How are the complex numbers $z_1$ and $z_2$ related if $\arg(z_1) = \arg(z_2)$?
My attempt :
Let $z_1$,$z_2\in \mathbb{C}$ such that $z_1=r_1(\cos\theta + i\sin\theta)=a+ib$ and $z_2=r_2(\cos\gamma+i\sin\gamma)=c+id$
Then, we have $\arg(z_1)=tan^{-1}(\frac{b}{a})$ and $\arg(z_2)=tan^{-1}(\frac{d}{c})$
As $\arg(z_1)=\arg(z_2)$ then $\tan^{-1}(\frac{b}{a})=\tan^{-1}(\frac{d}{c})\implies \frac{b}{a}=\frac{d}{c}\implies b.c=a.d$
This is correct, i have serious doubt about the interpretation of this. Can someone help me?
Best Answer
It is known that $\arg\left(\frac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)$, so this is equal to zero as $\arg(z_1)=\arg(z_2)$; so as $\frac{z_1}{z_2}$ has argument 0, it is a positive real number, so $z_1=kz_2$ for some positive real $k$.