Answer to Question #50229 in Complex Analysis for Salam

Question #50229

Use Residue theorem to compute
the integral from 0 to 2 pi
[ Sin theta ] / [ 4+ sin (theta) + cos (theta) ] dtheta

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Assignment Expert

12.01.15, 19:58

Dear vallle. We added more details to the solution.

vallle

06.01.15, 19:02

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

Assignment Expert

06.01.15, 16:45

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.

Alyaa

06.01.15, 00:57

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

Answer to Question #50229 in Complex Analysis for Salam

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