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

ln (e^{3i })

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- Mar 24th 2013, 07:08 AMszak1592Find ln z?
ln z = ln (mod of z) + i Arg z +- (2n pi) i

ln (e^{3i }) - Mar 24th 2013, 07:25 AMProve ItRe: Find ln z?
Can't you immediately read off what and are?

- Mar 24th 2013, 07:34 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, 10:29 AMPlatoRe: Find ln z?
- Mar 24th 2013, 05:03 PMProve ItRe: Find ln z?
- Mar 25th 2013, 02:51 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, 02:55 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, 12:09 PMSworDRe: 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.