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:

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



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

solving further we get:




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?


2 Answers 2


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 $-\frac{\pi}{2}<arg\frac{z_1}{z_2}<\frac{\pi}{2}$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.