. 4, Para. 1
No
Yes
Yes
(ii) connected
A3, Sect. 4, Para. 3
Yes
Yes
No
(iii) a region
A3, Sect. 4, Paras 6-8
No
Yes
No
(iv) bounded
A3, Sect. 5, Para. 4
No
Yes
Yes
(v) compact
A3, Sect. 5, Para. 5
No
No
No
2011 SOLUTIONS
Page 3
Question 3
(a) The standard parametrization for the line segment is (A2, Sect. 2, Para. 3)
( t ) = (1 t ) i + t1 = t + (1 t ) i
(t [0, 1])
Since is a smooth path and (Re z)(Im z) is continuous along the path then (B1,
Sect. 2, Para. 1)
.
(Re z )( Im z ) dz =
( Re ( t ) ) ( Im ( t ) ) ( t ) dt
/
t2 t3
1 i
= t (1 t )(1 i ) dt = (1 i ) =
6
2 30
0
1
f ( z) =
(b)
z2 1
is continuous on {-i, i} and hence on the circle C.
z2 + 1
Therefore the Estimation Theorem (Unit B1, Section 4, Para. 3) will be used to find
an upper limit.
The length of C is L = 2 * 2 = 4.
[[ We hve to find an upper limit for f(z). Sometimes I get confused over whether I
need to find an upper limit or lower limit for the numerator (top) and denominator of
f(z). I then think of a simple example. 2/2 < 3/2 < 3/1. So we need an upper limit for
the numerator (2 < 3) and a lower limit for the denominator (2 > 1). ]]
Using the Triangle Inequality (Unit A1, Section 5, Para. 3b) then, for z C, we have
2
z2 1 z2 + 1 = z + 1 = 4 + 1 = 5
Using the Backwards form of the Triangle Inequality (Unit A1, Section 5, Para. 3c)
then, for z C, we have
z2 + 1
Therefore M =
So
f ( z ) dz
z2 1 = 4 1 = 3
z2 1 5
z2 + 1 3
ML =
for z C.
5
20
* 4 =
.
3
3
2011 SOLUTIONS
Page 4
Question 4
Let R be the simply-connected region {z: |z| < 3}.
(a)
C is a closed contour in R, and
cos z
is analytic on R.
z
cos z
dz = 0 .
C z
(b)
As C is a simple-closed contour in R , f(z) = cos z is analytic on R , and = /3 is
inside C, then by Cauchys Integral Formula (B2, Sect. 2, Para. 1)
cos z
dz =
z /3
f ( z)
dz = 2 if ( / 3) = 2 i cos 3 = i .
z - /3
(c)
As C is a simple-closed contour in R , f(z) = cos z is analytic on R , and = /2 is
inside C, then by Cauchys n'th Derivative Formula (B2, Sect. 3, Para. 1) with n = 3
we have
cos z
( z / 2)
C
dz =
2 i ( 3 )
f (
3!
)=
i
i
sin =
.
3
2 3
2011 SOLUTIONS
Page 5
Question 5
(a)
f ( z) =
z+ 1
z+ 1
=
has simple poles at z = 0, and z = 2i.
2
z z + 4 z ( z 2i )( z + 2i )
Res ( f ,0 ) =
Res ( f ,2i ) =
lim
lim
z+ 1
1
( z 0) f ( z ) =
=
2
z 0
z 0 z + 4 4
lim
lim
z+ 1
1 + 2i
( z 2i ) f ( z ) =
=
z 2i
z 2i z ( z + 2i )
8
Res ( f , 2i ) =
lim
lim
z+ 1
1 2i
( z + 2i ) f ( z ) =
=
z 2i
z 2i z ( z 2i )
8
p(t )
t+ 1
dt
=
q( t ) t t 2 + 4 dt = 2 iS + iT
where S is the sum of the residues of f at the poles in the upper half-plane, and
T is the sum of the residues of f at the poles on the real axis.
t+ 1
1
1 + 2i
dt = 2 i
+ i =
2
+ 4
8
4 2
t (t
2011 SOLUTIONS
Page 6
Question 6
(a)
2011 SOLUTIONS
Page 7
Question 7
(a)
The conjugate velocity function q ( z )=i/ z 2 .
As q is a steady continuous 2-dimensional velocity function on the region {0} and
q is analytic on {0} then q is a model fluid flow (Unit D2, Section 1, Para. 14).
i
is a primitive of q . Therefore is a complex potential
z
function for the flow (Unit D2, Section 2, Para. 1).
(b)
On {0}, ( z )=
( )
(c) If C is the circulation of along and F is the flux of q across then (D1, Sect.
1, Para. 1 and D2, Sect. 1, Paras. 9 & 10)
i i
i
C + iF = q ( z ) dz = ( ( 4,0 ) ) ( ( 2,0) ) = =
4 2
4
Therefore the flux of q across is -.
[[ The normal is in the direction -i. ]]
2011 SOLUTIONS
Page 8
Question 8
(a)
Using the result in Unit D3, Section 2, Para. 1 then the iteration sequence
zn+1 = 2zn2 4 zn + 2 is conjugate to the iteration sequence
wn+1 = wn2 + (2*2 + (-4) / 2 - (-4)2 / 4) = wn2 - 2
and conjugating function h(z) = 2z - 2.
Therefore w0 = h(z0) = 2z0 - 2 = 2 - 2 = 0.
(b)
If is a fixed point of P-2 then P-2() = 2 - 2 = (D3, Sect. 1, Para 3).
As 2 - - 2 = ( + 1)( - 2) = 0 then P-2(z) has fixed points at z = -1 and z = 2.
P-2 '(z) = 2z.
As | P-2 '(-1)| = 2 > 1 and | P-2 '(2) | = 4 > 1 then both are repelling fixed points (D3,
Sect. 1, Para. 5).
(c) [[ If you have add coordinates on the axes of the diagram of the Mandelbrot set
then you will see that c is not in the Mandelbrot set.]]
The Mandelbrot set M can be specified as
c : Pcn ( 0 ) 2, for n = 1,2,
where Pc(z) = z2 + c. (D3, Sect. 4, Para. 5 and D3, Sect. 2, Para. 2).
Let c =
1
2
+i
Pc ( 0 ) = c =
Pc2 ( 0 ) =
1
2
+ i.
( 12 + i ) 2 + ( 12 + i ) =
1
4
1+ i +
1
2
+ i = 14 + 2i.
2
As Pc ( 0 ) > 2 then c does not lie in the Mandelbrot set.
2011 SOLUTIONS
Page 9
Question 9
(a)(i)
f ( x + iy ) = 2e i Re z z = 2e ix ( x iy ) = ( 2 cos x x ) + i ( 2 sin x + y ) = u ( x, y ) + iv( x, y )
where u(x,y) = 2 cos x - x, and v(x,y) = 2 sin x + y are real-valued functions .
(a)(ii)
f is defined on the region .
u
= 2 sin x 1 ,
x
u
= 0,
y
v
= 2 cos x ,
x
v
=1
y
u u v v
, , ,
x y x y
1. exist on
2. are continuous on S.
3. satisfy the Cauchy-Riemann equations on S
then, by the Cauchy-Riemann Converse Theorem (A4, Sect. 2, Para. 3), f is
differentiable on S.
(a)(iii) If (a, b) S, then using equations (A)
f / (a, b) =
u
( a, b ) + i v ( a, b ) = ( 2 sin a 1) + i 2 cos a = 1 (A4, Sect. 2, Para. 3).
x
x
Therefore f / is constant on S.
2011 SOLUTIONS
Page 10
(b)(i) The domain of g is (Unit A4, Section 1, Para. 7) and its derivative g/(z) = 2z
also has domain (Unit A4, Section 3, Para. 4) . Therefore g is analytic on - {0}.
Since g/(z) 0 on - {0}then g is conformal on - {0} (Unit A4, Section 4, Para. 6).
(b)(ii) As g is analytic on and g/ (2i) 0 then a small disc centred at 2i is mapped
approximately (Unit A4, Section 1, Para. 11) to a small disc centred at
g(2i) = -4 + 2 = -2. The disc is rotated by Arg (g / (2i)) = Arg 4i = /2, and scaled by a
factor | g / (2i)| = | 4i | = 4.
(b)(iii) 1 is in the domain of 1 and
.
1 (1) = 1 1 + 2i = 2i
0 is in the domain of 2 and 2 ( 0 ) = ( 0 + 2 ) i = 2i . Therefore 1 and 2 meet at
the point 2i.
/
As 1/ ( t ) = 2t + 2i then at 1, Arg 1 (1) = Arg ( 2(1 + i ) ) =
.
4
.
2
Therefore the angle from 1 to 2 at their point of intersection is /4.
( ( 0) ) =
As 2/ ( t ) = 3i then at 0, Arg
/
2
Arg ( 3i ) =
(b)(iv)
In diagram (b)(iv) the imaginary-axis is 2.
In diagram (b)(v) the horizontal line is g(2).
(b)(v)
2011 SOLUTIONS
Page 11
Question 10
(a)
The domain of f, A = {2n : n is an integer}.
[[ as cos(2n) = 1 ]]
(b)
cos z = 1
z2 z4 z6
+
...
2! 4! 6!
So 1 cos z =
, for z .
z2
2z 2 2z 4 z 2
z2 z4
1
1
+
... =
+
... .
2
4!
6! 2 12 360
z
2
z2 z4
= 1
+
...
So
1 cos z z 12 360
z2 z4
2 z 2 z 4
= 1 +
+ ... +
+ ... + ...
z 12 360
12 360
2
z2
1
1
= 1+
+ z4
+
+ ...
z 12
360 144
As
1
1
1
( 2 + 5) = 1 then the Laurent series about 0 for f is
+
=
360 144 720
240
3
2 z z
+ +
+ ... = an z n for 0 < |z| < 2
z 6 120
n=
f ( z ) dz = 2 ia
= 2 i ( 2) = 4 i. .
2011 SOLUTIONS
Page 12
1
n
) : n = 1,2,3,...}
f agrees with g throughout the set S A and S has a limit point which is in A. By the
Uniqueness theorem (B3, Sect. 5, Para. 7) f agrees with g throughout A. Hence f is the
iy
only analytic function with domain A such that f ( iy ) =
for y > 0.
1 cosh y
(d)
Since cos z = 1 when z = 0, z = 2, z = 4, . then f(z) has singularities at points
of the form 2k, k .
Singularity at z = 0 (k = 0).
At z = 0 we can use the Laurent series found in part (a). Since
lim
zf ( z ) = 2 then
z 0
z
z 2k
2k
=
+
1 cos z 1 cos( z 2k ) 1 cos( z 2k
Since
lim
( z 2k
z 2k
lim
f ( z) =
z 2k
2
( z 2k ) 2
2k ( z 2k )
+
= 0 + 4k
1 cos( z 2k ) 1 cos( z 2k )
2011 SOLUTIONS
Page 13
Question 11
(a)
Let Df = {z: |z| < 3} and Dg = {z: |z| > 3}. Df and Dg are regions.
Since Df Dg = then f and g are not direct analytic continuations of each other (C3,
Sect. 1, Para. 1).
z
When z Df then |z|/3 < 1 and the geometric series
n= 0 3
the sum
1
3
=
. (B3, Sect. 3, Para. 5)
z
1 3 3 z
3
When z Dg then 3/|z| < 1 and the geometric series is convergent and has
n= 0 z
n
n
1
z
3 3
3 z
3
3
=
the sum
. So = =
.
=
3
1 z z 3
z n= 0 z
z z 3 3 z
n= 1 z
Let h( z ) =
3
on Dh, where Dh = C {3}.
3 z
2011 SOLUTIONS
Page 14
(b)
Let f(z) = z2 exp(1 + z2) and R = {z : | z | < 2}.
[[ The boundary R is defined in A3, Sect. 5, Para. 10. ]]
As f is analytic on the bounded region R and continuous on its closure R (C2, Sect.
4, Para. 3) then, by the Maximum Principle (C2, Sect. 4, Para. 4), there exists an
R = { z : | z | = 2} such that
| f(z) | | f() | for z R .
When z R we can write it in the form z = 2 exp(i), where is real.
Then | z2 exp(1 + z2) | = |z2| | exp(1 + z2) |
= 4 exp( Re(1 + z2) )
= 4 exp( Re(1 + 4exp(i2) ) )
= 4 exp ( 1 + 4 cos 2).
2011 SOLUTIONS
Page 15
Question 12
(a)
Using the formula for a transformation mapping points to the standard triple (D1,
Sect. 2, Para. 11) then the Mbius transformation f1 which maps -1 - i, 0 , and 1 + i
to the standard triple of points 0, 1, and respectively is
( z ( 1 i ) ) ( 0 (1 + i ) ) =
( z (1 + i ) ) ( 0 ( 1 i ) )
f1 ( z ) =
z (1 + i )
z (1 + i )
(b)(i)
(b)(ii) Let C be the boundary of S. Then C is a generalised circle (D1, Sect. 1, Para.
10), and -1 i and 1 + i are inverse points of C since, when z C,
| z - (-1 i) | = | z - (1 + i) | (D1, Sect. 3, Para. 4)
Therefore f1 ( 1 i ) = 0 and f1 (1 + i ) = are inverse points with respect to f1 ( C ) (D1,
Sect. 3, Para. 6). So f ( C ) is a circle with centre 0 (D1, Sect. 3, Para. 5) and, as
1
f1 ( 0 ) = 1 , of radius 1. Therefore f1 ( S ) = T .
[[ Before I consulted the handbook I said a general point on the boundary of S was
a ia where a is real.
a + ia 1 i ( a + 1) + i ( a 1)
=
As f1 ( a ia ) =
a ia 1 i
( a 1) i( a + 1) then a ia is mapped to a point on the
unit circle as f ( a ia ) = 1 . As extended Mobius transformations map generalised
1
circles onto generalized circles that the boundary of S is mapped onto the unit circle.
As the point -1 i S and is mapped to 0 then 0 f1 ( S ) . Therefore f1 ( S ) = T .
]]
2011 SOLUTIONS
Page 16
< Im z <
< Im z <
} . ]]
}.
Therefore the image of R may be found by finding the image of { z : 2 < Im z < 2}
and then rotating it counter-clockwise about the origin by 5/4. Using the figure in
D1, Sect. 4, Para. 5 it is apparent that exp(R) = S.
[[ I find it easier to imagine a clockwise rotation by 3/4. ]]
So a conformal mapping from f(R) onto S is g(z) = exp(z).
Since the combination of conformal mapping is also conformal then a conformal
mapping from R to T is
e z (1 + i )
f ( z ) = ( f1 g )( z ) = f1 ( exp( z ) ) = z
e (1 + i )
f 1 ( w) = Log f1 1 ( w) + 2 i = Log
( 1 + i ) w + (1 + i )
w 1
+ 2 i = Log (1 + i )
+ 2 i
w 1
w + 1
(b)(v) Therefore
p = f 1 ( 0) = Log ( (1 + i ) ) + 2 i = log e 1 i + iArg ( 1 i ) + 2 i = log e 2 + i 54 .
Not every conformal mapping from R to T maps p to 0.
If a is real then, for any point w R, the point w + a R.
So the function g1(z) = exp(z + a) also maps R to S and (f1 o g1)(z) maps R to T.
Therefore if a is non-zero then the conformal mapping (f1 o g1) will map another point,
p a, to 0.