Nice! Tnx! --rwg Date: 2017-03-28 10:15 From: Dan Asimov <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Reply-To: math-fun <math-fun@mailman.xmission.com> I took this as a personal challenge. First note that 1/2 ((-I)^(1 + √2) + I^(1 + √2)) = cos(pi(1+sqrt(2))/2) . Let T = pi(1+sqrt(2))/2. Also note that 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2 = (cos T)^2 so that sqrt(-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2) = i sin T . Then the whole expression is (cos T - i sin T)^(1 + sqrt(2)) + (cos T + i sin T)^(1 + sqrt(2)) or exp( -i T (1 + sqrt(2)) + exp( i T (1 + sqrt(2)) = 2 cos(T(1 + sqrt(2))) = 2 cos(pi(1 + sqrt(2))^2 / 2) = 2 cos((pi/2) (3 + 2 sqrt(2)) = 2 cos(3 pi/2 + pi sqrt(2)) = 2 [cos(3 pi/2) cos(pi sqrt(2)) - sin(3 pi/2) sin(pi sqrt(2))] = 2 sin(pi sqrt(2)) Q.E.D. —Dan Bill Gosper wrote (March 27, 2017 at 5:25 pm): ----- In[69]:= ((1/2 ((-I)^(1 + √2) + I^(1 + √2)) - Sqrt[-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2])^(-1 + √2) + (1/2 ((-I)^(1 + √2) + I^(1 + √2)) + Sqrt[-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2])^(-1 + √2)) // FullSimplify Out[69]= 2 Sin[π 2√2] (How??) -----