# The Arg function

1. Dec 17, 2012

### wahaj

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 = $\pi$/8
z2 = 3$\pi$/4
find Arg (z1z2)

Answer is 7$\pi$/8
which just seems like simple addition so why go through all this trouble with Arg?

2. Dec 17, 2012

### lurflurf

$$\text{Arg}(z)=-\imath \log \left( \dfrac{z}{|{z}|} \right)=2 \arctan \left(\dfrac{\Im{(z)}}{|{z}|+\Re{(x)}}\right)$$

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)

Last edited: Dec 17, 2012
3. Dec 17, 2012

### wahaj

in the third expression what are those fancy symbols which look like J and R

4. Dec 17, 2012

### lurflurf

Those are the real and imaginary parts of z.

Arg(z1) = $\pi$/8
Arg(z2) = 3$\pi$/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. Dec 17, 2012

### wahaj

My bad about the question. Thanks for help I understand all this now.

6. Dec 17, 2012

### Curious3141

You missed out a factor of $-i$ there, I think.

7. Dec 17, 2012

### micromass

Staff Emeritus
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:

the number $\varphi$ 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

$$\tan \varphi= \frac{10}{1}=10$$

So then you can find $\varphi$ easily.

It might be also good to remark that the principal argument is always a number in $(-\pi,\pi]$.
So if you ever get an argument equal to $3\pi$, then the principal argument is $\pi$ (note that $3\pi$ and $\pi$ correspond to the same angle!).

8. Dec 17, 2012

### lurflurf

^yes thank you.

9. Dec 17, 2012

### micromass

Staff Emeritus
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. Dec 17, 2012

### wahaj

So basically Arg is just the angle $\varphi$ in rei$\varphi$ given by tan-1 b/a. a and b come from a+bi

11. Dec 17, 2012

### micromass

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

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

So don't just memorize $\tan^{-1}(\frac{b}{a})$, but rather try to think if your answer makes sense.

12. Dec 17, 2012

### wahaj

Alright. Thank you all for the help