The integral $\oint f\left(z\right)dz$ evaluated around the unit circle on the complex plane for $f\left(z\right)=\frac{\cos z}{z}$ is

Complex Integral Evaluation

The problem requires us to evaluate the integral $\oint f\left(z\right)dz$ for the function $f\left(z\right)=\frac{\cos z}{z}$ around the unit circle on the complex plane. The unit circle is a contour defined by $|z|=1$.

Singularity Identification

The first step in evaluating a complex integral using methods like Cauchy's Integral Formula or the Residue Theorem is to identify the singularities of the function $f(z)$. A singularity is a point where the function is not analytic (i.e., not differentiable).

- The given function is $f\left(z\right)=\frac{\cos z}{z}$.
- Singularities occur where the denominator is zero. In this case, the denominator is $z$.
- Setting the denominator to zero, we get $z=0$.
- Thus, $z=0$ is the only singularity of the function $f(z)$.

Contour Inclusion Check

Next, we determine if the identified singularity $z=0$ lies inside, on, or outside the given contour. The contour is the unit circle, $|z|=1$.

- The magnitude of the singularity is $|0|=0$.
- Since $0<1$, the singularity $z=0$ lies strictly inside the unit circle $|z|=1$.

Because the singularity is inside the contour, the integral will have a non-zero value, and we can use Cauchy's Integral Formula.

Cauchy's Integral Formula Application

Cauchy's Integral Formula is a powerful tool for evaluating integrals of analytic functions. It states that if $g(z)$ is analytic within and on a simple closed contour $C$, and $z_0$ is any point inside $C$, then:

$${\oint }_C\frac{g(z)}{z-z_0}dz=2\pi ig(z_0)$$

Let's compare our function $f\left(z\right)=\frac{\cos z}{z}$ with the form required by Cauchy's Integral Formula:

- Our function can be written as $\frac{\cos z}{z-0}$.
- By comparison, $g(z)=\cos z$.
- The point inside the contour is $z_0=0$.

We need to ensure that $g(z)=\cos z$ is analytic within and on the unit circle. The function $\cos z$ is an entire function, meaning it is analytic (differentiable) everywhere in the entire complex plane. Therefore, it is certainly analytic within and on the unit circle $|z|=1$.

Now, we can apply the formula:

$${\oint }_{|z|=1}\frac{\cos z}{z-0}dz=2\pi i\cdot g(0)$$

Calculate the value of $g(0)$:

$$g(0)=\cos (0)$$

We know that $\cos (0)=1$.

Substitute this value back into the formula:

$${\oint }_{|z|=1}\frac{\cos z}{z}dz=2\pi i\cdot (1)$$

$${\oint }_{|z|=1}\frac{\cos z}{z}dz=2\pi i$$

Integral Result

The evaluated integral around the unit circle for $f\left(z\right)=\frac{\cos z}{z}$ is $2\pi i$.