z1 and z2 are two complex numbers such that |z1| = |z2| and arg (z1) + arg (z2) = π, then show that z1 = – z̅2.

According to the question,

Let \[{{z}_{1}}~=\text{ }|{{z}_{1}}|\text{ }(cos\text{ }{{\theta }_{1}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{1}})\text{ }and\text{ }{{z}_{2}}~=\text{ }|{{z}_{2}}|\text{ }(cos\text{ }{{\theta }_{2}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{2}})\]

Given that |z₁| = |z₂|

And \[\begin{array}{*{35}{l}}

arg\text{ }({{z}_{1}})\text{ }+\text{ }arg\text{ }({{z}_{2}})\text{ }=\text{ }\pi   \\

\Rightarrow ~{{\theta }_{1}}~+\text{ }{{\theta }_{2}}~=\text{ }\pi   \\

\Rightarrow ~{{\theta }_{1}}~=\text{ }\pi-\text{ }\text{ }{{\theta }_{2}}  \\

Now,\text{ }{{z}_{1}}~=\text{ }|{{z}_{2}}|\text{ }(cos\text{ }(\pi-\text{ }\text{ }{{\theta }_{2}})\text{ }+\text{ }I\text{ }sin\text{ }(\pi- \text{ }\text{ }{{\theta }_{2}}))  \\

\Rightarrow ~{{z}_{1}}~=\text{ }|{{z}_{2}}|\text{ }(-cos\text{ }{{\theta }_{2}}~+\text{ }I\text{ }sin\text{ }{{\theta }_{2}})  \\

\Rightarrow ~{{z}_{1}}~=\text{ }-|{{z}_{2}}|\text{ }(cos\text{ }{{\theta }_{2}}~\text{ }I\text{ }+sin\text{ }{{\theta }_{2}})  \\

\Rightarrow ~{{z}_{1}}~=\text{ }\text{ }[|{{z}_{2}}|\text{ }(cos\text{ }{{\theta }_{2}}~\text{ }I\text{ }+sin\text{ }{{\theta }_{2}})]  \\

\end{array}\]

Hence, z₁ = -z̅₂

Hence proved.