I wanted to figure out whether the solution to the following exercise is wrong:
If $f(z)=\log(\eta(z))$ with $\eta(z)=z^2-4z+3$ and $\text{arg }\eta\in (0,2\pi]$, what is $$\lim_{\varepsilon\to 0^+}f(4-i\varepsilon)?$$
Solution says it is $\log 3-i2\pi$; I got $\log 3+i2\pi$:
$$ \begin{aligned} \eta(4-i\varepsilon)&=(4-i\varepsilon)^2-4(4-i\varepsilon)+3\\ &=16-8i\varepsilon+\varepsilon^2-16+4i\varepsilon+3\\ &=3-4i\varepsilon+\varepsilon^2\\ &\overset{\varepsilon\to 0}{\sim}3-4i\varepsilon=\sqrt{3^2+\varepsilon^2}e^{i\arctan\left(-\frac{4\varepsilon}{3}\right)}\sim 3e^{-i\frac{4\varepsilon}{3}}\overset{\small\arg\eta\in(0,2\pi]}{=\mathrel{\mkern-3mu}=}3e^{i2\pi} \end{aligned} $$ Thus, $$\lim_{\varepsilon\to 0^+}\log(\eta(4-i\varepsilon))=\log(3e^{i2\pi})=\log3+i2\pi$$