Prove-that-i-i-i-where-i-1-

Question Number 2930 by Filup last updated on 30/Nov/15

Prove that Γ (i) = - i (i)!

where i = \sqrt{- 1}

Commented by 123456 last updated on 02/Dec/15

Γ (z) Γ (1 - z) = \frac{π}{\sin π z}

Γ (z) Γ (- z) = - \frac{π}{z \sin π z}

Γ (\bar{z}) = \bar{Γ} (z)

z = x ı

∣ Γ (x ı) ∣^{2} = \frac{π}{x \sinh π x}

Answered by 123456 last updated on 02/Dec/15

x! = Γ (x + 1)

Γ (x + 1) = x Γ (x)

so

x! = x Γ (x)

x = i

i! = i Γ (i)

Γ (i) = \frac{i!}{i} = \frac{i! i}{i^{2}} = - i! i

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