√3+i=(a+ib)(c+id),thentan−1ba+tan−1dc
has the value
√3+i=(a+ib)(c+id)
∴ac−bd=√3 and ad+bc=1
Now tan−1(ba)+tan−1(dc)
= tan−1(ab+dc1−ba.dc)= tan−1(bc+adac−bd)= tan−1(1√3)
=nπ+π6, nϵI