Why is Arg(z) considered discontinuous at -1?

[tacit osprey]

I'm confused as to why Arg(z) is discontinuous if rotating 1 by π and -π both result in you being at the same point -1?

[worn willow]

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[tacit osprey]

Why is Arg(z) considered discontinuous at π?

Why is Arg(z) considered discontinuous at -1?

[rustic bane]

what is gamma 1,2

ts thing

can u define it

[rustic bane]

are you familiar with how the prinicpal argument is defined for -pi to pi

below the x axis its negitive above its positive

[tacit osprey]

I just learned it a few days ago so kinda?

[solar spruce]

Well, the principal value arg(z) can have values in (-π, π] by convention. You can also use the convention [0, 2π), or any other semi-closed interval of length 2π, really.

And since rotating by a full circle doesn't change the complex number, we have Arg(z) = arg(z) + 2πn.

[rustic bane]

but i thnkg the question he gave uses the prinicpal argument

u have to stick to it

also the theta in principal argument is as

-pi <theta<= pi

[solar spruce]

Oh, right. Sometimes upper and lowercase are swapped.
I'm more used to lowercase functions meaning principal value and uppercase ones being multivalued.

[rustic bane]

in high school they just use principal argument really, i dont think they even teach the other convneitons

[solar spruce]

Oh, you learn complex numbers in high school? Cool!

[rustic bane]

everyone here does

[solar spruce]

Hm, interesting. Depends on the country, I guess.

[rustic bane]

true

[tacit osprey]

Are pi and -pi different angles?

[rustic bane]

see the argument in ur question can never reach -pi sicne its out of its domain but teh defination of principal argument

[tacit osprey]

Oh

[rustic bane]

it can only approach

[tacit osprey]

Okay

Don’t they both take you to -1 if you rotate pi or -pi

[rustic bane]

well

-pi isnt even in domain of argument again

so u cant ask that

[tacit osprey]

Oh okay

[rustic bane]

but they both approach it yes

[tacit osprey]

Oh and what happens if you were to take sqrt(z) which has Arg(z) in it

And take limit as z->-1

[rustic bane]

if u take sqrt(z)

[tacit osprey]

So the point it approaches is not in the range?

[rustic bane]

then its argument will get halved

z ^1/2 = (|z| e^(i x))1/2

[tacit osprey]

Okay

[rustic bane]

say it agian i did not get what u mean its is not in the range

[tacit osprey]

Like it’s not part of the possible outputs of Arg(z)

[rustic bane]

well if u take sqrt of z then make z approach -1 its arg(sqrt(z)) will approach pi/2

notice i siad arg(sqrt(z)) not arg(z)

[tacit osprey]

Oh

So if you take the limit as z->-1 of Arg(z) it approaches -pi which is not within (-pi,pi]?

[rustic bane]

limit does not exist

LHL = -pi , RHL = pi , and arg(-1) = pi

its discontinius

so it approaches -pi

but not quite gets there

[tacit osprey]

So if you approach from above or below the real axis the limits are different?

[rustic bane]

yes if u use the prinicplal argument convention

[tacit osprey]

Oh okay

So left hand and right not equal so dne

[solar spruce]

Would it help if I showed the graph of the principal value? As in, in the form z = arg(x + iy).

[rustic bane]

if u want it to be continoius u can use least positive argument convention

[rustic bane]

limit does not exist how ever the argumetn has a value

[tacit osprey]

Oh

So what is principle_arg(-1)?

[rustic bane]

pi

[tacit osprey]

okay

What happens if it is arg with any angle

[rustic bane]

lorddarpinger asked smth btw

[rustic bane]

then we convert it into teh principal range i.e make theta b/w -pi and pi

[tacit osprey]

Okay

[rustic bane]

if u have 3pi u subtract 2pi to get into teh range

[tacit osprey]

Is arg(z) continuous?

[rustic bane]

like this

in its domain yes

and we only talk about the domain

so its continious

[tacit osprey]

Oh you can convert any angle to be within the (-pi,pi]

[rustic bane]

yes u do that in tirg too right

[tacit osprey]

Ye

So like you want 2 * pi * n + angle

And take the angle?

[rustic bane]

nooo

[tacit osprey]

No?

[rustic bane]

whys ther i

oh pi

no still

[tacit osprey]

Oh then how to get the angle

[rustic bane]

u gotta think

like 4/3 pi

[tacit osprey]

Ye

[rustic bane]

where do u think it lies

3rd quadrant

so it will be (1-4/3)pi

[tacit osprey]

Oh

[rustic bane]

u gotta make sure abt argument's sign and all

[solar spruce]

Here you go. Ignore the almost-vertical part, that's just an artifact.
You can see that arg(x + iy) is discontinuous on the ray {x ≤ 0, y = 0}.

[tacit osprey]

Oh ye I see

How would you convert to within the right range then

[solar spruce]

Well, suppose z = a + bi. Then if r = |z| = √(a^2 + b^2), then θ = arg(z) satisfies:
cos(θ) = a/r
sin(θ) = b/r
Usually you'd do it like this:

1. Look at the signs of a and b. Those correspond to the signs of cos(θ) and sin(θ), so you know which quarter the argument lies in.
2. Solve tan(θ) = b/a in the respective quarter.

Of course, you need to mind two things:

1. If a = b = 0, then the argument is undefined.
2. If a = 0, then tan(θ) is undefined, so you just look at the sign of b and deduce whether the argument is π/2 or -π/2.

[tacit osprey]

Oh okay

Thanks

🙂

Why is it undefined when a=0 and b=0

[solar spruce]

You can rephrase this question the following way: what direction does a vector with length equal to zero point to?

[tacit osprey]

No direction?

[solar spruce]

Yup.

[tacit osprey]

Oh so it’s undefined cause there is no direction

[solar spruce]

Yeah.

[tacit osprey]

Is there formula for Arg(z) that works for all points within the principle range

Or do some a and b values result in the formula not working like a=0

[solar spruce]

Not really. Depending on where the complex number lies, it could be arctan(b/a), arctan(b/a) - π, arctan(b/a) + π, -π/2, π/2 or undefined.

[tacit osprey]

Oh okay

[solar spruce]

Usually it's calculated using the algorithm I described above, with some slight variations, maybe.

[tacit osprey]

So like determine which quadrant then get the angle and if no angle look at b

[solar spruce]

Well, I mean, the angle either lies in a quarter, on a ray or nowhere.

[tacit osprey]

What do you mean by ray or nowhere

[solar spruce]

Well, rays lie between quadrants, after all.

And if z = 0, then the argument is just undefined.

[tacit osprey]

Oh okay makes sense

So do the 2 steps to get the argument?

And if one or both are 0 perform the other 2 steps

[solar spruce]

Yeah. Though, if at least one of a and b is zero, then it's pretty easy to see what the argument is, anyway.

[tacit osprey]

Oh okay

Thanks you two 🙂

[solar spruce]

You're welcome!

[junior wadi]

@tacit osprey

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[tacit osprey]

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[junior wadi]

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