# Thread: Find ln z?

1. ## Find ln z?

ln z = ln (mod of z) + i Arg z +- (2n pi) i

ln (e3i )

2. ## Re: Find ln z?

Can't you immediately read off what $\displaystyle \left| e^{3i} \right|$ and $\displaystyle \arg{\left( e^{3i} \right)}$ are?

3. ## Re: Find ln z?

I can get the mod but i dont how to find the Arg of e^3i, please teach me how to find the Arg of this kinda function

4. ## Re: Find ln z?

Originally Posted by szak1592
I can get the mod but i dont how to find the Arg of e^3i, please teach me how to find the Arg of this kinda function
$e^z=e^{x+yi}=e^x[\cos(y)+i\sin(y)]$ so $\text{arg}(e^z)=~?$

5. ## Re: Find ln z?

Originally Posted by szak1592
I can get the mod but i dont how to find the Arg of e^3i, please teach me how to find the Arg of this kinda function
Surely you can see it's written as $\displaystyle e^{\theta i}$. What is $\displaystyle \theta$?

6. ## Re: Find ln z?

okay it is 3, so cos 3 + i sin 3 so it means arg z = arc tan of sin3/cos3, rite???

7. ## Re: Find ln z?

You are overcomplicating things. The argument is the angle that is made. You already read off that this angle is 3 (radians).

Of course if you were going to do it your way, you'd note that sin(x)/cos(x) = tan(x) and so your arctangent cancels off anyway.

8. ## Re: Find ln z?

ln (e3i )
You should have noticed that the ln and e^ functions are inverses of each other, so in at least some branch, ln(e^(z)) = z. Then think about what the angle/magnitude of this would be, whether this result makes sense and if not, what you need to do to make it make sense.