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This is from a collection book of problems on complex variables (Volkovyskii, Lunts, Aramanovich).

I don't know how to tackle it without involving heavy unpromising calculations:

Prove that both values of $\sqrt{z^2-1}$ lie on the straight line passing through the coordinate origin and parellel to the bisector of the internal angle of the triangle with vertices at points $-1$ and $1$ and $z$, which passes through the vertex z.

What would be an elegant solution to this problem? Or a promising way to approach it?

Thank you.

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1 Answer 1

up vote 5 down vote accepted

This follows directly from the arguments of the numbers involved:

$$\arg\sqrt{z^2-1}=\frac{\arg\left(z^2-1\right)}2=\frac{\arg\left((z+1)(z-1)\right)}2=\frac{\arg(z+1)+\arg(z-1)}2\pmod{\pi}\;.$$

This is the angle at which the bisector at $z$ between the lines from $z$ to $1$ and $-1$ intersects the real axis, since these lines intersect the real axis at angles $\arg(z-1)$ and $\arg(z+1)$, respectively.

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