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

ln (e^{3i })

Printable View

- Mar 24th 2013, 06:08 AMszak1592Find ln z?
ln z = ln (mod of z) + i Arg z +- (2n pi) i

ln (e^{3i }) - Mar 24th 2013, 06:25 AMProve ItRe: Find ln z?
Can't you immediately read off what $\displaystyle \displaystyle \left| e^{3i} \right| $ and $\displaystyle \displaystyle \arg{\left( e^{3i} \right)}$ are?

- Mar 24th 2013, 06:34 AMszak1592Re: 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

- Mar 24th 2013, 09:29 AMPlatoRe: Find ln z?
- Mar 24th 2013, 04:03 PMProve ItRe: Find ln z?
- Mar 25th 2013, 01:51 AMszak1592Re: Find ln z?
okay it is 3, so cos 3 + i sin 3 so it means arg z = arc tan of sin3/cos3, rite???

- Mar 25th 2013, 01:55 AMProve ItRe: 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. - Mar 25th 2013, 11:09 AMSworDRe: Find ln z?Quote:

ln (e3i )

*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.