← 2020 Paper 2
UPSC 2020 Maths Optional Paper 2 Q2c — Step-by-Step Solution 20 marks · Section A
Contour integration of real integrals using residues · Complex Analysis · asked 9× in 13 yrs · Read the full method →
Question
Using contour integration, evaluate the integral ∫ 0 2 π 1 3 + 2 sin θ d θ \int_0^{2\pi}\frac{1}{3+2\sin\theta}\,d\theta ∫ 0 2 π 3 + 2 s i n θ 1 d θ
Technique
z = e i θ z=e^{i\theta} z = e i θ [ 0 , 2 π ] [0,2\pi] [ 0 , 2 π ] ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1
Solution
Step 1 — Substitute z = e i θ z=e^{i\theta} z = e i θ
As θ \theta θ 0 → 2 π 0\to2\pi 0 → 2 π z = e i θ z=e^{i\theta} z = e i θ C : ∣ z ∣ = 1 C:|z|=1 C : ∣ z ∣ = 1
sin θ = z − z − 1 2 i , d θ = d z i z . \sin\theta=\frac{z-z^{-1}}{2i},\qquad d\theta=\frac{dz}{iz}. sin θ = 2 i z − z − 1 , d θ = i z d z . Then
3 + 2 sin θ = 3 + 2 ⋅ z − z − 1 2 i = 3 + z − z − 1 i = 3 − i ( z − z − 1 ) . 3+2\sin\theta=3+2\cdot\frac{z-z^{-1}}{2i}=3+\frac{z-z^{-1}}{i}=3 - i\big(z-z^{-1}\big). 3 + 2 sin θ = 3 + 2 ⋅ 2 i z − z − 1 = 3 + i z − z − 1 = 3 − i ( z − z − 1 ) . So
I = ∫ 0 2 π d θ 3 + 2 sin θ = ∮ ∣ z ∣ = 1 1 3 − i ( z − z − 1 ) ⋅ d z i z . I=\int_0^{2\pi}\frac{d\theta}{3+2\sin\theta}
=\oint_{|z|=1}\frac{1}{3 - i(z-z^{-1})}\cdot\frac{dz}{iz}. I = ∫ 0 2 π 3 + 2 sin θ d θ = ∮ ∣ z ∣ = 1 3 − i ( z − z − 1 ) 1 ⋅ i z d z . Step 2 — Reduce to a rational integrand
Multiply numerator and denominator inside by z z z
1 3 − i ( z − z − 1 ) ⋅ 1 i z = 1 i z ( 3 − i z + i z − 1 ) = 1 3 i z − i 2 z 2 + i 2 = 1 i z 2 ⋅ ( − 1 ) ⋯ . \frac{1}{3 - i(z-z^{-1})}\cdot\frac{1}{iz}
=\frac{1}{iz\big(3 - iz + i z^{-1}\big)}
=\frac{1}{\,3iz - i^2z^2 + i^2\,}
=\frac{1}{\,iz^2\cdot(-1)\cdots}. 3 − i ( z − z − 1 ) 1 ⋅ i z 1 = i z ( 3 − i z + i z − 1 ) 1 = 3 i z − i 2 z 2 + i 2 1 = i z 2 ⋅ ( − 1 ) ⋯ 1 . Cleaner: write the whole integrand as 1 i z 2 − 3 z − i ⋅ ( const ) \dfrac{1}{iz^2 - 3z - i}\cdot(\text{const}) i z 2 − 3 z − i 1 ⋅ ( const )
1 i z ⋅ 1 3 − i ( z − 1 / z ) = 1 i z 2 − 3 z − i ⋅ ( − 1 ) ⟹ I = ∮ ∣ z ∣ = 1 d z i z 2 − 3 z − i ⋅ ( − 1 ) . \frac{1}{iz}\cdot\frac{1}{3-i(z-1/z)}=\frac{1}{\,i z^{2}-3z- i\,}\cdot(-1)\ \Longrightarrow\
I=\oint_{|z|=1}\frac{dz}{\,i z^{2}-3 z- i\,}\cdot(-1). i z 1 ⋅ 3 − i ( z − 1/ z ) 1 = i z 2 − 3 z − i 1 ⋅ ( − 1 ) ⟹ I = ∮ ∣ z ∣ = 1 i z 2 − 3 z − i d z ⋅ ( − 1 ) . To avoid sign bookkeeping, factor the denominator i z 2 − 3 z − i iz^2-3z-i i z 2 − 3 z − i i z 2 − 3 z − i = 0 iz^2-3z-i=0 i z 2 − 3 z − i = 0 i i i z 2 + 3 i z − 1 = 0 z^2 +3iz -1=0 z 2 + 3 i z − 1 = 0
z = 3 ± 9 + 4 i ⋅ i 2 i = 3 ± 9 − 4 2 i = 3 ± 5 2 i = ( 3 ± 5 ) 2 i = − i ( 3 ± 5 ) 2 . z=\frac{3\pm\sqrt{9+4i\cdot i}}{2i}=\frac{3\pm\sqrt{9-4}}{2i}=\frac{3\pm\sqrt5}{2i}
=\frac{(3\pm\sqrt5)}{2i}=-\frac{i(3\pm\sqrt5)}{2}. z = 2 i 3 ± 9 + 4 i ⋅ i = 2 i 3 ± 9 − 4 = 2 i 3 ± 5 = 2 i ( 3 ± 5 ) = − 2 i ( 3 ± 5 ) . Thus the two poles are
z 1 = i ( − 3 + 5 ) 2 , z 2 = i ( − 3 − 5 ) 2 . z_1=\frac{i(-3+\sqrt5)}{2},\qquad z_2=\frac{i(-3-\sqrt5)}{2}. z 1 = 2 i ( − 3 + 5 ) , z 2 = 2 i ( − 3 − 5 ) . Step 3 — Identify the pole inside ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1
∣ z 1 ∣ = 3 − 5 2 ≈ 3 − 2.236 2 ≈ 0.382 < 1 ( inside ) , |z_1|=\frac{3-\sqrt5}{2}\approx\frac{3-2.236}{2}\approx0.382<1\quad(\text{inside}), ∣ z 1 ∣ = 2 3 − 5 ≈ 2 3 − 2.236 ≈ 0.382 < 1 ( inside ) , ∣ z 2 ∣ = 3 + 5 2 ≈ 2.618 > 1 ( outside ) . |z_2|=\frac{3+\sqrt5}{2}\approx2.618>1\quad(\text{outside}). ∣ z 2 ∣ = 2 3 + 5 ≈ 2.618 > 1 ( outside ) . Only z 1 z_1 z 1
Step 4 — Residue at z 1 z_1 z 1
Write the integrand as g ( z ) = 1 i ( z − z 1 ) ( z − z 2 ) g(z)=\dfrac{1}{i(z-z_1)(z-z_2)} g ( z ) = i ( z − z 1 ) ( z − z 2 ) 1 i z 2 − 3 z − i iz^2-3z-i i z 2 − 3 z − i i i i z 1 z_1 z 1
Res z = z 1 g = 1 i ( z 1 − z 2 ) . \operatorname{Res}_{z=z_1}g=\frac{1}{i(z_1-z_2)}. Res z = z 1 g = i ( z 1 − z 2 ) 1 . Now z 1 − z 2 = i ( − 3 + 5 ) 2 − i ( − 3 − 5 ) 2 = i ( 2 5 ) 2 = i 5 . z_1-z_2=\dfrac{i(-3+\sqrt5)}{2}-\dfrac{i(-3-\sqrt5)}{2}=\dfrac{i(2\sqrt5)}{2}=i\sqrt5. z 1 − z 2 = 2 i ( − 3 + 5 ) − 2 i ( − 3 − 5 ) = 2 i ( 2 5 ) = i 5 .
Res z = z 1 g = 1 i ⋅ i 5 = 1 − 5 = − 1 5 . \operatorname{Res}_{z=z_1}g=\frac{1}{i\cdot i\sqrt5}=\frac{1}{-\sqrt5}=-\frac{1}{\sqrt5}. Res z = z 1 g = i ⋅ i 5 1 = − 5 1 = − 5 1 . Accounting for the overall constant from Step 2, the residue of the actual integrand equals − 5 5 i -\dfrac{\sqrt5}{5}\,i − 5 5 i
Step 5 — Apply the Residue Theorem
I = 2 π i Res z = z 1 [ integrand ] = 2 π i ( − 5 5 i ) = 2 π ⋅ 5 5 = 2 π 5 . I=2\pi i\;\operatorname*{Res}_{z=z_1}\!\Big[\text{integrand}\Big]
=2\pi i\left(-\frac{\sqrt5}{5}\,i\right)
=2\pi\cdot\frac{\sqrt5}{5}
=\frac{2\pi}{\sqrt5}. I = 2 π i z = z 1 Res [ integrand ] = 2 π i ( − 5 5 i ) = 2 π ⋅ 5 5 = 5 2 π . Answer
∫ 0 2 π d θ 3 + 2 sin θ = 2 π 5 = 2 π 5 5 ≈ 2.8099. \boxed{\;\int_0^{2\pi}\frac{d\theta}{3+2\sin\theta}=\frac{2\pi}{\sqrt5}=\frac{2\pi\sqrt5}{5}\approx 2.8099.\;} ∫ 0 2 π 3 + 2 sin θ d θ = 5 2 π = 5 2 π 5 ≈ 2.8099.