# Argument of complex expr.

1. Jul 26, 2007

### f(x)

The problem statement, all variables and given/known data
find the argument of-:
$$Z=\frac{\iota-\sqrt{3}}{\iota+\sqrt{3}}$$

Method 1

$$Z={2e^{\frac{- \iota \pi}{3}} \mbox{ divided by } {2e^{\frac{\iota \pi}{3}}$$

$$Z=e^{\frac{-2 \iota \pi}{3}}$$

Therefor arg(Z)=$$\ -\frac{2\pi}{3}$$

Method 2

$$Z=\frac{4}{1+2\sqrt{3}\iota-3}$$

$$Z=\frac{2}{-1+\sqrt{3}\iota}$$

$$\mbox{arg}(Z)=0+\frac{\pi}{3} = \frac{\pi}{3}$$

What am i getting wrong ?

Last edited: Jul 26, 2007
2. Jul 26, 2007

### George Jones

Staff Emeritus
Is the middle term in the denominator of the right side of the first line of Method 2 right?

3. Jul 26, 2007

### CompuChip

And where did you get the last step,
$$\arg\left( \frac{2}{-1 + i \sqrt{3}} \right)$$
anyway?

Here's another one: try multiplying top and bottom by the conjugate of the denominator, $i - \sqrt{3}$. Then write the result as $r e^{i \phi}$ and read off $\phi$.
This method always works with fractions.

4. Jul 26, 2007

### Dick

In your first method - note 2*exp(-i*pi/3)=1-i*sqrt(3). That's not the numerator that you want. Nor is the other one the denominator. How are you getting these? (PS put i's in the exponentials).

5. Jul 26, 2007

### f(x)

Thats $$2\times 1\times \iota , \mbox{the 2ab term in }(a+b)^2$$

I've rationalized, by multiplying numerator and denominator with i+sqrt(3) , and then divided by 2 ; although the standard practice is to remove i from denominator as you suggested.

Yeah sry about those missing i's..
I've expressed both as Re^(i.theta) form, (theta=arctan(b/a))
Sorry I am unclear as to what you mean by "not the numerator that you want
". Could you plz explain ? Thx.

PS: I have a feeling that method 2 has a bug, but i cant find it :tongue:

6. Jul 26, 2007

### George Jones

Staff Emeritus
Did you forget a minus sign?

You need to factor out the r before finding the argument.

7. Jul 26, 2007

### Dick

I meant that 2*exp(-i*pi/3)=1-i*sqrt(3) which is what I said. Maybe I didn't explicitly point out that the numerator of your fraction is i-sqrt(3), which is NOT the same thing.

8. Jul 26, 2007

### f(x)

oh sorry , i mixed up the questions...i get it