If `sqrt(5-12 i)+sqrt(5-12 i)=z`, then principal value of `a rgz`can be`pi/4`b. `pi/4`c. `(3pi)/4`d. `-(3pi)/4`A. `-(pi)/(4)`B. `(pi)/(4)`C. `(3pi)/(4)`D. `-(3pi)/(4)`

If `sqrt(5-12 i)+sqrt(5-12 i)=z`, then principal value of `a rgz`can b

1.	If `sqrt(5-12 i)+sqrt(5-12 i)=z`, then principal value of `a rgz`can be`pi/4`b. `pi/4`c. `(3pi)/4`d. `-(3pi)/4`A. `-(pi)/(4)`B. `(pi)/(4)`C. `(3pi)/(4)`D. `-(3pi)/(4)`
Answer» Correct Answer - A::B::C::D `sqrt(5-12i) = sqrt((3-2i)^(2)) = pm (3-2i)` `sqrt(-5-12i)= sqrt((2-3i)^(2)) = pm (2-3i)` `rArr z = sqrt(5-12i) + sqrt(5-12i) = sqrt((3-2i)^(2)) = pm (3-2i)` `sqrt(-5-12i)= sqrt((2-3i)^(2)) = pm (2-3i)` `rArr z = sqrt(5-12i) + sqrt(-5-12i)` `=- 1-i, -5 + 5i,5-5i, 1+i` Therefore, principal values of arg are `- 3pi//4, 3pi//4,-pi//4, pi//4`.

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If `sqrt(5-12 i)+sqrt(5-12 i)=z`, then principal value of `a rgz`can be`pi/4`b. `pi/4`c. `(3pi)/4`d. `-(3pi)/4`A. `-(pi)/(4)`B. `(pi)/(4)`C. `(3pi)/(4)`D. `-(3pi)/(4)`

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