If √5−12i+√−5−12i=z, then principal value of argz can be
A
−π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct options are A−π4 Bπ4
C3π4 D3π4 √5−12i=√(3−2i)2=±(3−2i) √−5−12i=√(2−3i)2=±(2−3i) ∴√5−12i+√−5−12i=−1−i,−5+5i,5−5i,1+i Therefore, the principal values of argz are −3π4,3π4,−π4,π4