If for complex numbers {z₁} and {z₂} , arg({z₁}/{z₂}) = 0, then |{z₁}

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4-1.Complex numbers

easy

If for complex numbers ${z_1}$ and ${z_2}$, $arg({z_1}/{z_2}) = 0,$ then $|{z_1} - {z_2}|$ is equal to

$|{z_1}| + |{z_2}|$

$|{z_1}| - |{z_2}|$

$||{z_1}| - |{z_2}||$

$0$

Solution

(c) We have $|{z_1} – {z_2}{|^2}$
$ = |{z_1}{|^2} + |{z_2}{|^2} – 2|{z_1}||{z_2}|\cos ({\theta _1} – {\theta _2})$
where ${\theta _1} = arg({z_1})$ and ${\theta _2} = arg({z_2})$
Since $arg\,{z_1} – arg\,{z_2} = 0$
$|{z_1} – {z_2}{|^2} = |{z_1}{|^2} + |{z_2}{|^2} – 2|{z_1}||{z_2}|$
$ = {(|{z_1}| – |{z_2}|)^2}$
==> $|{z_1} – {z_2}| = ||{z_1}| – |{z_2}||$

Standard 11

Mathematics

If ${z_1}{\rm{ and }}{z_2}$ be complex numbers such that ${z_1} \ne {z_2}$ and $|{z_1}|\, = \,|{z_2}|$. If ${z_1}$ has positive real part and ${z_2}$ has negative imaginary part, then $\frac{{({z_1} + {z_2})}}{{({z_1} – {z_2})}}$may be

hard

(IIT-1986)

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

easy

If $z$ is a complex number, then $z.\,\overline z = 0$ if and only if

normal

The real value of $\theta$ for which the expression $\frac{{1 + i\,\cos \theta }}{{1 – 2i\cos \theta }}$ is a real number is $\left( {n \in I} \right)$

normal

The conjugate of complex number $\frac{{2 – 3i}}{{4 – i}},$ is

easy

If for complex numbers ${z_1}$ and ${z_2}$, $arg({z_1}/{z_2}) = 0,$ then $|{z_1} - {z_2}|$ is equal to

Solution

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