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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
A
$0$
B
Purely imaginary
C
Purely real
D
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
