What is the arg() function in complex analysis?

[Peter VDD]

What is the arg() function? I can find no reference to it?

exp(z)=w => z=ln(|w|)+i*arg(w)+2*k*Pi*i

what's that arg()?

 

[dextercioby]

A complex # is characterized through modulus & argument.

z=\left|z\right| e^{i\varphi}

That \varphi is called argument.

The same real number appears as the argument (sic!) of the "sine" & "cosine",if u use Euler's formula in the exponential form written above.

Daniel.

 

[dextercioby]

And there's one more thing:

where does that 2\pi i k,k\in \mathbb{Z} come from...?Euler's formula explains it.It's called "multivaluedness" of the complex exponential (hence of the complex logarithm).

Daniel.

 

[Peter VDD]

Yes, I suspected something like that yet :) but the term is described nowhere in our course. {or I still have to find it}

Thx.

 

[Manchot]

So, basically, arg(z) = arccos(Re(z))?

 

[dextercioby]

Well,arccos returns a value in the interval [0,\pi],while that argument can be any #,complex even...

Daniel.

 

[Zurtex]

So, basically, arg(z) = arccos(Re(z))?

I don't see how that works, you saying that:

arg(70) = arg(109i + 70)?

Shouldn't there be something else in there?

 

[dextercioby]

No,he's saying something like

\arg (70+3i)=\arccos 70

which is ballooney.

Daniel.

 

[Data]

So, basically, arg(z) = arccos(Re(z))?

No. You can write

\arg z = \arccos \left( \mbox{Re}\left[ \frac{z}{|z|} \right] \right)

in a form similar to yours. The standard definition is if z = x + iy then

\arg z = \arctan \left(\frac{y}{x}\right)

though.