Let z be a purely imaginary number such that {mathop{rm Im}nolimits} ,(z) > 0 . Then arg(z) is eq

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4-1.Complex numbers

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Let $z$be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} \,(z) > 0$. Then $arg(z)$ is equal to

A

$\pi $

B

$\frac{\pi }{2}$

C

$0$

D

$ - \frac{\pi }{2}$

Solution

(b)Let $z = 0 + ib$, where $b > 0$. Then $z$lies on +ve $y$-axis and so $arg(z) = \frac{\pi }{2}$.

Standard 11

Mathematics

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