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For example:

The locus of satisfying $\arg(z)=\frac{\pi}{3}$ is ?

My Attempt: $$ z =x+iy = |z| \left[ \cos \left( \frac{\pi}{3} \right) + \sin \left( \frac{\pi}{3} \right) \right] = |z| \left[ \frac{1}{2} +i\frac{\sqrt{3}}{2} \right] \\ \implies \tan \left( \frac{\pi}{3} \right) = \sqrt{3}= \frac{y}{x} \implies y= \sqrt{3}x \\ |z|\ge 0 \implies x \ge 0 \quad \& \quad y \ge 0 \\ x=0 \quad \& \quad y=0 \implies \frac{y}{x}=\frac{0}{0} \ne \sqrt{3} \quad \text{($0/0$ is undefined)} $$ So the locus is a straight line $y=\sqrt{3}x$ in the first quadrant, except at origin.

If $\arg(z)=0$ then $$ z =x+iy= |z|[ \cos0+i\sin0 ]= |z|+i0 \\ \implies \tan0= \frac{0}{|z|}= 0= \frac{y}{x} \implies y=0 \\ |z| \ge 0 \implies x \ge 0 \ \ x=0 \quad \& \quad y=0 \implies \frac{y}{x} = \frac{0}{0} = \frac{0}{|z|} $$ Here the locus is $+x$ axis including the origin. For any value of $\arg(z) \ne 0$, $z$ is not defined at origin.

Does this mean that a complex number $z$ can be defined at origin only if $\arg(z)=0$, i.e. $\arg(0)=0$?

Is my approach correct ?

UmbQbify
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Sooraj S
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    The argument for $0+0i$ is undefined. – Kenny Lau May 12 '17 at 13:45
  • However, a complex number $z$ can be at the origin. – Kenny Lau May 12 '17 at 13:46
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    There's nothing wrong with the definition of complex numbers at the origin. A complex number $z=x+iy$ is defined for any real values of $x$ and $y$, and this includes $x=y=0$. The problem is with the definition of the arg function, which is not a well-defined function on the entire complex plane. $\text{arg}(z)$ only becomes well-defined if you put restrictions on its domain. One of the restrictions is that $\text{arg}(0)$ is undefined, so $0$ cannot be in the domain of the arg function. – Lee Mosher May 12 '17 at 13:46
  • @LeeMosher are you suggesting that $z$ is defined at origin in the example i have specified above ? since $y=0=\sqrt{3}x=\sqrt{3}*0=0$ – Sooraj S May 12 '17 at 13:57
  • @LeeMosher I hv just edited my original post. hope my doubt is clearly mentioned. – Sooraj S May 12 '17 at 14:16
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    In the real numbers we already have the matter of the sign of $0$, which allows for writing $0=+0=-0$. This is also $0=0e^{i\pi}=0e^{-i\pi}$. But complex arguments are not limited to $\pm\pi$, so we can say $0=0e^{i\phi}$, no matter the $\phi$. In particular, $0e^\frac{i\pi}{3}=0$. – Jaume Oliver Lafont May 12 '17 at 14:28
  • @JaumeOliverLafont thanx. so is it wrong to say that locus of $z$ is a straight line y=\sqrt{3}x in the first quadrant, except at origin ?, since $z$ is also defined at origin. – Sooraj S May 12 '17 at 15:22
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    Hmm... WolframAlpha seems to exclude 0 when $arg(z)=\frac{\pi}{3}$ http://www.wolframalpha.com/input/?i=arg(z)%3Dpi%2F3. but it includes 0 when $arg(z)=0$ http://www.wolframalpha.com/input/?i=arg(z)%3D0 Is that consistent? – Jaume Oliver Lafont May 12 '17 at 16:34
  • There is something tricky here... Forget about complex numbers for a while and check on WA what $sgn(0)$ is and the locus of $sgn(x)>0$ against $sgn(x)>=0$, which have the same result but it should be different according to the result of $sgn(0)$, shouldn't it? – Jaume Oliver Lafont May 12 '17 at 16:50
  • @JaumeOliverLafont I think as per WolframAlpha, for $\arg(z)\neq{0}$ excludes $0$, so it is a straight line $y=\sqrt{3}x$ in the first quadrant except at origin for $\arg(z)=\pi/3$. And if $arg(z)=0$ then it seems to include $0$. s that correct ?. Does this means that $z=0+0i$ is true only for $\arg(0)=0$ or it is undefined at $z=0$? is it due to my last argument in my original post ? – Sooraj S May 12 '17 at 18:10
  • @LeeMosher can't we say that $arg(0)=0$, since $z$ is undefined at the origin for any other value of $arg(z)$ ? – Sooraj S May 13 '17 at 01:54
  • Defining $\text{arg}(0)=0$ would guarantee that it is discontinuous unless the domain consists of only the non-negative $x$-axis. Which would make it a pretty useless definition. But yeah, you can define anything you like. The trick is making definitions that are useful. – Lee Mosher May 13 '17 at 13:52
  • @LeeMosher my comment was in regard to what WolframAlpha seems to provide as suggested by Jaume Oliver Lafont – Sooraj S May 13 '17 at 14:48
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    I would not use WolframAlpha as my guide to defining the arg function. – Lee Mosher May 13 '17 at 22:18

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