∣ i ! ∣ = sinh ( a ) a
If the above holds true for real a , find e a .
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@Dwaipayan Shikari I didn't consider that approach, so nice! I got the answer using the Euler form of the gamma function and then relating complex conjugates, which was much harder. Any idea how to calculate the real and imaginary parts?
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Did you mean Γ ( s ) = s e − γ s n = 1 ∏ ∞ e n s ( 1 − n s ) − 1
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No, I meant -
Γ ( z ) = z 1 n = 1 ∏ ∞ ( 1 + n 1 ) z ( 1 + n z ) − 1
You can prove that the conjugates are conserved under the 4 arithmetic operations.
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Γ ( 1 + i ) = i ! ⟹ ∣ i Γ ( i ) ∣ = ∣ i ! ∣ Γ ( 1 − i ) Γ ( i ) = sin ( i π ) π ⟹ − i ∣ Γ ( i ) ∣ 2 = e − π − e π 2 i π ∣ Γ ( i ) ∣ = 2 e π − e − π π ⟹ ∣ Γ ( i ) ∣ = sinh ( π ) π ∣ i Γ ( i ) ∣ = ∣ i ! ∣ = sinh ( π ) π Answer is e π = 1 . 1 5 5 EDIT ; ∣ i ∣ = 1