(A) Let $z = r(\cos \theta + i \sin \theta)$,where $\theta = \arg(z)$.Given $\arg(z) = \theta < 0$.Then $-z = -r(\cos \theta + i \sin \theta) = r(\cos

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(A) Let $z = r(\cos \theta + i \sin \theta)$,where $\theta = \arg(z)$.Given $\arg(z) = \theta Then $-z = -r(\cos \theta + i \sin \theta) = r(\cos(\theta + \pi) + i \sin(\theta + \pi))$.Thus,$\arg(-z) = \theta + \pi$ (if $\theta + \pi \le \pi$) or $\theta - \pi$ (if $\theta + \pi > \pi$).Since $-\pi Therefore,$\arg(-z) = \theta + \pi$.Now,$\arg(-z) - \arg(z) = (\theta + \pi) - \theta = \pi$.

Class 11 Mathematics 4-1.Complex numbers Argument of complex numbers

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If $\arg(z) < 0$,then $\arg(-z) - \arg(z)$ is equal to

A
$\pi$
B
$-\pi$
C
$-\frac{\pi}{2}$
D
$\frac{\pi}{2}$

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4-1.Complex numbers

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