Clarification on finding argument of complex number

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Homework Statement
See attached.
Relevant Equations
Complex numbers
I am going through this,quite straightforward ...just some bit of clarity from your end; the textbook indicates, the argument as ##-2.90## i have no problem with that as tan cycles at every ##\pi## radians. The problem is that i am not used to this approach.
My approach is more straightforward i.e

##\tan \alpha= 0.25##
##\alpha= \tan^{-1} 0.24 = 0.2449##
##\alpha_{required angle}= \pi + 0.2449=3.386=3.39 ##(2 dp) My question is' does it matter or there is a systematic approach to this; are both approaches valid. Of course, i can see that going backwards ( in terms of cycle) ##[3.386-\pi]=[0.244-\pi]=-2.8971##
Thanks.


or to answer my own question, it depends on the domain given.

1746310787234.webp
 
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Angles are usually measured counterclockwise from the positive real axis, so I would favor your answer, which is ##\pi + \alpha \approx 3.39##.

Of course, if the goal was to calculate the angle indicated by O in the drawing, then I would favor the answer they showed.
 
It depends on the branch. Here the book appears to be assuming -\pi < \arg z \leq \pi, which is the standard branch.
 
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chwala said:
##\alpha= \tan^{-1} 0.24 = 0.2449##
I know I'm just being picky but the above should be:
##\alpha= \tan^{-1} 0.25 = 0.2450##
 
Steve4Physics said:
I know I'm just being picky but the above should be:
##\alpha= \tan^{-1} 0.25 = 0.2450##
It looks like that was simply a typo. The line preceding that was:

##\displaystyle \quad\quad \tan \alpha= 0.25 ##
 
SammyS said:
It looks like that was simply a typo. The line preceding that was:

##\displaystyle \quad\quad \tan \alpha= 0.25 ##
I was being too picky. It’s just that there was a typo' and a rounding error in the same line. In retrospect, shouldn't have said anything!
 
Steve4Physics said:
I was being too picky. It’s just that there was a typo' and a rounding error in the same line. In retrospect, shouldn't have said anything!
You warned that you were being picky. I don't think you were being too picky.
I didn't notice that you did also notice the round-off error. That was worth mentioning.
... Maybe I'm being too picky? . . . No, not really.
 
Steve4Physics said:
I was being too picky. It’s just that there was a typo' and a rounding error in the same line. In retrospect, shouldn't have said anything!
@Steve4Physics that's a typo.
 
The standard branch −π<arg⁡z≤π is a principal branch of arg⁡z.
The solutions of arg⁡z, for which the arg⁡z value lies between −π<arg⁡z≤π, are called the principal values of arg⁡z.
 
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chwala said:
@Steve4Physics that's a typo.
Yes. I wouldn't have posted for only a typo'. But just in case you haven’t yet realised…

##\tan^{-1} 0.25 = 0.24497866...##

Rounded to four decimal places, this is ##0.2450##, not ##0.2449## (which is what you wrote in Post #1).
 
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Steve4Physics said:
Yes. I wouldn't have posted for only a typo'. But just in case you haven’t yet realised…

##\tan^{-1} 0.25 = 0.24497866...##

Rounded to four decimal places, this is ##0.2450##, not ##0.2449## (which is what you wrote in Post #1).
correct, in this post my focus was solely on how to determine the argument value... cheers.
 
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