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


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?

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2 Answers 2

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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 $\theta\in\left(-\frac\pi2,\frac\pi2\right)$.

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


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 $$-\frac{\pi}{2} \lt \tan^{-1}(x) \lt \frac{\pi}{2}$$

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