Confusion in mapping from s to F(s) plane

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When we map a close contour in s plane to F(s) plane where s is complex number ,we frequently use one rule i.e "number of encirclement of origin in F(s) plane is depend upon -

1.no. of zeroes of F(s) enclosed by close contour in s plane

2.no. of poles of F(s) enclosed by close contour in S plane

But 1st point (depends on zeroes) support intuation also because if close contour in s plane​ contain a zero of F(s) then it should contain origin

But I am confused in 2nd point

how pole containing contour in S plane is related to origin contained by contour in F(s) plane ?

What is intuation behind it?

Is there any rigrous mathematical proof of it so that we can show ,how pole and origin in both planes are related ?

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Here is a very informal answer.

The question is (I presume) about the relationship between poles, zeroes, angles and encirclements. The quantity of interest is ${f' \over f}$.

Informally, we have $\log f(s) = \log |f(s)| + i \operatorname{arg} f(s)$, so $\operatorname{im} \log f(s) =\operatorname{arg} f(s)$ and differentiating the left hand side gives $\operatorname{im} {f'(s) \over f(s)}$. So, from an intuition perspective, we expect that integrating $\operatorname{im} {f'(s) \over f(s)}$ over $\gamma$ (and taking the imaginary part) will give the change in $\operatorname{arg} f(s)$ as $f$ 'moves' along $\gamma$.

So, now suppose $f$ has the form $f(s) = (s-a)^p g(s)$ where $g$ is analytic at $a$ and $p$ is a (possibly negative) integer then ${f'(s) \over f(s)} = {p \over s-a} + {g'(s) \over g(s)}$. Then we see that (for sufficiently 'small' $\gamma$) that ${1 \over 2 \pi i} \int_\gamma {f' \over f} dz = p + {1 \over 2 \pi i} \int_\gamma {g' \over g} dz$, so the $(s-a)^p$ term contributes $p$ encirclements (if $p<0$ the count in the oppositive direction).

The relevant result is the argument principle ${1 \over 2 \pi i} \int_\gamma {f' \over f} dz = \sum_i \eta(\gamma, z_i) - \sum_j \eta(\gamma, p_j)$.