# What is the interval for possible values of the argument of a complex number?

It looks like there are different intervals in which the argument of a complex number can be.

1. Some say it goes from $-\pi$ to $+\pi$
2. others say it goes from $0$ to $2\pi$.

For the most part, both ways look compatible to each other.

However, if one states that $arg z \lt \pi$, the result appears to be different depending on what the interval of possible values looks like.For the first interval, that means all possible values, but for the second one only half of them.

The above statement is just a statement (my tautology club member number is my tautology club member number). It will hold true for some $z$ but not for others. It will also hold true for the same $z$ given the first definition, but won't hold true when using the other definition, even if it's the same number.

The numbers for which the statement holds true are different depending on what definition is used. I think this is a problem because it should be clear what numbers it's true for.

Edit: I'm not sure if I can deal with the deamons that I summoned. I added the complex-analysis tag, as this is apparently what I'm doing here. It was pointed out to me that this aesthetically pleasing for math majors.

I'm not a math major. I couldn't even order a pizza in analysis. Please keep it simple.

• a tautology club...how interesting! – hypergeometric Aug 21 '15 at 16:36
• @hypergeometric I've been told that it was not clear how I arrived at the statement, so I added the phrase to clarify that it is not derived from anything and is in fact just a statement of its own. As a sentence on its own however, this sentence doesn't add any information, I added the remark to clarify that I'm aware of the tautology it represents in an attempt to prevent users from spending time telling me about it. ref – null Aug 21 '15 at 16:49
• it was meant as a positive comment as that was interesting illustration :) – hypergeometric Aug 21 '15 at 20:59
• @hypergeometric I felt no offence by your comment and understood it as a positive one from the beginning. All good, no worries. thanks for stopping by =) – null Aug 21 '15 at 21:04

Welcome to complex analysis! I hope you stay around. It's quite a nice subject, and I think it's the most aesthetically pleasing undergraduate course offered to math majors.

You're not limited to those values. I can say the argument of $-i$ is $-\pi/2$, $3\pi/2$, $7\pi/2$, etc. So what's the answer? Well, I just choose whichever one I want and say, pick the other arguments that make my $Arg$ function continuous. But even then there's still another problem. In your example of choose $$0,2\pi\), our argument function won't be continuous for positive real numbers, if we choose \(-\pi,\pi$$ then argument function won't be continuous at any negative real number.

Each argument function you create by specifying a domain and range is a "branch" of the argument function. This choice occurs because argument is a "multi-valued" function (there are many reasonable answers as to what the argument of a complex number is).

There is a way around this problem for a lot of cases. Instead of thinking of the argument function as living on the complex-plane, we can create a complex manifold (a space with a weird shape the looks smooth) and define an argument function on that shape that is continuous everywhere and suddenly there's no more problem. In some sense this weird shape is where the argument function "naturally" lives.Here's the actual shape .

It's kind of like a spiral staircase, and whether the argument of $i$ is $3\pi/2$ or $-\pi/2$ is determined by what "floor" or "level" you're looking at.

This area isn't my strong suit and I haven't studied it in awhile. So I might be conflating terminology, and I'm sure there's someone who can explain this better.

• I'm afraid I don't understand how that solves the problem. From your answer I take away that there are infinitely many intervals, which isn't exactly helpful if "having more than one interval" is the problem. I don't see how having "many reasonable answers as to what the argument is" works with comparisons like <, because the value it is compared to is a constant value which doesn't change no matter how much you run around on the rainbow staircase. – null Aug 21 '15 at 17:51
• The interval is $(\infty,\infty)$. Say you start at the red line that's on the yellow part then Arg(1) = 0. Wind yourself up the stair case, Arg grows. Make a full revolution, you're on the red line in the blue shaded region directly above where you started. Arg there is $2\pi$. Go down the staircase until you reached a point directly below you started. Arg there would be $-2\pi$ The staircase is a space where Arg function can live, has a single value at any point, takes all values, and is continuous everywhere. The bad things you found when looking at Arg on the complex plane vanish – maxbaroi Aug 21 '15 at 19:51
• But it only has a single value within the interval between the two red lines. If another interval is picked, the value will be different (being further up or down the rainbow staircase). This looks like an overcomplicated way of saying what I stated in my question, but it doesn't solve anything. – null Aug 21 '15 at 20:11
• Okay, the answer to your original question is YOU decide the interval. The choice is yours and is completely arbitrary. The staircase is a shape where we can actually have an argument function that takes on all possible values, has a well defined value at any point on the staircase, and continuous everywhere. So in some sense, the argument function naturally lives on this staircase curve and not on the complex plane. – maxbaroi Aug 21 '15 at 20:16
• That means the set of numbers for which the statement $arg(z) \lt \pi$ holds true is not clearly specified, because it depends on in which floor of the magic mushroom factory I am, that is, to which height I climbed on the rainbow staircase. Thank you for the rather psychedelic insights. – null Aug 21 '15 at 20:40

I don't see how you arrive at that last statement. Of course, taking $\theta$ between $0$ and $\pi$ would certainly give only "half" the results of $0$ to $2\pi$. But the values for $-\pi$ to $0$ give the other half.

• This answer looks more like a comment. I edited my question to clarify the problem. – null Aug 21 '15 at 16:32