Is this a complex number at the second quadrant?

[mcastillo356]

TL;DR Summary

I am quite sure, but I don't manage with Geogebra. It involves the value of the ##\arg(w)## in the interval ##-\pi<\theta\leq{\pi}##, called the principal argument of ##w## and denote it ##\mbox{Arg(w)}##

Hi, PF, so long, I have a naive question: is ##\pi+\arctan{(2)}## a complex number at the second quadrant? To define a single-valued function, the principal argument of ##w## (denoted ##\mbox{Arg (w)}## is unique. This is because it is sometimes convenient to restric ##\theta=\arg{(w)}## to an interval of length ##2\pi##, say the interval ##0\leq{\theta}<2\pi##, or ##-\pi<\theta\leq{\pi}##. This last one is which I am concerned with.
PS: I post without preview
Regards!

 

[pbuk]

Hi, PF, so long, I have a naive question: is ##\pi+\arctan{(2)}## a complex number at the second quadrant?

No, it is a real number about equal to 4.25.

 

[mcastillo356]

Hi, PF

If ##a<0## and ##b>0##, are we at the second quadrant, therefore ##\arg{(a+bi)}=\pi+\arctan{(b/a)}##?.
If so, ##\pi+\arctan{(2)}## should fall at the second one. I'm quite sure of it. But then, which is the complex number that has ##\pi+\arctan{(2)}## as the principal value of the argument?

My attempt: ##w=-2+i##. But here comes the problem: ##\displaystyle\frac{1}{-2}\neq{2}##, so the statement ##\arg{(a+bi)}=\pi+\arctan{(b/a)}## must be false. The question is that is a true statement. The solution is, in my personal opinion, that the moduli ##|-2+i|=\sqrt{5}##, and the ##\mbox{Arg}## ought to be, ##\pi+\arctan{-2}##.

Regards!

 

[pbuk]

If ##a<0## and ##b>0##, are we at the second quadrant, therefore ##\arg{(a+bi)}=\pi+\arctan{(b/a)}##?

Yes.

If so, ##\pi+\arctan{(2)}## should fall at the second one. I'm quite sure of it.

No. If ##a<0## and ##b>0## then ## \frac b a \ne 2 ##.

But then, which is the complex number that has ##\pi+\arctan{(2)}## as the principal value of the argument?

As already discussed ##\pi+\arctan{(2)} \approx 4.25##. Is this between ## -\pi \text{ and } \pi ##?

My attempt: ##w=-2+i##. But here comes the problem: ##\displaystyle\frac{1}{-2}\neq{2}##, so the statement ##\arg{(a+bi)}=\pi+\arctan{(b/a)}## must be false. The question is that is a true statement. The solution is, in my personal opinion, that the moduli ##|-2+i|=\sqrt{5}##, and the ##\mbox{Arg}## ought to be, ##\pi+\arctan{-2}##.

I don't really understand any of that, but ## \arg(-2 + i) = \pi - \arctan \frac 1 2 ##.

 

[mcastillo356]

No. If ##a<0## and ##b>0## then ## \frac b a \ne 2 ##.

Oops!

As already discussed ##\pi+\arctan{(2)} \approx 4.25##. Is this between ## -\pi \text{ and } \pi ##?

Not at all

I don't really understand any of that, but ## \arg(-2 + i) = \pi - \arctan \frac 1 2 ##.

Need to revisit my background in complex numbers. Thanks for your attention, hope we will keep in touch. Regards!

 

[mcastillo356]

I don't really understand any of that, but ## \arg(-2 + i) = \pi - \arctan \frac 1 2 ##.

Understood. Let's see if my drawing agrees with your quote. ##\pi-\arctan\frac 1 2 ##, could be ##\mbox{Arg}## of ##w=(-2+i)##, ##\pi## counterclockwise, and ##\arctan \frac 1 2## clockwise?

I don`t manage with ##\pi+\arctan(-2)##. I've tried with a Casio fx-82MS, pretending to switch polar to rectangular coordinates: I won't bore the forums with my effort; I just declare that my attempt is ##\pi+\arctan(-2)=-0.1\mbox{radians}##.
I post with no preview
Regards!
PS: Edited. Reason: Mistake when describing the Geogebra file.

 

[mcastillo356]

I don't really understand any of that,

Neither me; how many days have I taken?

but ## \arg(-2 + i) = \pi - \arctan \frac 1 2 ##.

Thank you very much!
Greetings.

 

[mcastillo356]

Hi, PF, write to say how misled was I when saying that the ##\mbox{Arg}## corresponding to ##w=(-2 +i)## was ##\pi + \arctan(-2)=-60.29\;\mbox{degrees}##. Actually placing ##w## at the fourth quadrant