If |z₁| = |z₂| and arg (z₁/z₂) = $\mathrm{\pi }$, then z₁ + z₂ is equal to

We have

$\arg \left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}\right)=\mathrm{\pi }$

$\begin{array}{r}\mathrm{or} \arg \left({\mathrm{z}}_1\right)-\arg \left({\mathrm{z}}_2\right)=\mathrm{\pi }\\ \mathrm{or} \arg \left({\mathrm{z}}_1\right)=\arg \left({\mathrm{z}}_2\right)+\mathrm{\pi }\end{array}$

Let $\arg \left({\mathrm{z}}_2\right)=\mathrm{\theta }\text{. Then }\arg \left({\mathrm{z}}_1\right)=\mathrm{\pi }+\mathrm{\theta }\text{. }$

$\begin{array}{c}\therefore {\mathrm{z}}_1=\left|{\mathrm{z}}_1\right|[\cos (\mathrm{\pi }+\mathrm{\theta })+\mathrm{isin}(\mathrm{\pi }+\mathrm{\theta }\left)\right]\\ =\left|{\mathrm{z}}_1\right|(-\cos \mathrm{\theta }-\mathrm{isin}\mathrm{\theta })\end{array}$

$\begin{array}{c}\mathrm{and} {\mathrm{z}}_2=\left|{\mathrm{z}}_2\right|(\cos \mathrm{\theta }+\mathrm{isin}\mathrm{\theta })\\ =\left|{\mathrm{z}}_1\right|(\cos \mathrm{\theta }+\mathrm{isin}\mathrm{\theta }) \left(\because \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right|\right)\\ =-{\mathrm{z}}_1\end{array}$

$\Rightarrow {\mathrm{z}}_1+{\mathrm{z}}_2=0$