How can i evaluate the complex integral $\int _{ c } \frac { \cos(iz) }{ { z }^{ 2 }({ z }^{ 2 }+2i) } dz$ for a rectangle -3,3,-3i,i

Question

$$\int _{ c } \frac { \cos(iz) }{ { z }^{ 2 }({ z }^{ 2 }+2i) } dz$$

I used residue caculus for solving the problem but i am not pretty sure if the approach is right.

The attempt has been annexed in the pictures.

Answer 1

You can use partial fractions for this problem. You have: $$\int _{ c } \frac { \cos(iz) }{ { z }^{ 2 }({ z }^{ 2 }+2i) } dz $$ $$=\int _{c} \frac{\cosh(z)}{z^2(z-(1-i)(z+(1-i)))}dz$$ $$=\int _{c} \cosh(z)\frac{1}{z^2(z-(1-i)(z+(1-i)))}dz$$$$=\int _{c} \cosh(z)(\frac{A}{z}+\frac{B}{z^2}+\frac{C}{z-(1-i)}+\frac{D}{z+(1-i)})dz$$ Solve for A B C and D (left for you), distribute the $\cosh(z)$ and since the integral is a linear operator this implies $$\int _{ c } \frac { \cos(iz) }{ { z }^{ 2 }({ z }^{ 2 }+2i) } dz=\int _{c} \frac{A\cosh(z)}{z}dz+\int _{c}\frac{B\cosh(z)}{z^2}dz+\int _{c}\frac{C\cosh(z)}{z-(1-i)}dz+\int _{c}\frac{D\cosh(z)}{z-(i-1)}dz$$ I hope this is now clear you could apply the Cauchy and extended Cauchy integral formula to each integral. For example, the Cauchy intergral formula says $2\pi if(z_0)=\int_c \frac{f(z)}{z-z_0}dz$ where, in the first integral, $f(z)=A\cosh(z)$ and $ z_0=0 $. You would evaluate $2\pi if(z_0)$ for each integral and add up each term on the LHS (except for the second integral you would use extended Cauchy and evaluate $B\sinh(z_0)$). It doesn't look like you'll be getting $0$ like what you have in your answer, but maybe.

Please, let me know if you have questions with any part of this and please vote answer.

Answer 2

According to Wolfram alpha , you are correct: the residues add up to $0$. They are: $(\frac18-\frac i8)\cos (1+i), 0, (-\frac18+\frac i8)\cos (1+i)$ at $-1+i, 0, 1-i $ respectively.