← 2015 Paper 2

UPSC 2015 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Cauchy's residue theorem · Complex Analysis · asked 5× in 13 yrs · Read the full method →

Question

State Cauchy’s residue theorem. Using it, evaluate the integral

∫Cez+1z(z+1)(z−i)2 dz,C:∣z∣=2.\int_C\dfrac{e^z+1}{z(z+1)(z-i)^2}\,dz,\quad C:|z|=2.

Technique

Cauchy’s residue theorem; compute residues at three poles inside ∣z∣=2|z|=2: two simple poles (direct lim⁡(z−a)f(z)\lim(z-a)f(z)) and one double pole (derivative formula).

Solution

Cauchy’s residue theorem (statement)

Let ff be analytic on and inside a simple closed contour CC except at isolated singularities z1,…,znz_1,\dots,z_n inside CC. Then

∮Cf(z) dz=2πi∑k=1nRes⁡z=zkf(z).\oint_C f(z)\,dz=2\pi i\sum_{k=1}^n\operatorname{Res}_{z=z_k}f(z).

Step 1 — Singularities of integrand

f(z)=ez+1z(z+1)(z−i)2f(z)=\dfrac{e^z+1}{z(z+1)(z-i)^2} has:

All satisfy ∣z∣<2|z|<2, so all lie inside C:∣z∣=2C:|z|=2.

Step 2 — Residue at z=0z=0 (simple pole)

Res⁡z=0f=lim⁡z→0z⋅f(z)=e0+1(0+1)(0−i)2=21⋅(−1)=−2.\operatorname{Res}_{z=0}f=\lim_{z\to 0}z\cdot f(z)=\dfrac{e^0+1}{(0+1)(0-i)^2}=\dfrac{2}{1\cdot(-1)}=-2.

Step 3 — Residue at z=−1z=-1 (simple pole)

Res⁡z=−1f=lim⁡z→−1(z+1)f(z)=e−1+1(−1)(−1−i)2.\operatorname{Res}_{z=-1}f=\lim_{z\to -1}(z+1)f(z)=\dfrac{e^{-1}+1}{(-1)(-1-i)^2}.

Compute (−1−i)2=(1+i)2=1+2i−1=2i(-1-i)^2=(1+i)^2=1+2i-1=2i. So:

Res⁡z=−1f=e−1+1(−1)(2i)=e−1+1−2i=(e−1+1)⋅i2=i(1+1/e)2.\operatorname{Res}_{z=-1}f=\dfrac{e^{-1}+1}{(-1)(2i)}=\dfrac{e^{-1}+1}{-2i}=\dfrac{(e^{-1}+1)\cdot i}{2}=\dfrac{i(1+1/e)}{2}.

(Multiplied numerator and denominator by −i-i to rationalize: 1/(−2i)=i/21/(-2i)=i/2 since −i⋅(−2i)=2i2=−2-i\cdot(-2i)=2i^2=-2, so 1/(−2i)=i/21/(-2i)=i/2? Check: (−2i)(i/2)=−i2=1(-2i)(i/2)=-i^2=1? No, (−2i)(i/2)=−i2=1(-2i)(i/2)=-i^2=1 ✓. So 1/(−2i)=−(i/2)1/(-2i)=-(i/2)… let me redo.)

1−2i=1−2i⋅ii=i−2i2=i2\dfrac{1}{-2i}=\dfrac{1}{-2i}\cdot\dfrac{i}{i}=\dfrac{i}{-2i^2}=\dfrac{i}{2}. Hmm, −2i2=−2(−1)=2-2i^2=-2(-1)=2, so 1−2i=i2\dfrac{1}{-2i}=\dfrac{i}{2}. ✓

So Res⁡z=−1f=(e−1+1)⋅i2=i(1+1/e)2\operatorname{Res}_{z=-1}f=(e^{-1}+1)\cdot\dfrac{i}{2}=\dfrac{i(1+1/e)}{2}.

Wait — let me verify the sign: 1−2i\dfrac{1}{-2i}: multiply by ii=1\dfrac{i}{i}=1: i−2i⋅i=i−2(−1)=i2\dfrac{i}{-2i\cdot i}=\dfrac{i}{-2(-1)}=\dfrac{i}{2}. Yes, 1−2i=i2\dfrac{1}{-2i}=\dfrac{i}{2}.

But check: −2i⋅i2=−i2=1-2i\cdot\dfrac{i}{2}=-i^2=1. ✓

So Res⁡z=−1f=(e−1+1)−2i⋅ii\operatorname{Res}_{z=-1}f=\dfrac{(e^{-1}+1)}{-2i}\cdot\dfrac{i}{i}… I’m confusing myself. Let me just directly compute:

e−1+1(−1)(2i)=e−1+1−2i\dfrac{e^{-1}+1}{(-1)(2i)}=\dfrac{e^{-1}+1}{-2i}. Multiply top and bottom by ii: i(e−1+1)−2i2=i(e−1+1)2\dfrac{i(e^{-1}+1)}{-2i^2}=\dfrac{i(e^{-1}+1)}{2}.

So Res⁡z=−1f=i(e−1+1)2=i(1+1/e)2\operatorname{Res}_{z=-1}f=\dfrac{i(e^{-1}+1)}{2}=\dfrac{i(1+1/e)}{2}.

Step 4 — Residue at z=iz=i (double pole)

For a pole of order 2 at z=iz=i:

Res⁡z=if=lim⁡z→iddz[(z−i)2f(z)]=lim⁡z→iddz ⁣[ez+1z(z+1)].\operatorname{Res}_{z=i}f=\lim_{z\to i}\dfrac{d}{dz}\left[(z-i)^2 f(z)\right]=\lim_{z\to i}\dfrac{d}{dz}\!\left[\dfrac{e^z+1}{z(z+1)}\right].

Let g(z)=ez+1z(z+1)=ez+1z2+zg(z)=\dfrac{e^z+1}{z(z+1)}=\dfrac{e^z+1}{z^2+z}.

g′(z)=ez(z2+z)−(ez+1)(2z+1)(z2+z)2g'(z)=\dfrac{e^z(z^2+z)-(e^z+1)(2z+1)}{(z^2+z)^2}.

At z=iz=i: z2+z=i2+i=−1+i=i−1z^2+z=i^2+i=-1+i=i-1. So (z2+z)2=(i−1)2=i2−2i+1=−1−2i+1=−2i(z^2+z)^2=(i-1)^2=i^2-2i+1=-1-2i+1=-2i.

Numerator at z=iz=i: ei(i−1)−(ei+1)(2i+1)e^i(i-1)-(e^i+1)(2i+1).

Expand (ei+1)(2i+1)=ei(2i+1)+(2i+1)=2iei+ei+2i+1(e^i+1)(2i+1)=e^i(2i+1)+(2i+1)=2ie^i+e^i+2i+1.

So numerator =ei(i−1)−2iei−ei−2i−1=ei[i−1−2i−1]−2i−1=ei(−i−2)−2i−1=e^i(i-1)-2ie^i-e^i-2i-1=e^i[i-1-2i-1]-2i-1=e^i(-i-2)-2i-1.

Thus

Res⁡z=if=ei(−i−2)−2i−1−2i=−ei(i+2)−(2i+1)−2i=ei(i+2)+(2i+1)2i.\operatorname{Res}_{z=i}f=\dfrac{e^i(-i-2)-2i-1}{-2i}=\dfrac{-e^i(i+2)-(2i+1)}{-2i}=\dfrac{e^i(i+2)+(2i+1)}{2i}.

Multiply numerator and denominator by −i/(2)-i/(2)… easier: 12i=−i2\dfrac{1}{2i}=-\dfrac{i}{2} (since 2i⋅(−i/2)=−i2=12i\cdot(-i/2)=-i^2=1).

Res⁡z=if=−i2[ei(i+2)+(2i+1)]=−iei(i+2)2−i(2i+1)2.\operatorname{Res}_{z=i}f=-\dfrac{i}{2}\bigl[e^i(i+2)+(2i+1)\bigr]=-\dfrac{i e^i(i+2)}{2}-\dfrac{i(2i+1)}{2}.

Simplify:

So −iei(i+2)2=ei⋅(12−i)-\dfrac{ie^i(i+2)}{2}=e^i\cdot\bigl(\dfrac{1}{2}-i\bigr).

So

Res⁡z=if=ei ⁣(12−i)+1−i2.\operatorname{Res}_{z=i}f=e^i\!\left(\dfrac{1}{2}-i\right)+1-\dfrac{i}{2}.

Using ei=cos⁡1+isin⁡1e^i=\cos 1+i\sin 1:

ei(12−i)=(cos⁡1+isin⁡1)(12−i)=cos⁡12−icos⁡1+isin⁡12+sin⁡1=(cos⁡12+sin⁡1)+i(sin⁡12−cos⁡1).e^i(\tfrac{1}{2}-i)=(\cos 1+i\sin 1)(\tfrac{1}{2}-i)=\tfrac{\cos 1}{2}-i\cos 1+\tfrac{i\sin 1}{2}+\sin 1=\bigl(\tfrac{\cos 1}{2}+\sin 1\bigr)+i\bigl(\tfrac{\sin 1}{2}-\cos 1\bigr).

So

Res⁡z=if=(cos⁡12+sin⁡1+1)+i(sin⁡12−cos⁡1−12).\operatorname{Res}_{z=i}f=\bigl(\tfrac{\cos 1}{2}+\sin 1+1\bigr)+i\bigl(\tfrac{\sin 1}{2}-\cos 1-\tfrac{1}{2}\bigr).

Step 5 — Sum of residues

S=Res⁡0+Res⁡−1+Res⁡i.S=\operatorname{Res}_0+\operatorname{Res}_{-1}+\operatorname{Res}_i.

Real part of SS: −2+0+(cos⁡12+sin⁡1+1)=cos⁡12+sin⁡1−1-2+0+\bigl(\tfrac{\cos 1}{2}+\sin 1+1\bigr)=\tfrac{\cos 1}{2}+\sin 1-1.

Imaginary part of SS: 1+1/e2+sin⁡12−cos⁡1−12=1+1/e+sin⁡1−12−cos⁡1=1/e+sin⁡12−cos⁡1\tfrac{1+1/e}{2}+\tfrac{\sin 1}{2}-\cos 1-\tfrac{1}{2}=\tfrac{1+1/e+\sin 1-1}{2}-\cos 1=\tfrac{1/e+\sin 1}{2}-\cos 1.

Step 6 — Integral value

∫Cf dz=2πi⋅S=2πi ⁣[(cos⁡12+sin⁡1−1)+i(1/e+sin⁡12−cos⁡1)].\int_C f\,dz=2\pi i\cdot S=2\pi i\!\left[\bigl(\tfrac{\cos 1}{2}+\sin 1-1\bigr)+i\bigl(\tfrac{1/e+\sin 1}{2}-\cos 1\bigr)\right].

Distribute 2πi2\pi i (i⋅i=−1i\cdot i=-1):

=2πi(cos⁡12+sin⁡1−1)−2π(1/e+sin⁡12−cos⁡1).=2\pi i\bigl(\tfrac{\cos 1}{2}+\sin 1-1\bigr)-2\pi\bigl(\tfrac{1/e+\sin 1}{2}-\cos 1\bigr).

Answer

  ∫Cf dz=−2π ⁣(1/e+sin⁡12−cos⁡1)+2πi ⁣(cos⁡12+sin⁡1−1).  \boxed{\;\int_C f\,dz=-2\pi\!\left(\tfrac{1/e+\sin 1}{2}-\cos 1\right)+2\pi i\!\left(\tfrac{\cos 1}{2}+\sin 1-1\right).\;}
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