If 5−12i+−5−12i=z, then principal value of arg z can be

5−12i=(3−2i)2=±(3−2i)−5−12i=(2−3i)2=±(2−3i)⇒ z=5−12i+−5−12i=−1−i, −5+5i, 5−5i, 1+iTherefore, principal values of arg z are −3π/4, 3π/4, −π/4, π/4.