Continuity of principal complex arccos

Question

The principal branch of the complex arccos is defined by $$\operatorname{Arccos} z = -i \operatorname{Log}\left(z + i\sqrt{1-z^2}\right)$$ where $\sqrt{\cdot}$ and Log denote the principal branches of the complex square root and logarithm, respectively.

I'm trying to verify the standard fact that Arccos is continuous on the domain $D = \mathbb{C} \setminus ((-\infty, -1] \cup [1, \infty))$. Recall that $\sqrt{\cdot}$ and Log are both continuous everywhere except the negative real axis.

For $z \in (-1,1)$ it's easy: then $1-z^2 \in (0,1)$, where $\sqrt{\cdot}$ is continuous. And then $z+i\sqrt{1-z^2}$ has nonzero imaginary part, so Log is continuous there.

For $z$ with nonzero imaginary part, $z^2$ is not a positive real, so $1-z^2$ is not a negative real, hence $\sqrt{\cdot}$ is continuous at $1-z^2$. If moreover $z$ has positive imaginary part, then since the principal branch of $\sqrt{\cdot}$ has nonnegative real part, we see that $z + i \sqrt{1-z^2}$ has positive imaginary part, so Log is continuous there.

If $z$ has negative imaginary part, this argument doesn't work. How can we conclude that Arccos is continuous at such $z$?

Answer 1

Proof $\def\rr{\mathbb{R}}$ $\def\imp{\Rightarrow}$

Given any complex $z \notin \rr_{\le -1} \cup \rr_{\ge 1}$:

  $1-z^2 \notin \rr_{\le 0}$ because $z^2 \notin \rr_{\ge 1}$.

  Let $w=z+i\sqrt{1-z^2}$, which is continuous with respect to $z$.

  Then $(w-z)^2 = -(1-z^2)$ and hence $w^2 - 2z w + 1 = 0$.

  If $w \in \rr_{<0}$:

    $z = \frac{w^2+1}{2w} = \frac{|w|}{w} \frac{|w|^2+1}{2|w|} \in \rr_{\le -1} \cup \rr_{\ge 1}$ because $|w|^2+1 \ge 2|w|$, giving a contradiction.

  Therefore $w \notin \rr_{<0}$.

  Thus $\log_π$ is continuous at $w$.

  Thus $\arccos_π(z) = \log_π(z+i\sqrt{1-z^2})$ is continuous with respect to $z$.

Notes

In fact, we get not just continuity but holomorphicity since the principal square-root function and the principal logarithm function are both holomorphic on their domains.