Loading Document(a) Evaluate ∫γ(z)2dz from the point z=0 to the point z=1+i along the curve γ:z(t)=t2+it .
(b) Evaluate ∮γ(z−1)3eizdz ,where γ is given by ∣z∣=2 .
(c) Expand f(z)=(z+1)(z−4)z in Laurent's series valid for 1<∣z∣<4.
Answer.
(a) since f(z) is not analytic function we parameterize the integral based on path equation:
dz=(2t+i)dt and t goes from 0 to 1
∫01(t2−it)2×(2+i)dt=∫01(t4−t2−2it3)(2t+i)dt
=∫012t5+it4−2t3−it2−4it4+2t3dt=∫012t5−3it4−it2dt
=[3t6−i53t5−i3t3]01=31−(53+31)i=31−1514i
(b) f(z) has a pole at z=1 which lies inside the circle centered at the origin and of radius 2
so applying Cauchy's Integral yields:∮=2!2πi×[dz2d2eiz]z=1=πi×−ei=−iπei
(c) to get the expansion about z=0
first we do a partial fraction f(z)=(z+1)1/5+(z−4)4/5 and we expand each term
for (1+z)1/5 to converge at ∣z∣>1 we factor out z from the denominator
(1+z)1/5=5z1(1+1/z1)=5z1(1−z1+z21−z31+⋯)=51n=0∑∞zn+1(−1n)
for (z−4)4/5 to converge at ∣z∣<4 we factor out -4 from the denominator
−51(1−z/41)=−51(1+4z+(4z)2+⋯)=−51n=0∑∞(4z)n
combining the two expansions gives f(z)=51n=0∑∞zn+1(−1)n−(4z)n
(a) Locate and classify the singularities of f(z)=z(z+i)(z−1)2(z+3)sinz .
(b) Show that the function u(x,y)=x2−y2+sin2xcosh2y is harmonic everywhere.
Find its harmonic conjugate v(x,y) Then, express the analytic function u(x,y)+iv(x,y) in terms of the complex variable z and find f(1+i) .
(c) Prove that tan−1z=2ilni−zi+z , and hence or otherwise solve the equation tanz−(2+i)=0 .
Answer.
(a) f(z)=z(z+i)(z−1)2(z+3)sin(z) has the following singularities
(b) u(x,y)=x2−y2+sin2xcosh2y
ux=2x+2cosh2ycos2xuxx=2−4sin2xcosh2y
uy=−2y+2sin2xsinh2yuyy=−2+4sin2xcosh2y
uxx+uyy=0⟹u(x,y) is harmonic
ux=vy&uy=−vx
vx=2y−2sinh2ysin2x⟹v=2xy+cos2xsinh2y+C1
vy=2x+2cos2xcosh2y⟹v=2xy+cos2xsinh2y+C2
v=2xy+cos2xsinh2y+c
f(x,y)=x2−y2+i2xy+sin2xcosh2y+icos2xsinh2y+iC
f(z)=z2+sin2z+iC
f(1+i)=2i+sin2cosh2+icos2sinh2+iC
(c) ω=tan−1z⟹z=tanω=cosωsinω
z=2ieiω−e−iω×eiω+e−iω2=i(eiω+e−iω)eiω−e−iω
z=−i(eiω+e−iω)eiω−e−iω×eiωeiω=−ie2iω+1e2iω−1
ze2iω+z=−ie2iω+i
(z+i)e2iω=i−z
e2iω=i+zi−z
2iω=ln(i+zi−z)
ω=2i1(ln(1−z)−ln(1+z))=2−i(ln(1−z)−ln(1+z))=2i(ln(1+z)−ln(1−z))
ω=tan−1z=2iln(1−z1+z)
for the equation tanz=2+i⟹z=tan−1(2+i)=2iln(3−i3+i)
(a) Expand the function f(z)=z(z−3)1 in a Laurent series valid for the annular region 1<∣z−4∣<4 .
(b) Evaluate each of the following integrals:
(i) ∮γ1(z2+1)(z+3)1dz , where γ1 is the ellipse ∣z−i∣+∣z+i∣=6
(ii) ∮γ2(z4e2/z−(z−5i)3z4)dz,γ2:∣z∣=2
Answer.
(a) partial fraction f(z)=z−1/3+z−31/3=31(z−31−z1)
the expansion is at z=4 so we rewrite f(z)=31((z−4)+11−(z−4)+41)
to comply with convergence condition we factor out z−4 and 4 from the denominator of the two fractions respectively
f(z)=31(z−41×1+z−411−41×1+4z−41)
now we expand each term in the form 1+z1=n=0∑∞(−1)nzn
f(z)=31×(n=0∑∞(−1)n×(z−41)n+1+41×(−1)n+1×(4z−4n)
=n=0∑∞(−1)n(3(z−4)−n−1−121(4z−4)n)
(b)
(i) ∮γ1(z2+1)(z+3)1dz
f(z) has poles at i,−i,−3 of which ±i lies inside the ellipse
=∮z−i1/(z+i)(z+3)dz+∮z+i1/(z−i)(z+3)dz
=2πi×([(z+i)(z+3)1]z=i+[(z−i)(z+3)1]z=−i)=−5πi
(ii) ∮(z4e2/z−(z−5i)3z4)dz=∮z4e2/zdz−∮(z−5i)3z4dz
∮z4e2/zdz=0
∮(z−5i)3z4dz=0