Answer on Question #50229 – Math – Complex Analysis
Use Residue theorem to compute
the integral from 0 to 2 pi
[ Sin theta ] / [ 4+ sin (theta) + cos (theta) ] dtheta
Solution.
I = ∫ 0 2 π sin θ 4 + sin θ + cos θ d θ = ∣ z = e i θ , d θ = d z i z cos θ = 1 2 ( z + 1 z ) sin θ = 1 2 i ( z − 1 z ) ∣ = ∮ ∣ z ∣ = 1 1 2 i ( z − 1 z ) 4 + 1 2 i ( z − 1 z ) + 1 2 ( z + 1 z ) ⋅ d z i z = ∮ ∣ z ∣ = 1 f ( z ) d z , I = \int_{0}^{2\pi} \frac{\sin \theta}{4 + \sin \theta + \cos \theta} d\theta = \left| \begin{array}{l} z = e^{i\theta}, \, d\theta = \frac{dz}{iz} \\ \cos \theta = \frac{1}{2}\left(z + \frac{1}{z}\right) \\ \sin \theta = \frac{1}{2i}\left(z - \frac{1}{z}\right) \end{array} \right| = \oint_{|z| = 1} \frac{\frac{1}{2i}\left(z - \frac{1}{z}\right)}{4 + \frac{1}{2i}\left(z - \frac{1}{z}\right) + \frac{1}{2}\left(z + \frac{1}{z}\right)} \cdot \frac{dz}{iz} = \oint_{|z| = 1} f(z) dz, I = ∫ 0 2 π 4 + sin θ + cos θ sin θ d θ = ∣ ∣ z = e i θ , d θ = i z d z cos θ = 2 1 ( z + z 1 ) sin θ = 2 i 1 ( z − z 1 ) ∣ ∣ = ∮ ∣ z ∣ = 1 4 + 2 i 1 ( z − z 1 ) + 2 1 ( z + z 1 ) 2 i 1 ( z − z 1 ) ⋅ i z d z = ∮ ∣ z ∣ = 1 f ( z ) d z ,
where
f ( z ) = 1 2 i ( z − 1 z ) ( 4 + 1 2 i ( z − 1 z ) + 1 2 ( z + 1 z ) ) i z = − z 2 − 1 2 z ∗ z ( 4 − i 2 ( z − 1 z ) + 1 2 ( z + 1 z ) ) = − z 2 − 1 z ( 8 z − i ( z 2 − 1 ) + ( z 2 + 1 ) ) f(z) = \frac{\frac{1}{2i}\left(z - \frac{1}{z}\right)}{\left(4 + \frac{1}{2i}\left(z - \frac{1}{z}\right) + \frac{1}{2}\left(z + \frac{1}{z}\right)\right)iz} = -\frac{z^2 - 1}{2z * z\left(4 - \frac{i}{2}\left(z - \frac{1}{z}\right) + \frac{1}{2}\left(z + \frac{1}{z}\right)\right)} = -\frac{z^2 - 1}{z\left(8z - i\left(z^2 - 1\right) + \left(z^2 + 1\right)\right)} f ( z ) = ( 4 + 2 i 1 ( z − z 1 ) + 2 1 ( z + z 1 ) ) i z 2 i 1 ( z − z 1 ) = − 2 z ∗ z ( 4 − 2 i ( z − z 1 ) + 2 1 ( z + z 1 ) ) z 2 − 1 = − z ( 8 z − i ( z 2 − 1 ) + ( z 2 + 1 ) ) z 2 − 1 f ( z ) = z 2 − 1 z [ ( i − 1 ) z 2 − 8 z − ( 1 + i ) ] . f(z) = \frac{z^2 - 1}{z\left[(i - 1)z^2 - 8z - (1 + i)\right]}. f ( z ) = z [ ( i − 1 ) z 2 − 8 z − ( 1 + i ) ] z 2 − 1 .
Consider
( i − 1 ) z 2 − 8 z − ( 1 + i ) = 0 , hence D = 64 + 4 ( i − 1 ) ( i + 1 ) = 64 + 4 ( i 2 − 1 ) = 64 − 8 = 56 , so (i - 1)z^2 - 8z - (1 + i) = 0, \quad \text{hence} \quad D = 64 + 4(i - 1)(i + 1) = 64 + 4(i^2 - 1) = 64 - 8 = 56, \quad \text{so} ( i − 1 ) z 2 − 8 z − ( 1 + i ) = 0 , hence D = 64 + 4 ( i − 1 ) ( i + 1 ) = 64 + 4 ( i 2 − 1 ) = 64 − 8 = 56 , so z 1 = 8 + 56 2 ( i − 1 ) = i + 1 ( i − 1 ) ( i + 1 ) ( 4 + 14 ) = i + 1 i 2 − 1 ( 4 + 14 ) = i + 1 − 2 ( 4 + 14 ) = ( 1 + i ) ( − 2 − 7 2 ) z_1 = \frac{8 + \sqrt{56}}{2(i - 1)} = \frac{i + 1}{(i - 1)(i + 1)} \left(4 + \sqrt{14}\right) = \frac{i + 1}{i^2 - 1} \left(4 + \sqrt{14}\right) = \frac{i + 1}{-2} \left(4 + \sqrt{14}\right) = (1 + i)\left(-2 - \sqrt{\frac{7}{2}}\right) z 1 = 2 ( i − 1 ) 8 + 56 = ( i − 1 ) ( i + 1 ) i + 1 ( 4 + 14 ) = i 2 − 1 i + 1 ( 4 + 14 ) = − 2 i + 1 ( 4 + 14 ) = ( 1 + i ) ( − 2 − 2 7 ) z 2 = 8 − 56 2 ( i − 1 ) = i + 1 ( i − 1 ) ( i + 1 ) ( 4 − 14 ) = i + 1 i 2 − 1 ( 4 − 14 ) = i + 1 − 2 ( 4 − 14 ) = ( 1 + i ) ( − 2 + 7 2 ) z_2 = \frac{8 - \sqrt{56}}{2(i - 1)} = \frac{i + 1}{(i - 1)(i + 1)} \left(4 - \sqrt{14}\right) = \frac{i + 1}{i^2 - 1} \left(4 - \sqrt{14}\right) = \frac{i + 1}{-2} \left(4 - \sqrt{14}\right) = (1 + i)\left(-2 + \sqrt{\frac{7}{2}}\right) z 2 = 2 ( i − 1 ) 8 − 56 = ( i − 1 ) ( i + 1 ) i + 1 ( 4 − 14 ) = i 2 − 1 i + 1 ( 4 − 14 ) = − 2 i + 1 ( 4 − 14 ) = ( 1 + i ) ( − 2 + 2 7 )
The isolated singular points of the function f ( z ) f(z) f ( z ) z = { 0 ; ( 1 + i ) ( − 2 ± 7 2 ) } z = \left\{0; \left(1 + i\right)\left(-2 \pm \sqrt{\frac{7}{2}}\right)\right\} z = { 0 ; ( 1 + i ) ( − 2 ± 2 7 ) } z = 0 z = 0 z = 0 z = ( 1 + i ) ( 7 2 − 2 ) z = \left(1 + i\right)\left(\sqrt{\frac{7}{2}} - 2\right) z = ( 1 + i ) ( 2 7 − 2 ) ∣ z ∣ = 1 |z| = 1 ∣ z ∣ = 1
The residues in these points are the following:
r e s z = 0 f = lim z → 0 f ⋅ z = lim z → 0 z 2 − 1 ( i − 1 ) z 2 − 8 z − ( 1 + i ) = − 1 − ( 1 + i ) = 1 1 + i = 1 − i ( 1 + i ) ( 1 − i ) = 1 − i 1 − i 2 = 1 − i 1 − ( − 1 ) = 1 − i 2 res_{z=0} f = \lim_{z \to 0} f \cdot z = \lim_{z \to 0} \frac{z^2 - 1}{(i - 1)z^2 - 8z - (1 + i)} = \frac{-1}{-(1 + i)} = \frac{1}{1 + i} = \frac{1 - i}{(1 + i)(1 - i)} = \frac{1 - i}{1 - i^2} = \frac{1 - i}{1 - (-1)} = \frac{1 - i}{2} re s z = 0 f = z → 0 lim f ⋅ z = z → 0 lim ( i − 1 ) z 2 − 8 z − ( 1 + i ) z 2 − 1 = − ( 1 + i ) − 1 = 1 + i 1 = ( 1 + i ) ( 1 − i ) 1 − i = 1 − i 2 1 − i = 1 − ( − 1 ) 1 − i = 2 1 − i r e s z = ( 1 + i ) ( 7 2 − 2 ) f = lim z → ( 1 + i ) ( 7 2 − 2 ) f ⋅ [ z − ( 1 + i ) ( 7 2 − 2 ) ] = lim z → ( 1 + i ) ( 7 2 − 2 ) z 2 − 1 z ( i − 1 ) [ z + ( 1 + i ) ( 7 2 + 2 ) ] = res_{z = (1 + i)\left(\sqrt{\frac{7}{2}} - 2\right)} f = \lim_{z \to (1 + i)\left(\sqrt{\frac{7}{2}} - 2\right)} f \cdot \left[ z - (1 + i)\left(\sqrt{\frac{7}{2}} - 2\right) \right] = \lim_{z \to (1 + i)\left(\sqrt{\frac{7}{2}} - 2\right)} \frac{z^2 - 1}{z(i - 1)\left[ z + (1 + i)\left(\sqrt{\frac{7}{2}} + 2\right) \right]} = re s z = ( 1 + i ) ( 2 7 − 2 ) f = z → ( 1 + i ) ( 2 7 − 2 ) lim f ⋅ [ z − ( 1 + i ) ( 2 7 − 2 ) ] = z → ( 1 + i ) ( 2 7 − 2 ) lim z ( i − 1 ) [ z + ( 1 + i ) ( 2 7 + 2 ) ] z 2 − 1 = = ( 1 + i ) 2 ( 7 2 − 2 ) 2 − 1 ( i + 1 ) ( i − 1 ) ( 7 2 − 2 ) [ ( 1 + i ) ( 7 2 − 2 ) + ( 1 + i ) ( 7 2 + 2 ) ] = 2 i ( 7 2 − 2 ) 2 − 1 − 2 ( 7 2 − 2 ) [ 2 ( 1 + i ) ( 7 2 − 2 ) ] = = 2 i ( 7 2 − 2 ) 2 − 1 − 2 ( 7 2 − 2 ) [ 2 ( 1 + i ) ( 7 2 − 2 ) ] = ( 1 − i ) 2 i ( 7 2 − 2 ⋅ 2 7 2 + 4 ) − 1 − 4 ⋅ 2 ( 7 2 − 2 ⋅ 2 7 2 + 4 ) = ( 1 − i ) 2 i ( 15 2 − 4 7 2 ) − 1 − 4 ⋅ 2 ( 15 2 − 4 7 2 ) = = ( 1 − i ) 2 i ( 15 2 − 4 7 2 ) − 1 − 4 ⋅ 2 ( 15 2 − 4 7 2 ) ( 15 2 + 4 7 2 ) = ( 1 − i ) 2 i ( 15 2 − 4 7 2 ) ( 15 2 + 4 7 2 ) − ( 15 2 + 4 7 2 ) − 8 ( 225 4 − 16 ⋅ 7 2 ) = ( 1 − i ) 2 i ( 225 4 − 16 ⋅ 7 ⋅ 2 4 ) − ( 15 2 + 4 7 2 ) − 8 ( 225 4 − 16 ⋅ 7 ⋅ 2 4 ) = ( 1 − i ) i 2 − ( 15 2 + 4 7 2 ) − 2 = ( 1 − i ) ( − i 4 + 15 4 + 2 7 2 ) = ( 1 − i ) ( − i 4 + 15 4 + 4 14 4 ) = − i 4 + 15 4 + 4 14 4 − 1 4 − 15 4 i − 4 14 4 i = − ( 4 + 14 ) i + 7 2 + 14 . \begin{array}{l}
= \frac{(1 + i)^{2}\left(\sqrt{\frac{7}{2}} - 2\right)^{2} - 1}{(i + 1)(i - 1)\left(\sqrt{\frac{7}{2}} - 2\right)\left[(1 + i)\left(\sqrt{\frac{7}{2}} - 2\right) + (1 + i)\left(\sqrt{\frac{7}{2}} + 2\right)\right]}
= \frac{2i\left(\sqrt{\frac{7}{2}} - 2\right)^{2} - 1}{-2\left(\sqrt{\frac{7}{2}} - 2\right)\left[2(1 + i)\left(\sqrt{\frac{7}{2}} - 2\right)\right]}
= \\
= \frac{2i\left(\sqrt{\frac{7}{2}} - 2\right)^{2} - 1}{-2\left(\sqrt{\frac{7}{2}} - 2\right)\left[2(1 + i)\left(\sqrt{\frac{7}{2}} - 2\right)\right]}
= (1 - i)\frac{2i\left(\frac{7}{2} - 2 \cdot 2\sqrt{\frac{7}{2}} + 4\right) - 1}{-4 \cdot 2\left(\frac{7}{2} - 2 \cdot 2\sqrt{\frac{7}{2}} + 4\right)}
= (1 - i)\frac{2i\left(\frac{15}{2} - 4\sqrt{\frac{7}{2}}\right) - 1}{-4 \cdot 2\left(\frac{15}{2} - 4\sqrt{\frac{7}{2}}\right)}
= \\
= (1 - i) \frac{2i\left(\frac{15}{2} - 4\sqrt{\frac{7}{2}}\right) - 1}{-4 \cdot 2\left(\frac{15}{2} - 4\sqrt{\frac{7}{2}}\right)\left(\frac{15}{2} + 4\sqrt{\frac{7}{2}}\right)}
= (1 - i) \frac{2i\left(\frac{15}{2} - 4\sqrt{\frac{7}{2}}\right)\left(\frac{15}{2} + 4\sqrt{\frac{7}{2}}\right) - \left(\frac{15}{2} + 4\sqrt{\frac{7}{2}}\right)}{-8\left(\frac{225}{4} - \frac{16 \cdot 7}{2}\right)} \\
= (1 - i) \frac{2i\left(\frac{225}{4} - \frac{16 \cdot 7 \cdot 2}{4}\right) - \left(\frac{15}{2} + 4\sqrt{\frac{7}{2}}\right)}{-8\left(\frac{225}{4} - \frac{16 \cdot 7 \cdot 2}{4}\right)}
= (1 - i) \frac{\frac{i}{2} - \left(\frac{15}{2} + 4\sqrt{\frac{7}{2}}\right)}{-2}
= (1 - i)\left(-\frac{i}{4} + \frac{15}{4} + 2\sqrt{\frac{7}{2}}\right) \\
= (1 - i)\left(-\frac{i}{4} + \frac{15}{4} + \frac{4\sqrt{14}}{4}\right) = -\frac{i}{4} + \frac{15}{4} + \frac{4\sqrt{14}}{4} - \frac{1}{4} - \frac{15}{4}i - \frac{4\sqrt{14}}{4}i = -(4 + \sqrt{14})i + \frac{7}{2} + \sqrt{14}.
\end{array} = ( i + 1 ) ( i − 1 ) ( 2 7 − 2 ) [ ( 1 + i ) ( 2 7 − 2 ) + ( 1 + i ) ( 2 7 + 2 ) ] ( 1 + i ) 2 ( 2 7 − 2 ) 2 − 1 = − 2 ( 2 7 − 2 ) [ 2 ( 1 + i ) ( 2 7 − 2 ) ] 2 i ( 2 7 − 2 ) 2 − 1 = = − 2 ( 2 7 − 2 ) [ 2 ( 1 + i ) ( 2 7 − 2 ) ] 2 i ( 2 7 − 2 ) 2 − 1 = ( 1 − i ) − 4 ⋅ 2 ( 2 7 − 2 ⋅ 2 2 7 + 4 ) 2 i ( 2 7 − 2 ⋅ 2 2 7 + 4 ) − 1 = ( 1 − i ) − 4 ⋅ 2 ( 2 15 − 4 2 7 ) 2 i ( 2 15 − 4 2 7 ) − 1 = = ( 1 − i ) − 4 ⋅ 2 ( 2 15 − 4 2 7 ) ( 2 15 + 4 2 7 ) 2 i ( 2 15 − 4 2 7 ) − 1 = ( 1 − i ) − 8 ( 4 225 − 2 16 ⋅ 7 ) 2 i ( 2 15 − 4 2 7 ) ( 2 15 + 4 2 7 ) − ( 2 15 + 4 2 7 ) = ( 1 − i ) − 8 ( 4 225 − 4 16 ⋅ 7 ⋅ 2 ) 2 i ( 4 225 − 4 16 ⋅ 7 ⋅ 2 ) − ( 2 15 + 4 2 7 ) = ( 1 − i ) − 2 2 i − ( 2 15 + 4 2 7 ) = ( 1 − i ) ( − 4 i + 4 15 + 2 2 7 ) = ( 1 − i ) ( − 4 i + 4 15 + 4 4 14 ) = − 4 i + 4 15 + 4 4 14 − 4 1 − 4 15 i − 4 4 14 i = − ( 4 + 14 ) i + 2 7 + 14 .
According to the residue theorem,
I = 2 π i ⋅ [ res z = 0 f + res z = ( 1 + i ) ( 7 2 − 2 ) f ] = 2 π i ⋅ ( 1 − i 2 − ( 4 + 14 ) i + 7 2 + 14 ) = 2 π i ⋅ ( − ( 9 2 + 14 ) i + 4 + 14 ) = π ⋅ ( ( 9 + 2 14 ) i + ( 8 + 2 14 ) i ) \begin{array}{l}
I = 2\pi i \cdot \left[ \operatorname{res}_{z=0} f + \operatorname{res}_{z= (1+i)\left(\sqrt{\frac{7}{2}} - 2\right)} f \right] = 2\pi i \cdot \left(\frac{1 - i}{2} - (4 + \sqrt{14})i + \frac{7}{2} + \sqrt{14}\right) = 2\pi i \cdot \left(-\left(\frac{9}{2} + \sqrt{14}\right)i + 4 + \sqrt{14}\right) \\
= \pi \cdot \left((9 + 2\sqrt{14})i + (8 + 2\sqrt{14})i\right)
\end{array} I = 2 πi ⋅ [ res z = 0 f + res z = ( 1 + i ) ( 2 7 − 2 ) f ] = 2 πi ⋅ ( 2 1 − i − ( 4 + 14 ) i + 2 7 + 14 ) = 2 πi ⋅ ( − ( 2 9 + 14 ) i + 4 + 14 ) = π ⋅ ( ( 9 + 2 14 ) i + ( 8 + 2 14 ) i )
Answer: ( 9 + 2 14 ) π + ( 8 + 2 14 ) π i (9 + 2\sqrt{14})\pi + (8 + 2\sqrt{14})\pi i ( 9 + 2 14 ) π + ( 8 + 2 14 ) πi
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Comments
Dear vallle. We added more details to the solution.
Thanks Done about singualrites,but please about Residue at zero how get the final result =(1-i)/2 when we substitute zero what the final of res at zero, please appreciate more details. with the second res pllleaase
Dear Alyaa. To find singularities (1+i).-2 {-+}sqrt root (7/2)-2 we put (i-1)z^2-8z-(1+i)=0 and solve this quadratic equation, D=64+4(i-1)(i+1)=64+4(i^2-1)=64-4(-1-1)=64-8=56, next z1=(8+sqrt(56))/2(i-1)=(8+sqrt(56))*(i+1)/(2(i-1)(i+1))=(8+sqrt(56))*(i+1)/(2(i^2-1))=(8+sqrt(56))*(i+1)/(-4), z2=(8-sqrt(56))/2(i-1)=(8-sqrt(56))*(i+1)/(2(i-1)(i+1))=(8-sqrt(56))*(i+1)/(2(i^2-1))=(8-sqrt(56))*(i+1)/(-4), which can be simplified. Why do not you accept this method? Another method (via series expansion) is very lengthy.
please How found the second singularity ? (1+i).-2 {-+}sqrt root (7/2)-2 please Details Also About Res, can find it with another method please ,appreciate