if |z1+z2|>|z1-z2| then prove that -pi/2<arg(z1/z2)<pi/2

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \left|{\mathrm{z}}_1+{\mathrm{z}}_2\right|>\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right| \\ \Rightarrow \left|{\mathrm{z}}_2\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}+1\right)\right|>\left|{\mathrm{z}}_2\left(\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}-1\right)\right| \\ \Rightarrow \left|{\mathrm{z}}_2\right|\left|\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}+1\right|>\left|{\mathrm{z}}_2\right|\left|\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}-1\right| \left[\mathrm{As} \left|\mathrm{A}.\mathrm{B}\right|=\left|\mathrm{A}\right|\left|\mathrm{B}\right|\right] \\ \Rightarrow \left|\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}+1\right|>\left|\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}-1\right| \\ \mathrm{As} \mathrm{we} \mathrm{have} \frac{{\mathrm{z}}_1}{{\mathrm{z}}_2} \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{and} 1 \mathrm{and} -1 \mathrm{both} \mathrm{are} \mathrm{real} \mathrm{numbers}, \mathrm{hence} \mathrm{we} \mathrm{can} \mathrm{say} \\ \mathrm{that}: \\ \mathrm{Real}\left\{\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}+1\right\}>\mathrm{Real}\left\{\frac{{\mathrm{z}}_1}{{\mathrm{z}}_2}-1\right\} \\ \mathrm{Hence} \\ \mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{<}\mathbf{arg}\left(\frac{{\mathbf{z}}_{\mathbf{1}}}{{\mathbf{z}}_{\mathbf{2}}}\right)\mathbf{<}\frac{\mathbf{\pi }}{\mathbf{2}} \end{gathered}$$

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