Thread: Leibniz rule and findamental theorem of algebra

1. Leibniz rule and fundamental theorem of algebra

Hello,
If 0<a<b, we have $\frac12\displaystyle\int_0^{2\pi} \frac{d\theta}{(b+a*cos\theta)^2}= \frac12\displaystyle\int_0^{2\pi} \rho(\theta)^2d\theta$.

By diﬀerentiating under the integral sign we will deduce the fundamental theorem of algebra: a nonconstant polynomial $\mathbb{p}(\text{z})$ with coeﬃcients in$\mathbb{C}$ has a root in $\mathbb{C}$

Arguing by contradiction, assume $p(z)\not=0$ for all $z\in\mathbb {C}$. For $r\geq 0,$ consider the following integral around a circle of radius r centered at the origin:

$I(r)=\displaystyle\int_0^{2\pi} \frac{d\theta}{\rho(r*e^{i\theta})}$.

This integral makes sense since the denominator is never 0, so $\frac{1}{\text{p}(\text{z})}$ is continuous on $\mathbb{C}$. In case of area of ellipse, we can write $\rho=\frac{1}{b+a*cos \theta}$. But how to write $\rho$ in case of integral around a circle of radius r centered at the origin.

I didn't understand what is the meaning of this integral? If any member of this MHF knows it, may reply with correct answer.

2. Re: Leibniz rule and findamental theorem of algebra

Hello, Please read ''differentiating under integral sign".

3. Re: Leibniz rule and findamental theorem of algebra

There is a proof of the Fundamental Theorem of Algebra using Leibniz's rule for differentiation under the integral sign

Consider the integral

$$I(r)=\int_0^{2\pi } \frac{1}{p\left(r e^{i \theta }\right)} \, d\theta$$

That's $p$ not $\rho$

We take the derivative and show that $I(r)$ is constant and reach a contradiction.

Now, I don't understand the part with $\rho$ and the ellipse

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