Independent worked answers. These are prepared for revision and are not an official KNEC marking scheme. The questions and marks below follow the supplied June/July 2022 paper.

Question 1 (20 marks)

(a) (6 marks) Given f(z)=z²+3z−2, where z=x+jy, express f(z)=u+jv and verify the Cauchy–Riemann equations.

f(z)=(x+jy)²+3(x+jy)−2
f(z)=(x²−y²+3x−2)+j(2xy+3y)
u=x²−y²+3x−2,   v=2xy+3y
ux=2x+3,   uy=−2y,   vx=2y,   vy=2x+3
ux=vy   and   uy=−vx

Both Cauchy–Riemann equations hold.

(b) (14 marks) The circle |z|=3 is mapped to the w-plane by the transformation below. Find the image circle’s centre and radius.

w=(z−3j)/(z+2j),   |z|=3
w(z+2j)=z−3j
z=j(3+2w)/(1−w)

Taking moduli and writing w=u+jv gives:

|3+2w|=3|1−w|
(3+2u)²+4v²=9[(1−u)²+v²]
u²+v²−6u=0
(u−3)²+v²=9
Centre=(3,0),   radius=3

Question 2 (20 marks)

(a) (6 marks) Sketch the odd extension of f(t)=1−t/π, for 0<t<π, then determine its Fourier sine series.

fo(t)=−1−t/π,   −π<t<0
fo(t)=1−t/π,   0<t<π,   fo(t+2π)=fo(t)

The odd extension has no cosine terms. Its sine coefficients are:

bn=(2/π)∫0π(1−t/π)sin(nt)dt=2/(πn)
fo(t)=(2/π)∑n=1∞sin(nt)/n

(b) (14 marks) The graph in Figure 1 rises linearly from (−π,0) to (0,π), jumps to (0,0), then rises linearly to (π,π), and repeats with period 2π. Give its analytical description and Fourier series.

g(t)=t+π,   −π<t<0
g(t)=t,   0<t<π,   g(t+2π)=g(t)

The mean value and Fourier coefficients are:

a0/2=π/2,   an=0
bn=−[1+(−1)n]/n
bn=−2/n   (n even),   bn=0   (n odd)
g(t)=π/2−∑k=1∞sin(2kt)/k

Question 3 (20 marks)

(a) (9 marks) Use Newton–Raphson for x³+4x−16=0, with first approximation x0=1.5. Give the root to six decimal places.

xn+1=(2xn³+16)/(3xn²+4)
x0=1.5000000000
x1=2.1162790698
x2=2.0048330889
x3=2.0000087419
x4=2.0000000000
x=2.000000   (to six decimal places)

(b) (11 marks) Use Newton–Gregory forward interpolation for the table below to find f(1.2) and f(2.0), correct to three decimal places.

x:   1.0,   1.3,   1.6,   1.9,   2.2
f(x):   4.000,   6.187,   9.256,   13.690,   18.688
Δf:   2.187,   3.069,   4.434,   4.998
Δ²f:   0.882,   1.365,   0.564
Δ³f:   0.483,   −0.801
Δ⁴f:   −1.284
h=0.3,   p=(x−1.0)/0.3
f(x)=f0+pΔf0+[p(p−1)/2]Δ²f0+[p(p−1)(p−2)/6]Δ³f0+[p(p−1)(p−2)(p−3)/24]Δ⁴f0
p(1.2)=2/3
f(1.2)=4.000+1.458−0.098+0.02385185+0.03698765=5.42083951
f(1.2)=5.421
p(2.0)=10/3
f(2.0)=4.000+7.290+3.430+0.83481481−0.18493827=15.36987654
f(2.0)=15.370

Question 4 (20 marks)

(a) (6 marks) Evaluate the triple integral shown.

∫01∫01−x∫01−x−yy dz dy dx
∫01−x−yy dz=y(1−x−y)
∫01−xy(1−x−y)dy=(1−x)³/6
∫01(1−x)³/6 dx=1/24

(b) (14 marks) Verify Green’s theorem for the line integral around the parabola y=x² from (−1,1) to (1,1), followed by the line segment from (1,1) back to (−1,1).

∮C(xy² dx−x²y dy),   P=xy²,   Q=−x²y
Qx−Py=−2xy−2xy=−4xy
∬R−4xy dA=∫−11∫x²1−4xy dy dx=0

For the parabola, set x=t, y=t², with −1≤t≤1.

P dx+Q dy=(t⁵−2t⁵)dt=−t⁵dt
∫−11−t⁵dt=0

On the top segment, y=1 and x runs from 1 to −1.

∫1−1x dx=0
∮C(P dx+Q dy)=0=∬R(Qx−Py)dA

Question 5 (20 marks)

(a) (10 marks) Evaluate the integral in polar coordinates over the region bounded by the circle x²+y²=2y.

∬R[y/(x²+y²)]dA
x²+y²=2y   ⇒   r=2sinθ,   0≤θ≤π,   0≤r≤2sinθ
[y/(x²+y²)]dA=[r sinθ/r²]r dr dθ=sinθ dr dθ
∫0π∫02sinθsinθ dr dθ=2∫0πsin²θ dθ=π

(b) (10 marks) Verify Stokes’ theorem for the vector field on the upper unit hemisphere.

F=(2x−y)i−yz²j−y²zk,   x²+y²+z²=1,   z≥0
∇×F=(0,0,1)
∬S(∇×F)·n dS=∬Sz dS=∫02π∫0π/2cosφ sinφ dφ dθ=π

The positively oriented boundary is the equator, traversed counter-clockwise as viewed from above.

r(t)=(cos t, sin t, 0),   0≤t≤2π
F·dr=(2cos t−sin t)(−sin t)dt
∮∂SF·dr=∫02π(−2sin t cos t+sin²t)dt=π
∮∂SF·dr=∬S(∇×F)·n dS=π

Question 6 (20 marks)

(a) (10 marks) Given that λ1=−1 is an eigenvalue of the matrix below, find the positive value of x and the second eigenvalue.

A=[[1, x+1], [x+3, 3]]
det(A+I)=8−(x+1)(x+3)=0
x²+4x−5=(x−1)(x+5)=0
x=1   (positive value)
A=[[1,2],[4,3]],   tr(A)=4
λ1+λ2=4,   −1+λ2=4,   λ2=5

(b) (10 marks) Find the state transition matrix Φ(t) for the system below.

d x/dt=Cx,   C=[[0,1],[−14,−9]]
det(λI−C)=λ²+9λ+14=(λ+2)(λ+7)
λ1=−2,   λ2=−7
Φ(t)=(e−2t/5)[[7,1],[−14,−2]]+(e−7t/5)[[−2,−1],[14,7]]
Φ(0)=I

Question 7 (20 marks)

(a) (8 marks) Sketch the first-quadrant region bounded by y=x, y=2x, and x=1, then evaluate the integral.

D:   0≤x≤1,   x≤y≤2x
∬D(x²+y²)dy dx=∫01∫x2x(x²+y²)dy dx
∫x2x(x²+y²)dy=10x³/3
∫0110x³/3 dx=5/6

(b) (12 marks) Use the divergence theorem for F=4xi−2y²j+z²k on the first-octant region bounded by x²+y²=4, z=0, and z=3.

∇·F=4−4y+2z
∭V(∇·F)dV=∭V4dV−4∭Vy dV+2∭Vz dV
∭V4dV=4(π)(3)=12π
−4∭Vy dV=−12∫0π/2∫02r²sinθ dr dθ=−32
2∭Vz dV=π∫032z dz=9π
∯SF·n dS=12π−32+9π=21π−32

Question 8 (20 marks)

(a) (10 marks) Show that the line integral is path-independent, then evaluate it along the two stated segments.

∫[(2xy−y⁴+3)dx+(x²−4xy³)dy]
Py=2x−4y³=Qx

A potential function is:

φ(x,y)=x²y−xy⁴+3x
∫(1,0)(2,0)P dx=φ(2,0)−φ(1,0)=6−3=3
∫(2,0)(2,1)Q dy=φ(2,1)−φ(2,0)=8−6=2
∫(1,0)(2,1)F·dr=3+2=5

(b) (5 marks) Evaluate the double integral.

∫34∫121/(x+y)² dy dx
∫121/(x+y)² dy=1/(x+1)−1/(x+2)
∫34[1/(x+1)−1/(x+2)]dx=ln(25/24)

(c) (5 marks) Reverse the order of integration and evaluate.

∫01∫x²1x³ey³dy dx
0≤y≤1,   0≤x≤√y
∫01∫0√yx³ey³dx dy=(1/4)∫01y²ey³dy
(1/4)∫01y²ey³dy=(e−1)/12
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