Yes, I know that, the 2*Pi is the result of 2*round(12*log(2)/(2*Pi)), that is 2*{n*log(k)/(2*Pi)} in k^(1/2+I*n), where { } is the nearest integer. in 1/5^(1/2+12*I) with the same operations of convert, argument we get -arctan(4*sin(ln(5))*cos(ln(5))*(512*cos(ln(5))^10-1280*cos (ln(5))^8+1152*cos(ln(5))^6-448*cos(ln(5))^4+70*cos(ln(5))^ 2-3)/(2048*cos(ln(5))^12-6144*cos(ln(5))^10+6912*cos(ln(5)) ^8-3584*cos(ln(5))^6+840*cos(ln(5))^4-72*cos(ln(5))^2+1)) when the answer is really a combination of Pi and log(5). The terms in arctan( sin ( log( ))) and the same with cos vanishes. But this is a maple typical problem, what is the unexpected is the simplification of all the terms. Now, if you add 2 complex numbers, the answer is far from trivial, I mean, a simple mutiple of logs and Pi, and what ? just try any sum of 2 terms. like (1 + 3 I) (1 + 5 I) 2 + 3 we get : 2 sin(3 ln(2)) + 3 sin(5 ln(3)) arctan(-------------------------------) 2 cos(3 ln(2)) + 3 cos(5 ln(3)) which is NOT a combination of log(3) and log(2) and Pi. There is something missing isn't ? Best regards, simon plouffe Le 14/08/2013 23:04, Axel Vogt a écrit :

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For example, arg(1/2^(1/2+12*I) , if you use Maple or Mathematica or by hand you can expand this BUT, actually the expression is quite cumbersome and tricky.

The proper answer is 2*Pi - 12*log(2). But that answer does not appear at fist hand, the <12> comes from an operation with { } and [ ]. ...

Using Maple 17: (1/2^(1/2+12*I)); convert(%, exp); argument(%);

-12*ln(2)+2*Pi

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