heta_1 - heta_2 \equiv 60^\circ \pmod{180^\circ},

February 22, 2026

$heta_1 - heta_2 \equiv 60^\circ \pmod{180^\circ},$

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We are given that the angle between $ z_1 $ and $ z_2 $ is $ 60^\circ $, which corresponds to the smallest positive argument of $ z_1 \overline{z_2} $. Since both are on the unit circle, $ |z_1| = |z_2| = 1 $, and
z_1 \overline{z_2} = (\cos heta_1 + i \sin heta_1)(\cos heta_2 - i \sin heta_2) = \cos( heta_1 - heta_2) + i \sin( heta_1 - heta_2).
The argument of $ z_1 \overline{z_2} $ is $ heta_1 - heta_2 $ (modulo $ 360^\circ $), and the smallest positive argument is given to be $ 60^\circ $. Since cosine determines the angle uniquely within $ [0^\circ, 180^\circ] $ for this context (as it's the smallest angle between them), we have
and since the angle is acute and positive, $ heta_1 - heta_2 = 60^\circ $ (modulo $ 360^\circ $, but the minimal positive value is taken). Therefore,
\cos( heta_1 - heta_2) = \cos 60^\circ = rac{1}{2}.
oxed{rac{1}{2}}

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