Suppose that $0<|a|<1$. I'm interested in finding the (necessary and sufficient) conditions under which
$$|1\pm\sqrt{1-a^2}|<|a|$$
where $a$ is complex.
For $a$ real, one can easily determine this this holds/fails for plus/minus depending on the sign of $a$.
I'm struggling to see a nice way of understanding the general complex case. Does anyone have an intuition as to how to go about this? I've tried to think of it geometrically, but my plane geometry is rusty at best! The triangle inequality seems far too weak, unfortunately. Perhaps there's a change of variables that's enlightening? Alternatively, it may suffice to just consider the real part...
Note that obviously if the plus holds, the minus fails, since
$$|1+\sqrt{1-a^2}||1-\sqrt{1-a^2}|=|a|^2$$
Any hints or ideas would be most welcome - it's hopefully an interesting problem (that's sprung out of extending a homework question from an undergrad assignment I'm marking, for context)!
It is clear that $a$ can not be $0$, $1$, or $-1$ in $$ \tag{*} |1\pm\sqrt{1-a^2}|<|a| \quad . $$
Let $a \in \Bbb C \setminus \{ -1, 0, 1 \}$ and $b_{1,2}$ the two solutions of $a^2 + b^2 = 1$. Then $b_j \ne 1$ and $$ |1 - b_j| < |a| \\ \Longleftrightarrow |1 - b_j|^2 < |a|^2 = |1-b_j^2| \\ \Longleftrightarrow |1-b_j| < |1+b_j| \\ \Longleftrightarrow \operatorname{Re} b_j > 0 \quad . $$ for $j=1,2$. Therefore either
The second case occurs exactly if $a \in \Bbb R $ with $|a|\ge 1$.
Conclusion: $(*)$ has a solution exactly if $$ a \in \Bbb C \setminus \bigl((-\infty, -1] \cup \{ 0 \} \cup [1, \infty) \bigr) $$ and in particular if $0 < |a| < 1$. The solution is $$ |1-\sqrt{1-a^2}|<|a| $$ where the square root is chosen to have positive real part.