Question

Evaluate the integral $\oint$ e^z/z²+1 dz where c is the circle |z|=3.

Answer 1

$\int\limits_{|z| = 3} {\frac{{{e^z}}}{{{z^2} + 1}}dz = } \int\limits_{|z| = 3} {\frac{{{e^z}}}{{(z + i)(z - i)}}dz}$

Singular points of the integrand:

${z_1} = i,\,\,{z_2} = - i$ - simple poles.

Both points fall inside the circle $|z | = 3$ :

Let's find the residues at these points:

$\mathop {res}\limits_{z = i} f(z) = \mathop {\lim }\limits_{z \to i} \frac{{{e^z}}}{{(z + i)(z - i)}}(z - i) = \mathop {\lim }\limits_{z \to i} \frac{{{e^z}}}{{z + i}} = \frac{{{e^i}}}{{2i}}$

$\mathop {res}\limits_{z = - i} f(z) = \mathop {\lim }\limits_{z \to i} \frac{{{e^z}}}{{(z + i)(z - i)}}(z + i) = \mathop {\lim }\limits_{z \to i} \frac{{{e^z}}}{{z - i}} = \frac{{{e^{ - i}}}}{{ - 2i}}$

Then

$\int\limits_{|z| = 3} {\frac{{{e^z}}}{{{z^2} + 1}}dz = } = 2\pi i\sum {\mathop {res}\limits_{z = {z_k}} } f(z) = 2\pi i\left( {\frac{{{e^i}}}{{2i}} - \frac{{{e^{ - i}}}}{{2i}}} \right) = \pi \left( {{e^i} - {e^{ - i}}} \right)$

Answer:$\pi \left( {{e^i} - {e^{ - i}}} \right)$