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We define the (incomplete) elliptic integral of the first kind with elliptic modulus $0 < k < 1$ as

$$ F(z; k) = \int_{0}^{z} \frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}, \text{ Im}(z) > 0 $$

It is well-known that $F$ maps the complex upper half-plane conformally onto some rectangle $R(k)$. One method for proving this starts out with showing that $F$ maps the extended real line one-to-one onto the boundary $\delta R(K)$. But I do not quite understand how $F(z; k)$ accomplishes this with respect to the turning angles involved. I can see how 90 degree counterclockwise turns are "generated" at $w = -1$ and $w = 1$, but not at $w = -1/k$ and $w = 1/k$, where the root gets real again.

Should the denominator in the integrand not be $\sqrt{1-w^2}\sqrt{1-k^2w^2}$ for that to happen? And if so, why is the elliptic integral always shown without this additional separated square root? As it is now it seems not even to be analytic in $\mathbb{H}$. And furthermore, if we wish to have continuity in the closure of the upper half-plane $\overline{\mathbb{H}}$ --- something which is explicitly used in a couple of arguments I have seen ---, do we not really need the even further separated $k\sqrt{1+w}\sqrt{1-w}\sqrt{\frac{1}{k}+w}\sqrt{\frac{1}{k}-w}$ as denominator, like in the Schwarz-Christoffel mapping formula? For me this is all a bit confusing.

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This function is analytic for all points in the upper half plane, except for real values $\pm 1, \pm 1/k$, because these values are where the function is not differentiable. Along the positive real line, from $0$ to $1$, the integral is real, but as we traverse "over" 1, something must happen because it is a point of singularity.

We require the argument $\theta\in (-\pi,0)$ to define the branch we are on, since the roots in the denominator tell us we have multiple branches. Consider the above integrand defined along Re($z$)>0 and further, define $w=1-\rho e^{i\theta}$ for radius $\rho>0$. We define $w$ this way, because we'd like to see what happens as we "pass over" $z=1$. Then $\sqrt{1-w}=\sqrt{1-(1-\rho e^{i\theta})}=\rho^{1/2}e^{i\theta/2}$. For $\theta=-\pi$, we have $e^{-(\pi/2) i}=-i$ by Euler's formula. Thus $\rho^{1/2}e^{-(\pi/2)i}=-i\sqrt{\rho}=-i\sqrt{w-1}$ as $\theta:-\pi\rightarrow 0$. Thus $$\int\limits_0^{z}\frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}=\int\limits_0^{1}\frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}+\int\limits_1^{z}\frac{dw}{-i\sqrt{(1+w)}\sqrt{(w-1)}\sqrt{(1-k^2w^2)}}=\int\limits_0^{1}\frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}+i\int\limits_1^{z}\frac{dw}{\sqrt{(1+w)}\sqrt{(w-1)}\sqrt{(1-k^2w^2)}}$$ The second integral implies that for $1<\hbox{ Re(}z)$, we are now traversing a vertical line along the real value provided by the first integral. This is to say that you are right about a counter-clockwise turn at z=1, but also that a similar argument shows another counter-clockwise turn at z=1/k and not z=-1.

To see the clock-wise turns at $z=-1,-1/k$, take the integral along the real axis for Re($z$)<0 and note,

$\int\limits_0^{-z}\frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}=-\int\limits_0^{z}\frac{dw}{\sqrt{(1-w^2)(1-k^2w^2)}}$. In other words, $F(z;k)$ is symmetric about the real axis.

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