$$\lim_{z\to0}\ln\left\lvert\frac{iz-1}{iz+1}\right\rvert$$
The $ln$ is the natural logarithm on a real number since the inside is the norm of a complex number. My question is: can I just move the limit inside by saying that $ln$ is a continuous function?
i.e.
$$\lim_{z\to0}\ln\left\lvert\frac{iz-1}{iz+1}\right\rvert = \ln\left\lvert\frac{i(\lim_{z\to0} z)-1}{i(\lim_{z\to0} z)+1}\right\rvert=\ln1=0$$
Is this correct?
On
The expression inside the modulus delimiters $|\ldots|$ represents a continuous function $\mathbb{C} \setminus\{i\} \to \mathbb{C}$. The modulus itself is a continuous function $\mathbb{C} \to \mathbb{R}_{\geqslant0}$. The natural logarithm is a continuous function $\mathbb{R}_{>0} \to \mathbb{R}$. Because $|w| = 0$ if and only if $w = 0$, the expression inside the delimiters is zero if and only if $z = -i$. So, the whole expression is defined for all $z \in \mathbb{C} \setminus \{i, -i\}$, and represents the composite of three continuous functions: $$ \mathbb{C} \setminus \{i, -i\} \xrightarrow{f} \mathbb{C} \setminus \{0\} \xrightarrow{g} \mathbb{R}_{>0} \xrightarrow{\ln} \mathbb{R}. $$ Therefore, for $z \ne \pm i$, the limit symbol does almost literally "move" inside three pairs of implied parentheses: $$ \lim_{z\to0}\ln(g(f(z))) = \ln(\lim_{z\to0}g(f(z))) = \ln(g(\lim_{z\to0}f(z))) = \ln(g(f(0))). $$
Yes, though I might prefer $$\lim_{z\to0}\ln\left|\frac{iz-1}{iz+1}\right| = \ln\left(\lim_{z\to0}\left| \frac{iz-1}{iz+1}\right|\right) = \ln\left|\lim_{z\to0} \frac{iz-1}{iz+1}\right| = \ln1=0$$
Your logarithm is always of a positive real number and for small complex $z$ you will have $\frac{iz-1}{iz+1}$ close to $-1+i0$, and so $\left|\frac{iz-1}{iz+1}\right|$ close to $1$, leading to $\ln\left|\frac{iz-1}{iz+1}\right|$ close to $0$