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

Let $z _{1}$ and $z _{2}$ be two complex numbers such that $\overline{ z }_{1}=i \overline{ z }_{2}$ and $\arg \left(\frac{ z _{1}}{\overline{ z }_{2}}\right)=\pi$. Then

medium

(JEE MAIN-2022)

$|{z_1} + {z_2}|\, = \,|{z_1}| + |{z_2}|$ is possible if

easy

If ${(\sqrt 8 + i)^{50}} = {3^{49}}(a + ib)$ then ${a^2} + {b^2}$ is

medium

Which of the following are correct for any two complex numbers ${z_1}$ and ${z_2}$

normal

If $z$ is a complex number such that $| z | = 4$ and $arg \,(z) = \frac {5\pi }{6}$ , then $z$ is equal to