Are these version of Cauchy’s Theorem (complex analysis) equivalent?

I am looking through different complex analysis texts and comparing their statement and proofs of Cauchy’s theorem. One version I found is:

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A second version of Cauchy’s Theorem is:
enter image description here

Now I was wondering if the corollary of the second version is in fact equivalent to the first version. The biggest difference I see is that the first version makes it clear that we need f to be analytic inside the closed curve whereas the second version doesn’t say that. Is one of the four criterions equivalent to saying the interior of the curve is analytic?

Is this proof about Cauchy’s sequence correct?

Let $(a_n)$ be a sequence of real numbers and for each $nin N$, let $f_n=a_1+a_2+a_3+…+a_n$ and $g_n$=$|a_1|+|a_2|+…+|a_n|$. Prove that if $(g_n)$ is a Cauchy sequence, then so is $(f_n)$.

My attempt: Since $(g_n)$ is a Cauchy sequence, we have,

$||a_1|+|a_2|+|a_3|+…|a_n||<epsilon$ for all $ngeq m$

This implies $|a_1+a_2+…a_n|<epsilon$ for all $ngeq m$

Thus $(f_n)$ is a Cauchy sequence.

mg.metric geometry – Second case of induction in Cauchy’s arm lemma

What if in the second case of induction, vertices q1, q2 and qn are already collinear, angle αn > 90°, and α’n-1 is for example 90° . Then α*n-1 would not be larger than αn-1 and no edges will get stretched, thus |q1q*n| won’t be larger than |q1qn|. Did I miss something? Does Q have to be strictly convex, but Q’ only convex?

complex analysis – Prove an inequality by Cauchy’s Theorem

I am trying to solve the following question:

Let $f(z)=c_0+c_1z+…+c_nz^n$ be a polynomial,Prove that $$int_{-1}^1|f(x)|^2dxleqpiint_{0}^{2pi}|f(e^{itheta})|^2dfrac{dtheta}{2pi}$$

If all $c_k$ are real,it suffices to integral $f(z)^2$ along two semicircles,but I have no idea when $c_k$ are complex,can anyone help me?

real analysis – Prove Cauchy’s inequality: $(sum_{i=1}^n a_i b_i)^2 le sum_{i=1}^n a_i^2 sum_{i=1}^n b_i^2.$

Prove that for any real number $a_i, b_i, i = 1,2,…,n$:
$$(sum_{i=1}^n a_i b_i)^2 le sum_{i=1}^n a_i^2 sum_{i=$1}^n b_i^2.$$
Proof in my book goes:
$$sum_{i=1}^n a_i^2 sum_{i=$1}^n b_i^2-(sum_{i=1}^n a_i b_i)^2 = sum_{i,j=1}^na_i^2b_j^2-sum_{i,j=1}^na_ib_ia_jb_j =$$
$$=frac{1}{2}sum_{i,j=1}^n(a_ib_j-b_ia_j)^2ge0$$
But I can’t see why there is $frac{1}{2}$ and I would be glad if someone explained it.
Tia.

integration – Verification of an Cauchy’s contour Integral of Complementary Error function?

I tried to find an integral of the following,$DeclareMathOperator{erfc}{erfc}$

$intlimits_0^{2pi} erfc(a + bcos(theta))erfc(c + dsin(theta)),dtheta $

Where, $a,b,c,d in Bbb R$

Now, $cos(theta) = frac{e^{jtheta} + e^{-jtheta}}{2};quadsin(theta) = frac{e^{jtheta } – e^{-jtheta }}{2j}$

Let $z = {e^{jtheta }}$

Therefore, I can rewrite the integral as,
$ointlimits_{|z| = 1} {erfcleft( {a + bleft( {frac{{z + {z^*}}}{2}} right)} right)erfcleft( {c + dleft( {frac{{z – {z^*}}}{{2j}}} right)} right)frac{{dz}}{{iz}}} $

Again, let $f(z) = erfcleft(a + bleft(frac{z + z^*}{2}right)right)erfcleft(c + dleft(frac{z – z^*}{2j}right) right)$

Hence , the final integral can be written as,

$ointlimits_{|z| = 1}f(z)frac{dz}{iz} = 2pi f(0) = 2pierfc(a)erfc(c)$

As $0$ falls inside $|z| = 1$.

Can you tell me if I am correct or wrong or it needs more reasoning? Thank you.

ho.history overview – Did Euler ever use anything similar to Cauchy’s inequality?

This could be asked more provocatively, indeed how it arose, as “how did Euler do so much mathematics without using and/or knowing Cauchy’s inequality?”, something that came up in the context when reading a comment about how ubiquitous said inequality/variants is in various fields today.

But I think the title question is at least fact-based, and of historical interest.

Q. Did Euler ever use anything similar to Cauchy’s inequality?

Particularly I might have expected something like the integral version to appear somewhere in calculus of variations. Alternatively, something beyond simply a geometric-arithmetic mean might have occurred in number theory arguments. (I figure, as a first go, asking mathematicians is better than historians, especially as if there is “something similar” I’d like to hear it.)

cv.complex variables – Cauchy’s Integral with quadratic exponential term

As I was studying the Cauchy’s integral formula, I tried to do the integral:

begin{equation}
I(x) = int_{-infty}^{infty} frac{1}{x – a} e^{(i A x^2 + i B x)} dx
end{equation}

with $A>0, B>0$ and $a > 0$.

Consider an integral on a complex plan:
begin{equation}
J(z) = int_{C + C_R} frac{1}{z – a} e^{(i A z^2 + i B z)} dz
end{equation}

where $C$ is along the real axis $-infty rightarrow +infty$ and $C_R$ is the upper half circle $z = Re^{itheta}$ with $R rightarrow infty$ and $theta in (0, 2pi)$.

Naively, I would expect $C_R$ part of the integral gives zero and $C$ part of the integral gives $I(x)$, then the $I(x)$ can be derived by Cauchy’s integral formula.

However, as I tried to check the $C_R$ part of the integral, I found that:
begin{array}
$I_R(z = Re^{itheta}) &=& int_0^{pi} dtheta frac{iRe^{itheta}}{Re^{itheta} – a} exp(iAR^2e^{2itheta}+iBRe^{itheta})
end{array}

begin{array}
$|I_R| &leq & int_0^{pi} dtheta |frac{iRe^{itheta}}{Re^{itheta} – a}| |exp(iAR^2e^{2itheta}+iBRe^{itheta})|
end{array}

where the first term

begin{equation}
|frac{iRe^{itheta}}{Re^{itheta} – a}| leq frac{R}{R-a} rightarrow 1 as R rightarrow infty
end{equation}

and the second term
begin{equation}
|exp(iAR^2e^{2itheta}+iBRe^{itheta})| leq e^{-AR^2sin(2theta) – BRsin(theta)}
end{equation}

will not approach to zero because of $e^{-AR^2sin(2theta)}$.

Is there anything wrong in my approach? And is there any other way I can perform this integral I(x)?

Thanks a million for advises!