A calculus problem by gino varkey

Calculus Level 5

The magnitude of the complex number Γ ( 4 + 2 i ) = ( 3 + 2 i ) ! \Gamma(4+2i) = (3+2i)! can be expressed as A × B × cosech ( B ) \sqrt{A \times B \times \text{cosech}(B)} , where A A is an integer, and B B is a real number.

Submit your answer as A × B π \dfrac{A\times B}{\pi} .

Notations:

  • i = 1 i = \sqrt{-1} .
  • For real numbers x x and y y , the magnitude of x + y i x+yi is x + y i = x 2 + y 2 |x+yi| = \sqrt{x^2 + y^2} .
  • cosech ( ) \text{cosech}(\cdot) represents the hyperbolic cosecant function, cosech ( z ) = 2 e z e z \text{cosech}(z) = \dfrac2{e^z-e^{-z}} .


The answer is 1040.

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1 solution

Gino Varkey
Jul 11, 2018

(3+2i)!=(3+2i) (2+2i) (1+2i)*((2i)!)

||(3+2i)!||=||(3+2i)||×||(2+2i)||×||1+2i||×||(2i)!||,

||(3+2i)!||=√(520)*√(2π×cosech(2π))

||(3+2i)!||÷√(520)=√(2π×cosech(2π))

So A = 520, B = 2pi

Additional information Let Gamma function=G(x) And © denote complex conjugate

G(ni)×G(1-ni)=π÷sin(niπ)=π÷(i×sinh(nπ))

So ni×G(ni)×G(1-ni)=nπ×cosech(nπ)

G(1+ni)×G(1-ni)=(ni)!×(-ni)!=(ni)!×(ni)!©

=||(ni)!||^2=(nπ×cosech(nπ))

So ||(2i)!||=√(2π×cosech(2π))

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