What is the Argument of a Complex Expression?

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Homework Help Overview

The discussion revolves around finding the argument of a complex expression involving the variable Z, defined in two different methods. The subject area includes complex numbers and their properties, particularly focusing on the argument and representation in exponential form.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore two methods for determining the argument of Z, questioning the correctness of steps taken in both methods. There are inquiries about specific terms in the expressions and suggestions to use the conjugate of the denominator for simplification.

Discussion Status

Several participants are actively questioning assumptions and clarifying steps in the calculations. There is a recognition of potential errors in the methods presented, particularly regarding the terms used in the expressions. Guidance is offered on how to approach the problem using different methods, but no consensus has been reached on the correct approach.

Contextual Notes

Participants note potential mistakes in the calculations and express uncertainty about specific terms in the expressions. There is mention of standard practices in handling complex fractions, indicating a focus on proper mathematical conventions.

f(x)
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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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