Product of Complex Numbers with cosh and sin

Calculus Level 5

k = 1 cosh ( k 2 + k + 1 2 ) + i sin ( k + 1 2 ) cosh ( k 2 + k + 1 2 ) i sin ( k + 1 2 ) . \prod_{k=1}^\infty \frac{\cosh\left(k^2+k+\frac{1}{2}\right) + i \sin\left(k+\frac{1}{2}\right)}{\cosh\left(k^2+k+\frac{1}{2}\right) - i \sin\left(k+\frac{1}{2}\right)}. Evaluate the product above. The answer should be of the form a 2 e 2 + a 1 e + a 0 + i ( b 2 e 2 + b 1 e + b 0 ) c 2 e 2 + c 1 e + c 0 . \frac{a_2 e^2 + a_1 e + a_0 + i \left( b_2 e^2 + b_1 e + b_0\right)}{c_2 e^2 + c_1 e + c_0}. Enter 2 2 a 2 + 2 1 a 1 + 2 0 a 0 + 3 2 b 2 + 3 1 b 1 + 3 0 b 0 + 5 2 c 2 + 5 1 c 1 + 5 0 c 0 2^2 a_2 + 2^1 a_1 + 2^0 a_0 + 3^2 b_2 + 3^1 b_1 + 3^0 b_0 + 5^2 c_2 + 5^1 c_1 + 5^0 c_0 for the answer.


The answer is 35.

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1 solution

Sayeed Tasnim
Jan 18, 2016

I believe the answer is e 2 1 + 2 e i e 2 + 1 \frac{e^2 - 1 + 2e i}{e^2+1} . However, I'm still looking for a solution. :(

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