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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
$2\, cos\alpha ,\, \alpha $
$-2\, cos\alpha ,\, \alpha $
$-2\, cos\alpha ,\, \alpha - \pi $
None of these
Solution
$z=(1+\cos 2 \alpha)+i \sin 2 \alpha$
$z=2 \cos ^{2} \alpha+2 i \sin \alpha \cos \alpha$
$z=2 \cos \alpha[\cos \alpha+i \sin \alpha]$
$z=-2 \cos \alpha[-\cos \alpha-i \sin \alpha]$
$z=2 \cos \alpha[\cos (\alpha-\pi)+i \sin (\alpha-\pi)]$
[ $\because $$\frac{\pi }{2} < \alpha < \frac{{3\pi }}{2}$]
Thus, $|z|=-2 \cos \alpha$ and $\arg (z)=\alpha-\pi$
Similar Questions
Let $z$ be complex number satisfying $|z|^3+2 z^2+4 z-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero.
Match each entry in List-$I$ to the correct entries in List-$II$.
| List-$I$ | List-$II$ |
| ($P$) $|z|^2$ is equal to | ($1$) $12$ |
| ($Q$) $|z-\bar{z}|^2$ is equal to | ($2$) $4$ |
| ($R$) $|z|^2+|z+\bar{z}|^2$ is equal to | ($3$) $8$ |
| ($S$) $|z+1|^2$ is equal to | ($4$) $10$ |
| ($5$) $7$ |
The correct option is:
