Does $\arg(z_1-z_2)=0$ imply $|z_1-z_2|=||z_1|-|z_2||$

Question

If for $z_1,z_2\in\mathbb{C}$, $\arg(z_1-z_2)=0$ then prove that $|z_1-z_2|=||z_1|-|z_2||$

If it was for finding the condition for $|z_1+z_2|=|z_1|+|z_2|$ I'd easily find both $z_1$ and $z_2$ lie on the same direction, ie. $\arg z_1=\arg z_2\implies z_2=kz_1, k\in\mathcal{R}$. How do I find this one ?

My Attempt $$ z_1-z_2=|z_1-z_2|\\ |z_1-z_2|^2=|z_1|^2+|z_2|^2-2\Re(z_1\bar{z}_2)\\ \Big(|z_1|-|z_2|\Big)^2=|z_1|^2+|z_2|^2-2|z_1||z_2|\\ \Re(z_1\bar{z}_2)=|z_1||z_2|\implies Re\big(|z|e^{\alpha}.|w|e^{-\beta}\big)=Re\big(|z||w|e^{i(\alpha-\beta)}\big)=|z||w|\\ |z||w|\cos(\alpha-\beta)=|z||w|\implies\cos(\alpha-\beta)=1\\ \implies\alpha-\beta=0\implies\arg z=\arg w\implies z=nw $$ But how does $\arg(z_1-z_2)=0\implies\arg(z_1)=\arg(z_2)$ ?

Answer 1

What the OP has set out to prove simply isn't true. Consider, for example, the values $z_1=1+i$ and $z_2=i$. Then $z_1-z_2=1$, so $\arg(z_1-z_2)=\arg(1)=0$, but $|z_1-z_2|=1$ while $||z_1|-|z_2||=|\sqrt2-1|\not=1$.

The fact is, "arg" is not linear. In particular, $\arg(z_1-z_2)=0$ does not imply that $\arg(z_1)=\arg(z_2)$.

Answer 2

Let $z_1=a+ib,z_2=c+id$

We have $$(a-c)^2+(b-d)^2=a^2+b^2+c^2+d^2-2\sqrt{(a^2+b^2)(c^2+d^2)}$$

Squaring both sides $$(ca+bd)^2=(a^2+b^2)(c^2+d^2)$$

By Brahmagupta-Fibonacci Identity, $$(a^2+b^2)(c^2+d^2)=(ca+bd)^2+(ad-bc)^2$$

So,we have $ad=bc\iff \dfrac ba=\dfrac dc$

Answer 3

$\arg(z_1-z_2) = 0$ means a ray starting from $z_1$, and making angle $0$ with $x$ axis, so both point on same side of origin: now, from here the proof geometrically follows.