What is the Argument of a Complex Expression?

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The discussion focuses on finding the argument of the complex expression Z = (i - √3) / (i + √3) using two methods. Method 1 leads to an argument of -2π/3, while Method 2 results in π/3, raising questions about the accuracy of the calculations. Participants highlight potential errors in the numerator and denominator, particularly regarding the placement of the imaginary unit 'i' and the need for rationalization. Suggestions include using the conjugate of the denominator to simplify the expression and ensure correct argument extraction. The conversation emphasizes the importance of careful manipulation of complex numbers to avoid mistakes in finding arguments.
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Homework Statement
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 ?
 
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Is the middle term in the denominator of the right side of the first line of Method 2 right?
 
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.
 
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).
 
George Jones said:
Is the middle term in the denominator of the right side of the first line of Method 2 right?

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

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

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.

Dick said:
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).

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 please explain ? Thx.

PS: I have a feeling that method 2 has a bug, but i can't find it :-p
 
f(x) said:
Thats 2\times 1\times \iota , \mbox{the 2ab term in }(a+b)^2

Did you forget a minus sign?

Sorry I am unclear as to what you mean by "not the numerator that you want
". Could you please explain ? Thx.

You need to factor out the r before finding the argument.
 
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.
 
oh sorry , i mixed up the questions...i get it
Thank you and sorry about this :blushing:
 
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