Clarification on finding argument of complex number

[chwala]

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.

 

[Mark44]

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.

 

[pasmith]

It depends on the branch. Here the book appears to be assuming -\pi < \arg z \leq \pi, which is the standard branch.

 

[Steve4Physics]

$\alpha= \tan^{-1} 0.24 = 0.2449$

I know I'm just being picky but the above should be:
$\alpha= \tan^{-1} 0.25 = 0.2450$

 

[SammyS]

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:

$\displaystyle \quad\quad \tan \alpha= 0.25$

 

[Steve4Physics]

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!

 

[SammyS]

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.

 

[chwala]

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.

 

[Gavran]

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.

 

[Steve4Physics]

@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).

 

[chwala]

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.