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Oskar45 avatar
Oskar45
Who joined Dec. 5, 2005, 2:35 a.m.
and authored 1949 notes
Wrote the following at Aug. 5, 2013, 10:25 a.m...
i pi = ln(-1)

is well known. Replacing (-1) by (i^2) and dividing by i, I get

pi = ln(i^2)/i

I really can't phantom how the right side of the last equation can evaluate to pi. What am I doing wrongly here?

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robespierre
Who joined Sept. 12, 2011, 1:28 p.m.
and authored 318 notes
Wrote the following at Aug. 5, 2013, 10:58 a.m...
bring out the 2:
π = 2ln(i)/i

π/2 = ln(i)/i

all correct. The function e^(iφ) is the unit circle starting from (1,0) and running CCW with period 2π. So iπ/2 is the power of e that maps to the point (0,1).

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jan-jaap avatar
jan-jaap
Who joined June 17, 2004, 10:35 a.m.
and authored 3659 notes
Wrote the following at Aug. 5, 2013, 12:04 p.m...
http://www.physicsforums.com/showthread.php?t=468510

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Oskar45 avatar
Oskar45
Who joined Dec. 5, 2005, 2:35 a.m.
and authored 1949 notes
Wrote the following at Aug. 7, 2013, 3:21 a.m...
jan-jaap wrote:
http://www.physicsforums.com/showthread.php?t=468510
Thanks - that cleared it up.

Actually, seems I'd started this thread quite prematurely - blame it on the default iPhone calculator (which doesn't handle complex stuff). Too late I remembered my HP-15C (30+ years old) - everything would have worked out nicely in the first place ;-)

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