Amplitude of sin frac{π }{5} + i,left( {1

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4-1.Complex numbers

easy

The amplitude of $\sin \frac{\pi }{5} + i\,\left( {1 - \cos \frac{\pi }{5}} \right)$

$\pi /5$

$2\pi /5$

$\pi /10$

$\pi /15$

Solution

(c) $\sin \frac{\pi }{5} + i\,\left( {1 – \cos \frac{\pi }{5}} \right)$ $ = 2\sin \frac{\pi }{{10}}\cos \frac{\pi }{{10}} + i2{\sin ^2}\frac{\pi }{{10}}$
$ = 2\sin \frac{\pi }{{10}}\left( {\cos \frac{\pi }{{10}} + i\sin \,\frac{\pi }{{10}}} \right)$
For amplitude, $\tan \theta = \,\frac{{\sin \frac{\pi }{{10}}}}{{\cos \frac{\pi }{{10}}}} = \tan \frac{\pi }{{10}}$ ==> $\theta = \frac{\pi }{{10}}$.

Standard 11

Mathematics

The maximum value of $|z|$ where z satisfies the condition $\left| {z + \frac{2}{z}} \right| = 2$ is

hard

If $|{z_1}|\, = \,|{z_2}|$ and $amp\,{z_1} + amp\,\,{z_2} = 0$, then

normal

If $|z|\, = 1,(z \ne – 1)$and $z = x + iy,$then $\left( {\frac{{z – 1}}{{z + 1}}} \right)$ is

easy

If $a > 0$ and $z = \frac{{{{\left( {1 + i} \right)}^2}}}{{a – i}}$, has magnitude $\sqrt {\frac{2}{5}} $, then $\bar z$ is equal to:

hard

(JEE MAIN-2019)

If $z$ and $\omega $ are two non-zero complex numbers such that $|z\omega |\, = 1$ and $arg(z) – arg(\omega ) = \frac{\pi }{2},$ then $\bar z\omega $ is equal to

medium

(AIEEE-2003)