Let, $z_1 = x_1 + iy_1 \ and \ z_2 = x_2 + iy_2$

By data $\big |z_1 + z_2 \big | = \big |z_1 - z_2 \big |$

$\big |x_1 + iy_1 + x_2 + iy_2 \big | = \big |x_1 + iy_1 - x_2 - i y_2 \big |$

$\therefore \big | (x_1 + x_2) + i (y_1 + y_2 ) \big | = \big | (x_1 -x_2) + i (y_1 - y_2) \big |$

$\therefore (x_1 + x_2 )^2 + (y_1 + y_2)^2 = (x_1 - x_2)^2 + (y_1 - y_2 )^2$

$\therefore x_1x_2 + y_1 y_2 = 0$

$\arg z_1 - \arg z_2 = \tan ^{-1} \dfrac {y_1}{x_1} - \tan^{-1} \dfrac {y_2}{x_2}$

$=\tan^{-1} \left [ \dfrac {\left ( \frac{y_1}{x_1} \right ) - \left ( \frac {y_2}{x_2} \right )}{1+ \left ( \frac {y_1}{x_1} \right ) \left ( \frac {y_2}{x_2} \right )} \right ]=\tan^{-1} \left [ \dfrac {x_2y_1 - y_2 x_1}{x_1 x_2 + y_1 y_2 } \right ] = \tan^{-1} \left [ \dfrac {x_2 y _1 - y_2 x_1}{0} \right ] = \dfrac {\pi}{2} $

Hence,

$\arg z_1 - \arg z_2 = \dfrac {\pi}{2}$