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I am having trouble understanding arguments in the complex plane. For example, my book talks about branchs and one particular branch they discussed is the principal value of the argument. What exactly does this particular branch tell us about the plane and what exactly do they mean by branch? Also, another branch the book discusses is that any branch of $argz$ must have a jump of $2\pi$ somewhere which takes values on the interval $(0,2\pi]$. What do they mean by that and why is it discontinuous?

They drew a graph, that I don't quite understand, which corresponds to the principal value of the argument:

enter image description here

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I wanted to let you know that my answer indeed contained several mistakes which is why I deleted it (or at least voted to delete it). I'm trying to correct them. But maybe It'll take too long and I won't do it. –  k.stm Jan 23 '13 at 9:41
    
@K.Stm. its okay thanks anyways –  Q.matin Jan 24 '13 at 3:01
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2 Answers

up vote 3 down vote accepted

The $\arg(z)$ is a function which returns the angle that $z = a +ib$, seen as vector $[a ~ b] \in \mathbb{R}^2$, forms with the positive $x$-axis.

Suppose now that $\arg(z) = \theta$. Then, for the properties of the angles, you can say that also $\arg(z) = \theta + 2\pi$, or in general that $\arg(z) = \theta + 2n\pi$, with $n \in \mathbb{Z}$. So there are an infinity of number which represent the angle of a given complex number $z$.

We have an ambiguity now. As a function, $\arg(z)$ should return only a value, not an infinity.

So we have the "branches". We define more than one $\arg()$ functions, namely $$\arg_n(z) : \mathbb{C} \rightarrow (2(n-1)\pi, ~ 2n\pi] ~ \forall n \in \mathbb{Z}$$

Each branch is labeled by a $n$. So in particular, for $n=1$, we have the so-called "principal value"

$$\arg_1(z) : \mathbb{C} \rightarrow (0, ~ 2\pi]$$

(NOTE: in my case we have the principal value for $n=1$ but maybe some book can use the case with $n=0$, but that's just a translation of $n$).

Now, what does happen to complex number which are very close to positive $x$-axis? Let $z_1 = a + ib$ and $z_2 = a - ib$, with $a>0$ and $b>0$ very small. These points are very close ($|z_1 - z_2| = 2b$), so we expect that, if $\arg(z)$ is a continuous function, then $\arg(z_1)$ should be very close to $\arg(z_2)$. But we have the following:

$$ \lim_{b \rightarrow 0^+} \arg(z_1) = 0$$ and $$ \lim_{b \rightarrow 0^+} \arg(z_2) = 2\pi$$

So we have a discontinuity.

The figure you posted is not really well-explained...

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Thanks a lot but i am still having trouble understanding. What does the principal value tell us? Why is it useful? –  Q.matin Jan 24 '13 at 2:59
    
For an example, it deals with multivalued functions. What do they mean by that? –  Q.matin Jan 24 '13 at 3:21
    
@Q.matin The value of the angle (or of the $\arg$) depends by real and imaginary parts of $z$, namely $a$ and $b$. On the plane, you have $a$ on the $x$-axis and $b$ on the $y$-axis. Suppose to deal with $\arg_1(z)$. Then, if $a>0$ and $b>0$, then $\arg_1(z) \in (0, \pi/2)$. If $a<0$ and $b>0$, then $\arg_1(z) \in (\pi/2, \pi)$. If $a<0$ and $b<0$, $\arg_1(z) \in (\pi, 3\pi/2)$ and finally if $a>0$ and $b<0$ then $\arg_1(z) \in (3\pi/2, 2\pi)$. –  the_candyman Jan 24 '13 at 8:46
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Thanks a lot, Candyman!! Very useful! Also, love your username. –  Q.matin Jan 24 '13 at 23:48
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@Q.matin :D :D :D –  the_candyman Jan 25 '13 at 12:57
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Your question is not about complex analysis but about one of the most basic facts of mathematics: That there is no such thing as a "continuous real-valued polar angle" valid in all of $\dot{\mathbb R}^2$. The abstract mathematical expression of this fact can be formulated as follows: The fundamental group of $S^1$ is ${\mathbb Z}$.

Now we have to deal with polar angles every day, and we don't want to mess up our (e.g., trigonometric) computations with clumsy notation for equivalence classes. Therefore when dealing with a problem that "does not go around the origin" we slit the plane along a suitably chosen ray emanating from the origin, e.g., along the ray $L:=\{(x,0)\ |\ x\leq 0\}$. In the slit set $${}^-{\mathbb R}^2:={\mathbb R}^2\setminus L$$ we now can define a "continuous real-valued polar angle" $\phi$. This $\phi$ is called the principal value of the argument in ${}^-{\mathbb R}^2$ and is actually a function of $(x,y)$ (or of $z=x+iy$): $${\rm Arg}:\quad {}^-{\mathbb R}^2\to\ ]{-\pi},\pi[\ ,\qquad (x,y)\mapsto\phi={\rm Arg}(x,y)\ .$$ It is possible to express ${\rm Arg}$ in terms of functions available on a pocket calculator: One has $${\rm Arg}(x,y)=\arctan{y\over x}\qquad(x>0)$$ and $${\rm Arg}(x,y)=2\arctan{y\over x+\sqrt{x^2+y^2}}\qquad\bigl((x,y)\in{}^-{\mathbb R}^2\bigr)\ .$$

If you are working on a problem where you cannot make a slit $L$, e.g., when you are studying a spiral $\sigma$ centered at the origin, you have to invent other means in order to get out of the described difficulty. One way is to consider the polar angle as a set-valued, or multivalued, function. This function is called $\arg$ and is defined as follows: $$\arg:\quad\dot{\mathbb R}^2\to{\mathbb R}/(2\pi{\mathbb Z}),\qquad (x,y)\mapsto\left\{\phi\in{\mathbb R}\ \Biggm|\ \cos\phi={x\over\sqrt{x^2+y^2}},\ \sin\phi= {y\over\sqrt{x^2+y^2}}\right\}\ .$$ In the problem with the spiral $\sigma$ you could work with a real-valued polar angle $\phi$ that goes from $-\infty$ to $\infty$ and is attached to the "running point" on the spiral, not to the points $(x,y)$ lying on the ground.

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Thanks a lot but im still having trouble understanding. Why do we use the principal value, what does it tell us? –  Q.matin Jan 24 '13 at 3:00
    
For an example, it deals with multivalued functions. What do they mean by that? –  Q.matin Jan 24 '13 at 3:22
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