Equality of absolute values of complex integrals It was pretty hard finding a short and precise title. Here is my problem:
The equation $$\bigg|\int_\gamma f(z)\text{d}z\bigg|\le\int_\gamma\big|f(z)||\text{d}z|$$ holds if $f$ is integrable (where $\gamma$ is a differentiable path $[0,1]\rightarrow\mathbb{C}$).
However, in what cases does "$=$" apply? Some cases are clear; for example if $f(z)=$const, $\forall z\in {\rm Img}(\gamma)$.
Our second idea was that both sides are equal if (or even iff?) $f(z)$ stays within one quadrant of the complex plane for all points in ${\rm Img}(\gamma)$.
So basically, the question is, when does "$=$" apply and is our second idea correct? Use "if" or "iff" (which is if and only if). Is there a nice proof to whatever answer there is? (No need to print the full proof; the basic idea suffices)
EDIT: instead of using the path definition of the integral, here is the corrected version:
$$\bigg|\int_0^1f(\gamma(t))\gamma'(t)\text{d}t\bigg|\le\int_0^1|f(\gamma(t))||\gamma'(t)|dt.$$
 A: It suffices to consider the following reduced question: if $g:[0,1]\to\mathbb{C}$ is Riemann/Lebesgue integrable, in the following inequality 
$$\left|\int_0^1g(t)dt\right|\le \int_0^1|g(t)|dt,\tag{1}$$
when does "$=$" holds?
To answer this question, let us recall how to deduce $(1)$. Since if $g$ is  Riemann integrable on $[0,1]$, it is automatically Lebesgue integrable, so let us focus on the Lebesgue integrable case. Denote $I=\int_0^1g(t)dt$. If $I=0$, then $(1)$ is trivially true, and "$=$" holds if and only if $g=0$ almost everywhere. If $I\ne 0$, then there exists a unique $\alpha\in\mathbb{C}$ with $|\alpha|=1$, such that $I=|I|\alpha$. Therefore,
$$|I|=\overline{\alpha}I=\mathrm{Re}(\overline{\alpha}I)=\mathrm{Re}\int_0^1\overline{\alpha}g(t)dt=\int_0^1\mathrm{Re}(\overline{\alpha}g(t))dt\le\int_0^1|g(t)|dt.$$
Note that for the last inequaltiy, "$=$" holds if and only if $\mathrm{Re}(\overline{\alpha}g(t))=|g(t)|$ for almost every $t\in[0,1]$, i.e. $g(t)=\alpha|g(t)|$ for almost every $t\in[0,1]$. To sum up, "$=$" in $(1)$ holds if and only if there exists $\alpha\in\mathbb{C}$ with $|\alpha|=1$, such that $g(t)=\alpha|g(t)|$ for almost every $t\in[0,1]$. To go back to your original question, simply let $g(t)=f(\gamma(t))\gamma'(t)$.
A: If you write
$$\bigg|\int_0^1f(\gamma(t))\gamma'(t)\text{d}t\bigg|\le\int_0^1|f(\gamma(t))||\gamma'(t)|dt$$
then it should be clear that you are asking this other question: take $a,b:[0,1]\to\mathbb{R}$ and write
$$A=\int_0^1a(t)dt\qquad B=\int_{0}^1b(t)dt\;,$$
when does one have
$$\sqrt{A^2+B^2}=\int_0^1(a^2(t)+b^2(t))^{1/2}dt$$
?
By squaring you get
$$A^2+B^2= \left(\int_{0}^{1}(a^2(t)+b^2(t))^{1/2}dt\right)^2=\|a^2+b^2\|_{1/2}$$
and then
$$A^2+B^2=\left(\int_0^1 a(t)dt\right)^2+\left(\int_{0}^1b(t)dt\right)^2=\left(\int_0^1 \sqrt{a^2(t)}dt\right)^2+\left(\int_0^1 \sqrt{b^2(t)}dt\right)^2=\left(\int_{0}^{1}(a^2(t)+b^2(t))^{1/2}dt\right)^2$$
because $$A^2+B^2\le\|a^2\|_{1/2}+\|b^2\|_{1/2}\le\|a^2+b^2\|_{1/2}$$
which is the reverse Minkowski inequality for $p=1/2$ (Minkowski and Holder hold for $0<p<1$ with the opposite inequality).
The only way to have equality in the first one is $a=\pm|a|$, $b=\pm|b|$ to hold for a.e. $t$;for the second, like in the usual Minkowsky inequality, is to have $a^2=\lambda^2 b^2$ for a fixed real $\lambda$, that is $a+ib=(\lambda + i)b$. 
Now, in your setting this means that $f(\gamma(t))\gamma'(t)$ has to be of the form $e^{i\alpha}\phi(t)$, where $\alpha\in\mathbb{R}$ is fixed and $\phi:[0,1]\to\mathbb{R^+}$, i.e.
$$f(\gamma(t))=\frac{e^{i\alpha}\phi(t)}{\gamma'(t)}\;.$$
If, wlog, you take $\gamma$ a $\mathcal{C}^1$ curve, parametrized with respect to arc length, then $|\gamma'(t)|=1$ and $\gamma'(t)=e^{i\beta(t)}$; then you have
$$f(\gamma(t))=e^{i\alpha-i\beta(t)}\phi(t)$$
which, upon rescaling $f$ by a constant factor, can be reduced to
$$f(\gamma(t))=e^{-i\beta(t)}\phi(t)$$
which is not as neat as before (errors always improve results!) but yields nonetheless some link with curvature:
$$\frac{df}{dz}(\gamma(t))e^{i\beta(t)}=f'(\gamma(t))\gamma'(t)=\frac{d}{dt}f(\gamma(t))=-i\beta'(t)e^{-i\beta(t)}\phi(t)+e^{-i\beta(t)}\phi'(t)$$
that is
$$\frac{df}{dz}(\gamma(t))=e^{-2i\beta(t)}(-i\beta'(t)\phi(t)+\phi'(t))$$
and $\beta'(t)$ is related to the curvature of $\gamma$.
A: For simplicity's sake define
$$
g(t)=f(\gamma(t))\gamma'(t)\tag{1}
$$
Note that
$$
\int_0^1|g(t)|\,\mathrm{d}t
=\sup_{|u(z)|=1}\mathrm{Re}\left(\int_0^1g(t)u(t)\,\mathrm{d}t\right)\tag{2}
$$
and
$$
\left|\,\int_0^1g(t)\,\mathrm{d}t\,\right|
=\sup_{\substack{|u(z)|=1\\u\text{ constant}}}\mathrm{Re}\left(\int_0^1g(t)u(t)\,\mathrm{d}t\right)\tag{3}
$$
By the principle that the sup over a larger set is larger, $(2)\ge(3)$.
Suppose that $(3)\ge(2)$. Then, by the compactness of the unit circle, there is a $u_0\in\partial D$ so that for any function $u:[0,1]\mapsto\partial D$, we have
$$
\mathrm{Re}\left(\int_0^1g(t)u(t)\,\mathrm{d}t\right)\le\mathrm{Re}\left(\int_0^1g(t)u_0\,\mathrm{d}t\right)\tag{4}
$$
Set
$$
u(t)=\left\{\begin{array}{}
\frac{|g(t)|}{g(t)}&\text{when }g(t)\ne0\\
1&\text{when }g(t)=0
\end{array}\right.\tag{5}
$$
Then we get from $(4)$ that
$$
\begin{align}
\int_0^1|g(t)|\,\mathrm{Re}\left(1-\frac1{u(t)}u_0\,\right)\,\mathrm{d}t
&=\int_0^1\,\mathrm{Re}\left(\,g(t)u(t)-g(t)u_0\,\right)\,\mathrm{d}t\\
&\le0\tag{6}
\end{align}
$$
However, since $\left|\frac1{u(t)}\right|=1$ and $|u_0|=1$, we have $\mathrm{Re}\left(1-\frac1{u(t)}u_0\,\right)\ge0$. Since $(6)$ represents the integral of two non-negative functions, yet is non-positive, we get that for almost all $t\in[0,1]$, either $g(t)=0$ or $u(t)=u_0$. Thus,
$$
g(t)=\frac{|g(t)|}{u_0}\tag{7}
$$
Therefore, equality can hold only when $g(t)$ has the same argument for all $t\in[0,1]$.
