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4-1.Complex numbers
normal
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
A
$2\, cos\alpha ,\, \alpha $
B
$-2\, cos\alpha ,\, \alpha $
C
$-2\, cos\alpha ,\, \alpha - \pi $
D
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 $\arg (z)=\alpha-\pi$
Standard 11
Mathematics
