How can we show that $\mathrm{Re}(z + \sqrt{1+z^2}) \ge 0$ for all complex $z$? Here, the square root is the principal value (i.e., $-\pi/2 < \mathrm{Arg} \sqrt{z} \le \pi/2$).  The result is clear when Re $z$ is nonnegative, and it appears to be true numerically.
This is a followup to What is the principal branch of the complex arcsine, in simple terms? .
 A: Consider the numbers
$$
 w_1 = z + \sqrt{1+z^2} \, , \,
 w_2 = z - \sqrt{1+z^2} \, .
$$
Then
$$ \tag{1}
 w_1 - w_2 = 2 \sqrt{1+z^2} \implies \operatorname{Re}{w_1} \ge \operatorname{Re}{w_2}
$$
from the choice of the square root, and
$$ \tag{2}
 w_1 w_2 = -1 \implies w_1 = - \frac{\overline{w_2}}{|w_2^2|} 
\implies \operatorname{Re}w_1 = - \frac{\operatorname{Re}w_2}{|w_2^2|}\, .
$$
If we assume that $\operatorname{Re}w_1 < 0$ then
$$
\operatorname{Re}w_1 < 0
\underset{(1)}{\implies}
\operatorname{Re}w_2 < 0
\underset{(2)}{\implies}
\operatorname{Re}w_1 > 0
$$
gives a contradiction, so that $\operatorname{Re} w_1 \ge 0$ must hold (and $\operatorname{Re}w_2 \le 0$).

Alternative (but equivalent) approach: $w = z + \sqrt{1+z^2}$ satisfies
$$
 w + \frac 1w = 2  \sqrt{1+z^2} \, .
$$
It follows that
$$
 0 \le \operatorname{Re}\left( w + \frac 1w \right) = 
\operatorname{Re} w \cdot \left( 1 + \frac 1{|w|^2}\right)
$$
and therefore $\operatorname{Re}w \ge 0$.
One can also see that $\operatorname{Re}w = 0$ if and only if $\operatorname{Re}(\sqrt{1+z^2}) = 0$, and that is exactly for numbers of the form $z = ia$ with $a \in \Bbb R$, $|a| \ge 1$.
