$\arg(\overline{z}), \arg(z^2)$ A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.42,43



Exer 3.42 multiple-valued: NO in general.
$$\arg(1+i)=\frac \pi 4 + 2k \pi$$
$$\arg(1-i)=\frac {-\pi} 4 - 2l \pi$$
Choose $k=1$,$l=17$ to arrive at a contradiction.
Exer 3.42 principal: NO, but YES iff $z \notin \mathbb R_{< 0}$
Pf:
$$Arg(\overline{z}) = -Arg(z) \iff -Arg(z) \in (-\pi,\pi] \iff Arg(z) \in [-\pi,\pi]) \iff Arg(z) \in (-\pi,\pi) \iff z \notin \mathbb R_{< 0}$$
QED
Exer 3.43 multiple-valued: NO in general.
For $z = 0$, yes vacuously. For $z \ne 0$, 
Observe that $\ln|z^2| = \ln(|z|^2) = 2\ln|z|$
$$\therefore, \ln(z^2) = 2\ln(z) \iff \arg(z^2) = 2\arg(z)$$
Consider
$$\arg(1+i)=\frac \pi 4 + 2k \pi$$
$$\arg(2i) = \frac \pi 2 + 4l \pi$$
Choose $k=1$,$l=17$ to arrive at a contradiction.
Exer 3.43 single-valued: NO, but YES iff $Arg(z) \in (-\frac \pi 2,\frac \pi 2]$.
Pf:
For $z = 0$, yes vacuously. For $z \ne 0$, 
Observe that $Ln|z^2| = Ln(|z|^2) = 2Ln|z|$
$$\therefore, Ln(z^2) = 2Ln(z) \iff Arg(z^2) = 2Arg(z)$$
$$Arg(z^2) = 2Arg(z) \iff 2Arg(z) \in (-\pi,\pi] \iff Arg(z) \in (-\frac \pi 2,\frac \pi 2]$$
QED



*

*Where have I gone wrong for above solutions?

*Which are wrong? I believe they're all right, but I might have missed something (For $z=0$, I guess the relevant statements are vacuously true).


$$\arg(\overline{z}) \equiv -\arg(z) \mod 2 \pi \ \forall z \in \mathbb R_{<0}$$
$$Arg(\overline{z}) \equiv -Arg(z) \mod 2 \pi \ \forall z \in \mathbb R_{<0}$$
$$\arg(\overline{z}) \equiv -\arg(z) \mod 2 \pi \ \forall z \notin \mathbb R_{<0}$$
$$Arg(\overline{z}) \equiv -Arg(z) \mod 2 \pi \ \forall z \notin \mathbb R_{<0}$$
$$\arg(z^2) \equiv 2\arg(z) \mod 2 \pi \ \forall z \in \mathbb R_{<0}$$
$$Arg(z^2) \equiv 2Arg(z) \mod 2 \pi \ \forall z \in \mathbb R_{<0}$$
$$\arg(z^2) \equiv 2\arg(z) \mod 2 \pi \ \forall z \notin \mathbb R_{<0}$$
$$Arg(z^2) \equiv 2Arg(z) \mod 2 \pi \ \forall z \notin \mathbb R_{<0}$$
It seems 2.6 and 2.8 are right by Wiki. What about the others?

(Partial answer for 3.42 below, 3.43 can be worked out in a similar way.)  The following assumes that $z \ne 0$ and $\arg(ab)=\arg(a)+\arg(b)$ was already established for the multi-valued $\arg$.

Exer 3.42 multiple-valued: NO in general.

YES, in general, since $\arg(z\bar z) = \arg(z)+\arg(\bar z)$, but on the other hand $\arg(z \bar z) = \arg(|z|^2) = 0$ because $|z|^2$ is a positive real number, so $\,\arg(\bar z) = -\arg(z)\,$.

$$\arg(1+i)=\frac \pi 4 + 2k \pi$$
  $$\arg(1-i)=\frac {-\pi} 4 - 2l \pi$$
  Choose $k=1$,$l=17$ to arrive at a contradiction.

You don't get to choose $k,l$ here. The multi-valued $\arg$ equality holds if there exists any pair $k,l$ for which it holds, in this case $k=l=0$ for example.

Exer 3.42 principal: NO, but YES iff $z \notin \mathbb R_{< 0}$

It looks like this uses the (common) definition of the principal value of $\operatorname{Arg}$ where the range is $(-\pi,\pi]$. In that case the answer is YES, except if $\,\operatorname{Arg}(z)=\pi\,$ (why?).
If using the other (less common) definition for the principal value of $\operatorname{Arg}$ where the range is $[0,2 \pi)$ then the answer is of course NO, except if $\,\operatorname{Arg}(z)=0\,$ (why?).
