Following is from Complex Analysis by Freitag :
My questions:
1- The text gives an example for $w_r(M,N)$ but it doesn't explain it or give a clear definition of it. For example where does $w_1(-S,-S) = \dfrac{I_1(-E,i)}{I_1(-S,i)^2}$ come from? How $w_r(M,N)$ is defined exactly?
2- On the meaning of equation: For example suppose the second row of matrices M and N have (c,d) and (c',d') in the entries, respectively. Then the mentioned equation means $$((ca'+dc')z+(cb'+dd'))^{r/2} = w_r(M,N) (c \frac{a'z+b'}{c'z+d'}+d)^{r/2} (c'z+d')^{r/2}$$ which no way to be correct for $w_r(M,N) \in {\{\pm 1}\}.$ What does the equation $I_r(MN, z) = w_r(M,N) I_r(M,Nz) I_r(N, z)$ mean?
$M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $N=\begin{pmatrix}a'&b'\\c'&d'\end{pmatrix}$.
The thing is the square root map $z\mapsto \sqrt{z}$ for a complex number $z$ does not satisfy $\sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2}$. This is due to the fact that we need a branch cut in $\Bbb{C}$ to define $\sqrt{z}$, and the product $z_1z_2$ may take the product across the boundary (i.e., when $\arg z_1+\arg z_2$ and $\arg(z_1z_2)$ are unequal). So, the proper equality is $$\sqrt{z_1z_2}=\pm\sqrt{z_1}\sqrt{z_2},\tag{1}$$ but what determine the choice of signs $\pm$ depends on the values of $z_1,z_2$ and on how you select your branch cut, and I don't know which branch cut your book is using. However, that is precisely why you have the equality $$I_r(MN,z)=\pm I_r(M,Nz) I_r(N,z).$$ The number $w_r(M,N) =\pm1$ is just something to handle the signs so that the reader can do more definite operations (i.e., no guessing which sign it is going to be, as the signs are already encoded in $w_r(M,N)$), and so the author can write $$I_r(MN,z)=w_r(M,N) I_r(M,Nz)I_r(N,z).$$ It has to be verified, though, that $w_r(M,N)$ does not depend on $z$, and I disagree, as the text seems to be suggesting, that $w_r(M,N)$ does not depend on $z$. In the book's example, we have \begin{align}\sqrt{-1}=I_1(-E,z)&=w_1(-S,-S)\ I_1(-S,-Sz)\ I_1(-S,z)\\&=(-1)\ \sqrt{\frac{1}{z}}\ \sqrt{-z}.\tag{2}\end{align} If $z=1$ with the usual convention $\sqrt{1}=1$, we have $\sqrt{-1}=-\sqrt{-1}$, which is false. So, I would write $w_r(M,N,z)$, rather than $w_r(M,N)$. I am certain that no matter which branch cut you use, there are always $z_1,z_2$ that violate (2). However, I should like to mention that $w_r(M,N)$ only depends weakly on $z$ (that is, for fixed $M,N$, and for almost all $z$, $w_r(M,N)$ is constant in an open neighbourhood of $z$).
I don't know why you think the expression is "in no way to be correct." Note from (1) that \begin{align}\left(c \frac{a'z+b'}{c'z+d'}+d\right)^{r/2}&=\left(\frac{(ca'+dc')z+(cb'+dd')}{c'z+d'}\right)^{r/2}\\&=\left(\pm\frac{\sqrt{(ca'+dc')z+(cb'+dd')}}{\sqrt{c'z+d'}}\right)^{r} \\&=\pm\frac{\left(\sqrt{(ca'+dc')z+(cb'+dd')}\right)^r}{\left(\sqrt{c'z+d'}\right)^r}.\end{align} Therefore, \begin{align}\left(c \frac{a'z+b'}{c'z+d'}+d\right)^{r/2}\left(c'z+d'\right)^{r/2}&=\left(\pm\frac{\left(\sqrt{(ca'+dc')z+(cb'+dd')}\right)^r}{\left(\sqrt{c'z+d'}\right)^r}\right)\left(\sqrt{c'z+d'}\right)^{r} \\&=\pm\left(\sqrt{(ca'+dc')z+(cb'+dd')}\right)^r \\&=\pm\big((ca'+dc')z+(cb'+dd')\big)^{r/2}.\end{align} So, you see that $$\big((ca'+dc')z+(cb'+dd')\big)^{r/2}=\pm\left(c \frac{a'z+b'}{c'z+d'}+d\right)^{r/2}\left(c'z+d'\right)^{r/2}.$$ This equation simply means that you have to worry about the branch cut, and taking $(r/2)$th power is not a multiplicative operation, but it is multiplicative up to sign change. (Taking $(r/2)$th power is multiplicative only when $r$ is even.)