If 𝐶 is the unit circle in the complex plane with its center at the origin, then the value of 𝑛 in the equation given below is __________ (rounded off to 1 decimal place).

${\oint }_C\frac{z^3}{(z^2+4)(z^2-4)}dz=2\pi in$

The correct answer is 0.

To find the value of $n$ in the given equation, we need to evaluate the contour integral:
$$I={\oint }_C\frac{z^3}{(z^2+4)(z^2-4)}dz$$
where $C$ is the unit circle in the complex plane centered at the origin, represented by $\left|z\right|=1$.

First, we find the singularities (poles) of the integrand:
$$f\left(z\right)=\frac{z^3}{(z^2+4)(z^2-4)}$$
The poles are the roots of the denominator:
$$(z^2+4)(z^2-4)=0$$

This gives two sets of roots:
1. From $z^2+4=0$, we have $z=\pm 2i$.
2. From $z^2-4=0$, we have $z=\pm 2$.
Therefore, the function $f\left(z\right)$ has four simple poles at $z=2,-2,2i,-2i$.

Next, we check which of these poles lie inside the contour of integration $C$ (the unit circle $\left|z\right|=1$):
- For $z=\pm 2$, the absolute value is $|\pm 2|=2>1$, so these poles lie outside the unit circle.
- For $z=\pm 2i$, the absolute value is $|\pm 2i|=2>1$, so these poles also lie outside the unit circle.

Since all the singularities of the integrand lie strictly outside the unit circle $C$, the function $f\left(z\right)$ is analytic everywhere inside and on the boundary of the contour $C$.

According to Cauchy's Integral Theorem, if a function is analytic within and on a closed contour, the line integral of the function along that contour is zero:
$${\oint }_{C}f\left(z\right)dz=0$$

Comparing this result with the given equation:
$${\oint }_{C}f\left(z\right)dz=2\pi in=0$$
We get:
$$n=0$$