# Given two complex numbers, how can we prove the given inequality?

If $$|{z_1+z_2}|>|z_1-z_2|$$ then prove that $$-\frac{\pi}{2}

Here is my progress:

squaring on both sides of the given inequality, and using the property of complex numbers:

let z be any complex number then $$|z|^2=z .\bar{z}$$ , where $$\bar{z}$$ denotes the conjugate of z

so, we get:

$$(z_1+z_2)(\bar{z_1}+\bar{z_2})> (z_1-z_2)(\bar{z_1}-\bar{z_2})$$

simplifying,

$$2z_1\bar{z_2}+2z_2\bar{z_1}>0$$
$$\frac{z_1}{\bar{z_2}}>-\frac{z_2}{\bar{z_2}}$$

Using the euler form and assuming argument of $$z_1$$ and $$z_2$$ to be $$\alpha$$ and $$\beta$$ respectively, we get after simplifying:

$$e^{2i\alpha}>-e^{2i\beta}$$
solving further we get:

$$e^{2i(\alpha-\beta)}>e^{-i\pi}$$

Hence,

$$\alpha-\beta>-\frac{\pi}{2}$$

which gives us half of the part of the inequality that we need to prove (as $$\alpha-\beta$$ is the argument of $$\frac{z_1}{z_2}$$) But how would we prove the other part of the inequality?

If $$|z_1+z_2|>|z_1-z_2|$$, then $$\left|\frac{z_1}{z_2}+1\right|>\left|\frac{z_1}{z_2}-1\right|$$. It other words, $$\frac{z_1}{z_2}$$ is closer to $$1$$ than to $$-1$$; so, it can be written as $$\rho\exp(i\theta)$$, with $$\theta\in\left(-\frac\pi2,\frac\pi2\right)$$.
This can be justified analytically as follows: if $$\frac{z_1}{z_2}=a+bi$$, with $$a,b\in\Bbb R$$, then\begin{align}|a+bi+1|>|a+bi-1|&\iff(a+1)^2+b^2>(a-1)^2+b^2\\&\iff2a>-2a\\&\iff a>0.\end{align}So, if $$a+bi=\cos(\theta)+i\sin(\theta)$$, you can always pick $$\theta\in\left(-\frac\pi2,\frac\pi2\right)$$.
If $$|{z_1+z_2}|>|z_1-z_2|$$ then prove that $$-\frac{\pi}{2}
Let's defined the complex numbers $$z_1 = a+b i$$ $$z_2 = p+q i$$ $$|{z_1+z_2}| \gt |z_1-z_2|$$ $$| (a+p) +(b+q)i | \gt | (a-p)+(b-q)i |$$ $$\sqrt{ (a+p)^2+(b+q)^2 } \gt \sqrt{ (a-p)^2+(b-q)^2}$$ $$(a+p)^2+(b+q)^2 \gt (a-p)^2+(b-q)^2$$ $$a^2+2ap+p^2+b^2+2bq+q^2 \gt a^2-2ap+p^2+b^2-2bq+q^2$$ $$2ap +2bq \gt -2ap -2bq$$ See that $$|z_1+z_2| \gt |z_1-z_2|$$ is not a condition, but already a fact
Now for the argument of the complex number $$arg( \frac{z_1}{z_2} ) = arg( \frac{ap+bq}{p^2+q^2} +\frac{bp-aq}{p^2+q^2} \cdot i )$$ $$= \tan^{-1}( \frac{bp-aq}{ap+bq} )$$ We already know the properties of the arctan function and it range is $$-\frac{\pi}{2} \lt \tan^{-1}(x) \lt \frac{\pi}{2}$$