Let z be a complex number. Then the angle bet | vedclass

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(C) Let $z = r(\cos 	heta + i \sin 	heta) = re^{i	heta}$.Then $-iz = -i(re^{i	heta}) = e^{-i\pi/2} (re^{i	heta}) = re^{i(	heta - \pi/2)}$.The ar

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

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(C) Let $z = r(\cos 	heta + i \sin 	heta) = re^{i	heta}$.Then $-iz = -i(re^{i	heta}) = e^{-i\pi/2} (re^{i	heta}) = re^{i(	heta - \pi/2)}$.The argument of $z$ is $	heta$ and the argument of $-iz$ is $	heta - \frac{\pi}{2}$.The angle between the vectors is the difference of their arguments:$	ext{Angle} = (	heta - \frac{\pi}{2}) - 	heta = -\frac{\pi}{2}$.Thus,the angle is $-\frac{\pi}{2}$.

Let $z$ be a complex number. Then the angle between vectors $z$ and $-iz$ is

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