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How to prove $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$ for $ z_1, z_2 \neq 0$ ?

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  1. If you know about the polar representation of complex numbers, you can write $z_1 = r_1 e^{i \theta_1}$, $z_2 = r_2 e^{i \theta_2}$ where $r_j = |z_j|$, $\theta_j = \arg(z_j)$ for $j=1,2$. Then $z_1/z_2 = (r_1/r_2) e^{i(\theta_1-\theta_2)}$, so $\theta_1-\theta_2$ is an argument for $z_1/z_2$.

  2. If you don't know yet about polar representation, you can use trigonometric identities. Say $z_j = x_j + i y_j$. Then $$ \frac{z_1}{z_2} = \frac{(x_1+iy_1)(x_2-iy_2)}{x_2^2+y_2^2}=\frac{(x_1x_2+y_1y_2)+i(y_1x_2-x_1y_2)}{x_2^2+y_2^2}. $$ Write $x_j = r_j \cos \theta_j$, $y_j = r_j \sin \theta_j$ for $j=1,2$. After pulling out some common factors of $r_1 r_2$, you will see some terms like $\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2$, which by a trigonometric identity is equal to $\cos(\theta_1-\theta_2)$; and $\sin \theta_1 \cos \theta_2 - \cos \theta_1 \sin \theta_2$, which by another identity is equal to $\sin(\theta_1-\theta_2)$. From this you can conclude again that $\theta_1-\theta_2$ is an argument for $z_1/z_2$ (well, after you have filled in some steps that I skipped, including taking care of $r_j$ factors).

  3. Finally, if you really want to prove that $\arg(z_1/z_2) = \arg(z_1) - \arg(z_2)$, then, then please be careful! For example if you follow the convention that arguments should be single values in the interval $[0,2\pi)$, then the statement you want to prove is false! If $\arg(z_1) < \arg(z_2)$, then you will have $\arg(z_1/z_2) = \arg(z_1)-\arg(z_2)+2\pi$.

It is okay if you think of argument as a multivalued function (then the statement you want to prove is an equality of sets; it is true, but you might want to think carefully about what it actually means, including what the subtraction $\arg z_1 - \arg z_2$ means). [Thanks for the comment below that brought this possible interpretation to my attention, I overlooked it before.]

So please work with a more careful formulation of the statement. Then you should be able to find a proof using either one of the ideas suggested above.

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    $\begingroup$ If $\arg(z)$ denotes the muli-valued argument of $z$, then $\arg(z_1/z_2)=\arg(z_1)-\arg(z_2)$. The equality is interpreted as a set equality. $\endgroup$ – Mark Viola Jul 22 '17 at 4:13
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This is equivalent to $arg(z_1) + arg(z_2) = arg(z_1 z_2)$. Write $z_1 = r_1 e^{i arg(z_1)}$ and $z_2 = r_2 e^{i arg(z_2)}$.

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