Question

Show that integral dz/(z^2-1)^2+3=π/2√2, where path of integration is unit circle in the positive sense

Answer 1

Answer on Question #67557 – Math – Complex Analysis

Question

Show that integral

$$\oint \frac{dz}{(z^2 - 1)^2 + 3} = \frac{\pi}{2} \sqrt{2},$$

where path of integration is unit circle in the positive sense.

Solution

Let's find zeros of the denominator:

$$(z^2 - 1)^2 + 3 = 0$$ $$(z^2 - 1)^2 = -3$$ $$z^2 - 1 = \pm i\sqrt{3}$$ $$z^2 = |1 \pm i\sqrt{3}|$$ $$z = \pm \sqrt{1 \pm i\sqrt{3}}$$

4 singular points

The module of each of the singular points is greater than one.

So that neither of them lies inside the unit circle.

Integral of a function over a closed contour, if the domain bounded by the contour does not contain singular points is equal to zero. (The Cauchy theorem)

Hence the value of the integral is zero

Answer: 0.

Answer provided by https://www.AssignmentExpert.com