Question

Use Cauchy’s integral formula to evaluate sin(z) /((2z + 1)*3e*z) dz

Answer 1

Cauchy’s integral formula:

$f(z_0)=\frac{1}{2\pi i}\int_C \frac{f(z)}{z-z_0}dz$

then:

$\frac{sinz}{ 3ez(2z + 1)}=\frac{sinz}{ 3e}\cdot\frac{1}{ z(2z + 1)}$

$\frac{1}{ z(2z + 1)}=\frac{1}{ z}-\frac{2}{ 2z + 1}$

$\int_C \frac{sinz}{ 3ez(2z + 1)}dz=\int_C \frac{sinz}{ 3ez}dz-\int_C \frac{2sinz}{ 3e(2z + 1)}dz$

for $\int_C \frac{sinz}{ 3ez}dz$ :

$f(z)=\frac{sinz}{ 3e},z_0=0$

$f(0)=0\implies \int_C \frac{sinz}{ 3ez}dz=0$

for $\int_C \frac{2sinz}{ 3e(2z + 1)}dz$ :

$f(z)=\frac{sinz}{ 3e},z_0=-1/2$

$f(-1/2)=-0.059$

so,

$\int_C \frac{sinz}{ 3ez(2z + 1)}dz=0.059\cdot2\pi i =0.369i$