If |{z₁}|, = ,|{z₂}| and arg,,left( {frac{{{z₁}}}{{{z₂}}}} right) = π , then {z₁} + {z₂

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4-1.Complex numbers

easy

If $|{z_1}|\, = \,|{z_2}|$ and $arg\,\,\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \pi $, then ${z_1} + {z_2}$ is equal to

$0$

Purely imaginary

Purely real

None of these

Solution

(a) We have $arg\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \pi $
==> $arg({z_1}) – arg({z_2}) = \pi $ ==>$arg\,\,({z_1}) = arg\,\,({z_2}) + \pi $
Let $arg\,\,({z_2}) = \theta $, then $arg$$({z_1}) = \pi + \theta $
${z_1} = |{z_1}|[\cos (\pi + \theta ) + i\sin (\pi + \theta )]$
$ = |{z_1}|( – \cos \theta – i\sin \theta )$
and ${z_2} = |{z_2}|(\cos \theta + i\sin \theta )$$ = |{z_1}|(\cos \theta + i\sin \theta )$
$(\because |{z_1}| = |{z_2}|)$
Hence ${z_1} + {z_2} = 0$.

Standard 11

Mathematics

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|,$then arg $({z_1}) – $arg $({z_2})$ is equal to

easy

(IIT-1987)

If $z $ is a complex number of unit modulus and argument $\theta$, then ${\rm{arg}}\left( {\frac{{1 + z}}{{1 + (\bar z)}}} \right)$ equals.

medium

(JEE MAIN-2013)

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

easy

If $\alpha $ and $\beta $ are different complex numbers with $|\beta | = 1$, then $\left| {\frac{{\beta – \alpha }}{{1 – \overline \alpha \beta }}} \right|$ is equal to

medium

(IIT-1992)

If $z_1$ is a point on $z\bar{z} = 1$ and $z_2$ is another point on $(4 -3i)z + (4 + 3i)z -15 = 0$, then $|z_1 -z_2|_{min}$ is (where $ i = \sqrt { – 1}$ )

normal

If $|{z_1}|\, = \,|{z_2}|$ and $arg\,\,\left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \pi $, then ${z_1} + {z_2}$ is equal to

Solution

Similar Questions

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|,$then arg $({z_1}) – $arg $({z_2})$ is equal to

If $z $ is a complex number of unit modulus and argument $\theta$, then ${\rm{arg}}\left( {\frac{{1 + z}}{{1 + (\bar z)}}} \right)$ equals.

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If $z_1$ is a point on $z\bar{z} = 1$ and $z_2$ is another point on $(4 -3i)z + (4 + 3i)z -15 = 0$, then $|z_1 -z_2|_{min}$ is (where $ i = \sqrt { – 1}$ )

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