Understanding the Arg Function: Evaluating, Usage & Properties

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  • #1
wahaj
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We are doing complex numbers, polar coordinates, Euler's formula etc in class and there is this thing called the Argument (Arg) function. I am having trouble understanding what this is. This is a new concept and sadly my notes aren't much help. So could someone kindly answer the following questions.
what is this function?
How do I evaluate it?
Where is it used?
Are there any properties of this function I should know about?
Anything for further reading is also appreciated but keep in mind that this is the first I am seeing all this stuff. No wikipedia please because math on wikipedia doesn't make sense to me with all those complicated equations and symbols that I have never seen before.

Here is a sample question from my assignment just to make sure I am getting my message across properly. Feel free to use it as an example if you want.
z1 = [itex]\pi[/itex]/8
z2 = 3[itex]\pi[/itex]/4
find Arg (z1z2)

Answer is 7[itex]\pi[/itex]/8
which just seems like simple addition so why go through all this trouble with Arg?
 
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  • #2
[tex]\text{Arg}(z)=-\imath \log \left( \dfrac{z}{|{z}|} \right)=2 \arctan \left(\dfrac{\Im{(z)}}{|{z}|+\Re{(x)}}\right)[/tex]

The reason for introducing it is that complex numbers in geometric form are easier to multiply. This is because arg is a logarithm of the unit scaled numer. In particular

Arg(a b)=Arg(a)+Arg(b)
 
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  • #3
in the third expression what are those fancy symbols which look like J and R
 
  • #4
Those are the real and imaginary parts of z.

In your example, I believe it should read
Arg(z1) = [itex]\pi[/itex]/8
Arg(z2) = 3[itex]\pi[/itex]/4
find Arg (z1z2)

There are different ways to represent complex numbers. If we will be adding them the algebraic form is best
z1=a1+i b1
z2=a2+i b2
z1+z2=(a1+a2)+i (b1+b2)

If we are multiplying, the trigonometric (which is related to exponential) form is best
z1=r1(cos θ1+sin θ1)
z2=r2(cos θ2+sin θ2)
z1 z2=(r1 r2)((cos (θ1+θ2)+sin (θ1+θ2)))

notice that if
z=r(cos θ+sin θ)
r=z/|z|
θ=Arg(z)
 
  • #5
My bad about the question. Thanks for help I understand all this now.
 
  • #6
lurflurf said:
[tex]\text{Arg}(z)=\log \left( \dfrac{z}{|{z}|} \right)[/tex]

You missed out a factor of ##-i## there, I think.
 
  • #7
There is no need to memorize all those logarithm functions if you don't want to.

The way I define Arg is just as the angle between a point and the x-axis. So in the following picture:

220px-Complex_number_illustration_modarg.svg.png


the number [itex]\varphi[/itex] would be the argument.

Now, using trigonometric formulas, you can find easy formulas for the argument. For example, let's say that you have the point 1+10i. This corresponds to (1,10). So if we work in the triangle (0,0), (1,0), (1,10), then we can find the argument by

[tex]\tan \varphi= \frac{10}{1}=10[/tex]

So then you can find [itex]\varphi[/itex] easily.

It might be also good to remark that the principal argument is always a number in [itex](-\pi,\pi][/itex].
So if you ever get an argument equal to [itex]3\pi[/itex], then the principal argument is [itex]\pi[/itex] (note that [itex]3\pi[/itex] and [itex]\pi[/itex] correspond to the same angle!).
 
  • #8
^yes thank you.
 
  • #9
lurflurf said:
In particular

Arg(a b)=Arg(a)+Arg(b)

This is not true for general a and b. At least, not if you take the principal argument. If you define argument as a multivalued function, then it is true.
 
  • #10
So basically Arg is just the angle [itex]\varphi[/itex] in rei[itex]\varphi[/itex] given by tan-1 b/a. a and b come from a+bi
 
  • #11
wahaj said:
So basically Arg is just the angle [itex]\varphi[/itex] in rei[itex]\varphi[/itex] given by tan-1 b/a. a and b come from a+bi

Yes, but you need to be careful with that. For example, when calculating the argument of i that way. You would get [itex]\tan^{-1}(\frac{1}{0})[/itex] which is meaningless. On the other hand, [itex]Arg(i)=\frac{pi}{2}[/itex] is easy to see.

Also, you need to take care that [itex]Arg(z)[/itex] is always in the interval [itex](-\pi,\pi][/itex] (or whatever standard you have defined in class). The [itex]\tan^{-1}[/itex] function might not always do the right job for that (depending on how things are defined).

So don't just memorize [itex]\tan^{-1}(\frac{b}{a})[/itex], but rather try to think if your answer makes sense.
 
  • #12
Alright. Thank you all for the help
 

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