$\arg(\overline{z}), \arg(z^2)$

Question

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

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1. Where have I gone wrong for above solutions?
2. 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?

Answer 1

(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?).

Answer 2

problem 1

This is controversial problem, the answer depends on its definition.

See also

problem 2

take $z=re^{i\theta}$, $$\log z^2=2\log |z|+i(2\theta+2n\pi)$$, $$2\log z=2\log|z|+i(2\theta+4n\pi )$$

Thus $\log z^2$ contains $2\log z$.