# Question about the range of a complex function

Define $f:\mathbb{\mathbb{C}}^{2}\rightarrow\mathbb{R}$ by

$$f(z_{1},z_{2})=\text{Arg}(z_{1})+\text{Arg}(z_{2})-\text{Arg}(z_{1}z_{2})\;,$$

where $\text{Arg}$ denotes the principal argument. What is the range of the function $f$? (The answer is $\{-2\pi,0,2\pi\}$.)

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Where are you stuck? –  Stefan Walter May 30 '11 at 13:41
@amWhy: I undid your edit, since Arg was correct, both in being in roman font and in being uppercase -- at least according to en.wikipedia.org/wiki/Argument_(complex_analysis). –  joriki May 30 '11 at 13:53
@joriki, no problem, sorry for that! –  amWhy May 30 '11 at 16:01

$$z_1=r_1\mathrm e^{\mathrm i\phi_1}\;,$$ $$z_2=r_2\mathrm e^{\mathrm i\phi_2}\;,$$
where $r_1$ and $r_2$ are the magnitudes and $\phi_1$ and $\phi_2$ are the arguments of $z_1$ and $z_2$, respectively, the multiplication takes the simple form
$$z_1z_2=r_1r_2\mathrm e^{\mathrm i(\phi_1+\phi_2)}\;.$$
Now you can read off the argument of $z_1z_2$ from the exponential, but you have to take into account that this might wrap around. The usual choice for the range of the argument is $(-\pi,\pi]$ (this is what's called the "principal argument"), so $\phi_1+\phi_2\in(2\pi,2\pi]$. Thus you may have to add $2\pi$ or subtract $2\pi$ to get this back into the range $(-\pi,\pi]$, so the difference between the argument of $z_1z_2$ and the sum $\phi_1+\phi_2$ of the individual arguments can be $-2\pi$, $0$ or $2\pi$.