# $|z_1+z_2|>|z_1-z_2|$ implies $-\frac{\pi}{2}<arg\big(\frac{z_1}{z_2}\big)<\frac{\pi}{2}$

For two complex numbers $z_1$ and $z_2$, it is given that:

$$|z_1+z_2|>|z_1-z_2|$$

How could we prove that $-\frac{\pi}{2}<arg\big(\frac{z_1}{z_2}\big)<\frac{\pi}{2}$

If I take $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ I get $x_1x_2+i y_1y_2=0$ but it does help me in proving $-\frac{\pi}{2}<arg\big(\frac{z_1}{z_2}\big)<\frac{\pi}{2}$. How should I proceed?

• The title question is different from the question below. – Dietrich Burde Apr 7 '16 at 13:55

factor $z_2$ in each side and get: $$|z_1+z_2|>|z_1-z_2|=|z_2(\frac{z_1}{z_2}+1)|>|z_2(\frac{z_1}{z_2}-1)|=|z_2|\frac{z_1}{z_2}+1|>|z_2||\frac{z_1}{z_2}+1|=|\frac{z_1}{z_2}+1|>|\frac{z_1}{z_2}-1|$$ this statement means that $|\text{Real}\{\frac{z_1}{z_2}+1\}|>|\text{Real}\{\frac{z_1}{z_2}-1\}|$ which means $-\frac{\pi}{2}<arg\big(\frac{z_1}{z_2}\big)<\frac{\pi}{2}$
because if you take these slack variables: $$\text{Real}\{\frac{z_1}{z_2}+1\}=a_r\;,\;\text{Imag}\{\frac{z_1}{z_2}+1\}=a_i \;,\; \text{Real}\{\frac{z_1}{z_2}-1\}=b_r\;,\;\text{Imag}\{\frac{z_1}{z_2}-1\}=b_i$$ note that $a_i=b_i$ .then you can write the first expression like bellow: $$|\frac{z_1}{z_2}+1|>|\frac{z_1}{z_2}-1|\Rightarrow \sqrt{a_r^2+a_i^2}>\sqrt{b_r^2+b_i^2} \Rightarrow |a_r|>|b_r|$$ also for concluding the last part that $-\frac{\pi}{2}<arg\big(\frac{z_1}{z_2}\big)<\frac{\pi}{2}$ we have to solve for: $$|\text{Real}\{\frac{z_1}{z_2}+1\}|>|\text{Real}\{\frac{z_1}{z_2}-1\}|$$ which means: $$|\text{Real}\{\frac{z_1}{z_2}\}+1|>|\text{Real}\{\frac{z_1}{z_2}\}-1|$$ which can be solved easily with conditioning on $\text{Real}\{\frac{z_1}{z_2}\} \in \mathbb R$ by considering $\text{Real}\{\frac{z_1}{z_2}\}=r$ \begin{cases} r \ge 1 \; \quad \qquad \Rightarrow |r+1|>|r-1| \Rightarrow r+1>r-1 \quad \text{always true}\\ -1 \le r \le +1 \Rightarrow |r+1|>|r-1| \Rightarrow r+1>1-r \Rightarrow r > 0\\ r \le -1 \; \; \qquad \Rightarrow |r+1|>|r-1| \Rightarrow -r-1>1-r \quad \text{always false} \end{cases} so we see that only $\text{Real}\{\frac{z_1}{z_2}\} > 0$ satisfies inequality.
• How did you conclude final result after you got $\text{Real}\{\frac{z_1}{z_2}+1\}>\text{Real}\{\frac{z_1}{z_2}-1\}$ – Ananya Apr 7 '16 at 15:44
Hint 1: $|z_1 + z_2| > |z_1 - z_2|$ means that $z_1$ and $z_2$ are closer to one another in the complex plane than $z_1$ and $-z_2$ are.
Hint 2: What is the geometric significance of $\operatorname{arg}(z_1/z_2)$?
Hint we can write $arg(z_1/z_2)=arg(z_1)-arg(z_2)=\tan^{-1}(y_1/x_1)-\tan^{-1}(y_2/x_2)$ so now by inverse trigo we have $arg(z_1)-arg(z_2)=\tan^{-1}(\frac{x-y}{1+xy})$ so now we know that range of $\tan^{-1}(l)$ is $(\frac{-\pi}{2},\frac{\pi}{2})$ also note that here $xy>-1$ you can now prove it for other cases also.