How do I find the arguement of z, when z=(1+(sqrt3)i)^m?

Question

anonymous · Answer

So like this:
$$z=(1+(\sqrt3)i)^m$$

anonymous · Answer

@Loser66 ?

loser66 · Answer

arg (z) = arctan $\dfrac{y}{x} =arctan\dfrac{\sqrt3}{1}= arctan\sqrt3 = 60 ~~or ~~=\dfrac{\pi}{3}$

anonymous · Answer

but what if the power is not an integer, but a variable m? I'm asked to provide the general formula, not just just for a specific case

loser66 · Answer

oh yea, you are right.

anonymous · Answer

Do you have any other ideas?

anonymous · Answer

Because I'm pretty lost

loser66 · Answer

I don't know, I am sorry. let tag other

anonymous · Answer

That's alright, don't worry about it :)

asnaseer · Answer

@Teinis -- do you know that any complex number can be written as:$$z=|z|e^{Arg(z)}$$

anonymous · Answer

Yes I'm familiar with that equation

asnaseer · Answer

ok, so we can then say that:$$z^m=|z|^m(e^{Arg(z)})^m=|z|^me^{mArg(z)}$$Therefore:$$Arg(z^m)=mArg(z)$$

anonymous · Answer

Although isn't it $$z=\left| z \right| e ^{iArgz}$$

asnaseer · Answer

yes - sorry I missed the i

asnaseer · Answer

Did you follow the argument above for $z^m$?

anonymous · Answer

I don't really get how you got from your first equation to your second. Can you explain it?

asnaseer · Answer

|dw:1379858101504:dw|