Arg,(5 - √ 3 i) =

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4-1.Complex numbers

normal

$arg\,(5 - \sqrt 3 i) = $

${\tan ^{ - 1}}\frac{5}{{\sqrt 3 }}$

${\tan ^{ - 1}}\left( { - \,\frac{5}{{\sqrt 3 }}} \right)$

${\tan ^{ - 1}}\frac{{\sqrt 3 }}{5}$

${\tan ^{ - 1}}\left( { - \frac{{\sqrt 3 }}{5}} \right)$

Solution

(d)Let $z = 5 – \sqrt 3 \,i$
==> $r(\cos \theta + i\sin \theta ) = 5 – \sqrt 3 \,i$
==> $r\cos \theta = 5$and $r\sin \theta = – \sqrt 3 $
$\therefore $ $\tan \theta = – \frac{{\sqrt 3 }}{5} \Rightarrow \theta = {\tan ^{ – 1}}\left( { – \frac{{\sqrt 3 }}{5}} \right)$

Standard 11

Mathematics

The amplitude of $0$ is

normal

If $z$ and $\omega$ are two complex numbers such that $|z \omega|=1$ and $\arg (z)-\arg (\omega)=\frac{3 \pi}{2}$, then $\arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$ is:

(Here arg(z) denotes the principal argument of complex number $z$ )

hard

(JEE MAIN-2021)

If ${z_1},{z_2},{z_3}$be three non-zero complex number, such that ${z_2} \ne {z_1},a = |{z_1}|,b = |{z_2}|$ and $c = |{z_3}|$ suppose that $\left| {\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = 0$, then $arg\left( {\frac{{{z_3}}}{{{z_2}}}} \right)$ is equal to

hard

If ${z_1},{z_2},{z_3}$ are complex numbers such that $|{z_1}|\, = \,|{z_2}|\, = $ $\,|{z_3}|\, = $ $\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}}} \right| = 1\,,$ then${\rm{ }}|{z_1} + {z_2} + {z_3}|$ is

medium

(IIT-2000)

The values of $z$for which $|z + i|\, = \,|z – i|$ are

easy