If -5-12i-5-12i-z- then principal value of z can be

√(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 z are 3π4,3π4, π4,π4 ...

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Byju's Answer

Standard XI

Mathematics

Argument of a Complex Number

If √5-12i+√-5...

Question

If $\sqrt{5 - 12 i} + \sqrt{- 5 - 12 i} = z$ , then principal value of $arg z$ can be

- \frac{π}{4}

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\frac{π}{4}

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\frac{3 π}{4}

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\frac{3 π}{4}

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Solution

The correct option is D $\frac{3 π}{4}$
$\sqrt{5 - 12 i} = \sqrt{(3 - 2 i)^{2}} = \pm (3 - 2 i)$
$\sqrt{- 5 - 12 i} = \sqrt{(2 - 3 i)^{2}} = \pm (2 - 3 i)$
$∴ \sqrt{5 - 12 i} + \sqrt{- 5 - 12 i} = - 1 - i, - 5 + 5 i, 5 - 5 i, 1 + i$
Therefore, the principal values of $arg z$ are $- \frac{3 π}{4}, \frac{3 π}{4}, - \frac{π}{4}, \frac{π}{4}$

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Find the modulus, argument and the principal argument of the complex numbers.

z = \frac{\sqrt{5 + 12 i} + \sqrt{5 - 12 i}}{\sqrt{5 + 12 i} - \sqrt{5 - 12 i}}

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

The correct option is D 3π4√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

If $\sqrt{5 - 12 i} + \sqrt{- 5 - 12 i} = z$ , then principal value of $arg z$ can be