Question

If |z₁| = |z₂| and arg $\left(\frac{z_1}{z_2}\right)=\mathrm{\pi },$ then z₁ + z₂ = ____________.

Answer 1

Let |z₁| = |z₂| = r
Let arg z₁ = θ₁ and arg z₂ = θ₂
$$\begin{gathered} \Rightarrow z_1=r{e}^{i{\theta }_1}\mathrm{and}{z}_2=r{e}^{i{\theta }_2} \\ \mathrm{Since}\arg \left(\frac{z_1}{z_2}\right)=\mathrm{\pi } \\ \mathrm{i}.\mathrm{e}\arg z_1-\arg z_2=\mathrm{\pi } \\ \mathrm{i}.e{\theta }_1-{\theta }_2=\mathrm{\pi } \\ \mathrm{i}.\mathrm{e}{\theta }_1={\theta }_2+\mathrm{\pi } \\ {e}^{i{\theta }_1}=e^{i\left({\theta }_2+\mathrm{\pi }\right)} \\ =e^{i\mathrm{\pi }}{e}^{i{\theta }_2} \\ =\left(\mathrm{cos\pi }+i\mathrm{sin\pi }\right)\left(\cos {\theta }_2+i\sin {\theta }_2\right) \\ =-1\left(\cos {\theta }_2+i\sin {\theta }_2\right) \\ {e}^{i{\theta }_1}=-e^{i{\theta }_2} \\ \Rightarrow z_1+z_2=r{e}^{i{\theta }_1}+r{e}^{i{\theta }_2} \\ \mathrm{Hence},z_1+z_2=r\left(e^{i{\theta }_1}+e^{i{\theta }_2}\right)=0 \end{gathered}$$