SEC.
INTEGRATION DIFFERENTIATION
AND
OF POWER
SERIES
59
207
represents a continuous function at each point interior to its circle of convergence. In
this section, we prove that the sum S ( z ) is actually analytic within that circle. Our
proof depends on
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MATHEMATICS 1
MONDAY 21 JUNE 2010
9465
Afternoon
Time: 3 hours
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2010
Mathematics
STEP 9465/9470/9475
October 2010
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MATHEMATICS 3
MONDAY 21 JUNE 2010
9475
Afternoon
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Sixth Term Examination Papers
9470
MATHEMATICS 2
Wednesday 23 JUNE 2010
Morning
Time: 3 hours
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Section A:
1
Pure Mathematics
The points S , T , U and V have coordinates (s, ms), (t, mt), (u, nu) and (v, nv ), respectively.
The lines SV and U T meet the line y = 0 at the points with coordinates (p, 0) and (q, 0),
respectively. Show that
(m n)sv
p=
,
Section A:
1
Pure Mathematics
Two curves have equations x4 + y 4 = u and xy = v , where u and v are positive constants.
State the equations of the lines of symmetry of each curve.
The curves intersect at the distinct points A, B , C and D (taken anticlock
Section A:
1
Pure Mathematics
A proper factor of an integer N is a positive integer, not 1 or N , that divides N .
(i)
(ii)
2
Show that 32 53 has exactly 10 proper factors. Determine how many other integers of
the form 3m 5n (where m and n are integers) h
STEP Solutions
2009
Mathematics
STEP 9465, 9470, 9475
The Cambridge Assessment Group is Europe's largest assessment agency
and plays a leading role in researching, developing and delivering
assessment across the globe. Our qualifications are delivered in
STEP Mathematics I 2007: Report
General comments
There were significantly more candidates attempting this paper this year (an increase
of nearly 50%), but many found it to be very difficult and only achieved low scores.
In particular, the level of algebra
STEP
Mathematics
STEP 9465, 9470, 9475
STEP Solutions
June 2007
STEP2007
Oxford Cambridge and RSA Examinations
OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body, established by the
University of Cambridge Local Examinations Syndicate
PROOF THE THEOREM
OF
SEC. 45
147
Continuing with a function f which is analytic throughout a region R consisting
of a positively oriented simple closed contour C and points interior to it, we are now
ready to prove the CauchyGoursat theorem, namely that
142
INTEGRALS
CHAP.
4
I
3. Use the theorem in Sec. 42 to show that
when Co is any closed contour which does not pass through the point zo. [Compare
Exercise 10(b), Sec. 40.1

4. Find an antiderivative F z ( z ) of the branch h ( z ) of
in Example 4, Sec.
212
SERIES
CHAP.
5
The method of proof here is similar to the one used in proving Theorem 1. The
hypothesis of this theorem tells us that there is an annular domain about zo such that
for each point z in it. Let g(z) be as defined by equation (4), but now
202
CHAP. 5
SERIES
converges in the open disk lz 1 < 1z 1 ; and the theorem is proved when zo = 0.
When zo is any nonzero number, we assume that series (1) converges at z = z l
(zl # zo). If we write w = z  zo, series (1) becomes
and this series converge
The representation in Dl is a Maclaurin series. To find it, we write
and observe that, since Jz( < 1 and 12/2 1 < 1 in Dl,
As for the representation in 4,we write
Since I l/zl c 1 and 12/21 < 1 when 1 < lzl < 2, it follows that
If we replace the index of
192
SERIES
CHAP. 5
expansion (1) reduces to a Taylor series about zo.
If, however, f fails to be analytic at zo but is otherwise analytic in the disk
lz  sol < R2, the radius RI can be chosen arbitrarily small. Representation (1) is
then valid in the pun
SEC.
54
Term by term differentiation will be justified in Sec. 59. Using that procedure
here, we differentiate each side of equation (2) and write
That is,
00
cos z =
z2n
C(1)"(2n)!
(Izl < w .
n=O
EXAMPLE 3. Because sinh z = i sin(iz) (Sec. 34), we need
182
SERIES
(b) Write z, = x,
CHAP.
+ iy,
5
and recall from the theory of sequences of real numbers that
the convergence of x, and y, (n = 1, 2, .) implies that Ix,l I 1and ly,l 5 M 2
M
(n = 1,2, . . .) for some positive numbers M 1and M2.
.
53. TAYLOR SER
SEC.
CONVERGENCESEQUENCES177
OF
51
Conversely, if we start with condition (4), we know that, for each positive number
E, there exists a positive integer no such that
1 (xn+ iyn)  (x + iy) I < E whenever n > no.
But
and
and this means that
IxnXI
<E
and
l
172
INTEGRALS
CHAP.
4
2. Suppose that f ( z ) is entire and that the harmonic function u ( x , y) = Re[ f ( z ) ]has an
upper bound ug; that is, u ( x , y) 5 uo for all points ( x , y) in the xy plane. Show that
u (x , y ) must be constant throughout the
SEC. 50
and this enables us to write
(5)
Ian l
IP(z)l=la,+wl/znl> lzlnzRnla, l
2
2
whenever / z l > R .
Evidently, then,
If(z)l=
1
IP(z)I
2
<  whenever
lanlRn
lz 1
> R.
So f is bounded in the region exterior to the disk lz 1 5 R. But f is continuous i
162
INTEGRALS
CHAP.
4
EXAMPLE 1. If C is the positively oriented unit circle lzl = 1 and
f (z) = exp(ZZ),
then
EXAMPLE 2. Let zo be any point interior to a positively oriented simple closed
contour C . When f (z) = 1, expression (5) shows that
and
(Compar
CHAP.
4
For a verification, we use Theorem 2 to write
f ( z ) dz = 0;
and we note that this is just a different form of equation (3).
Corollary 2 is known as the principle of deformation ofpaths since it tells us that
if CI is continuously deformed into C
CAUCHY
INTEGRAL
FORMULA157
SEC. 47
47. CAUCHY INTEGRAL FORMULA
Another fundamental result will now be established.
Theorem. Let f be analytic everywhere inside and on a simple closed contour C ,
taken in the positive sense. I f z o is any point interior t
STEP I, 2002
Section A:
1
2
Pure Mathematics
Show that the equation of any circle passing through the points of intersection of the ellipse
(x + 2)2 + 2y 2 = 18 and the ellipse 9(x 1)2 + 16y 2 = 25 can be written in the form
x2 2ax + y 2 = 5 4a .
2
Let f