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Question
If z1 and z2 are two complex numbers such that \[\left| z_1 \right| = \left| z_2 \right|\] and arg(z1) + arg(z2) = \[\pi\] then show that \[z_1 = - \bar{{z_2}}\].
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Solution
Let θ1 be the arg(z1) and θ2 be the arg(z2).
It is given that
\[\left| z_1 \right| = \left| z_2 \right|\] and arg(z1) + arg(z2) = \[\pi\].
Since, z1 is a complex number.
\[z_1 = \left| z_1 \right|\left( \cos \theta_1 + i\sin \theta_1 \right)\]
\[ = \left| z_2 \right|\left[ \cos\left( \pi - \theta_2 \right) + i\sin\left( \pi - \theta_2 \right) \right]\]
\[ = \left| z_2 \right|\left[ - \cos\left( \theta_2 \right) + i\sin\left( \theta_2 \right) \right]\]
\[ = - \left| z_2 \right|\left[ \cos\left( \theta_2 \right) - i\sin\left( \theta_2 \right) \right]\]
\[ = - \bar{{z_2}}\]
Hence,
\[z_1 = - \bar{{z_2}}\].
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