If frac{pi }{2}

Question

$z=(1+\cos 2 \alpha)+i \sin 2 \alpha$

$z=2 \cos ^{2} \alpha+2 i \sin \alpha \cos \alpha$

$z=2 \cos \alpha[\cos \alpha+i \sin \alpha]$

$z=-2 \

Vedclass · Accepted Answer

$-2\, cos\alpha ,\, \alpha - \pi $

Vedclass · Answer

$z=(1+\cos 2 \alpha)+i \sin 2 \alpha$

$z=2 \cos ^{2} \alpha+2 i \sin \alpha \cos \alpha$

$z=2 \cos \alpha[\cos \alpha+i \sin \alpha]$

$z=-2 \cos \alpha[-\cos \alpha-i \sin \alpha]$

$z=2 \cos \alpha[\cos (\alpha-\pi)+i \sin (\alpha-\pi)]$

                                                [  $\because $$\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{2}$]

Thus, $|z|=-2 \cos \alpha$ and $\arg (z)=\alpha-\pi$

If $\frac{\pi }{2} < \alpha < \frac{3}{2}\pi $ , then the modulus and argument of $(1 + cos\, 2\alpha ) + i\, sin\, 2\alpha $ is respectively

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