If √5−12i+√−5−12i=z, then principal value of argz can be
A
−π4
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B
π4
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C
3π4
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D
3π4
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Solution
The correct option is 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