A Complex Infinite Product

Calculus Level 5

If z = k = 1 ( 1 + i 2 k 2 ) \displaystyle z = \prod_{k=1}^{\infty} \left( 1 + \dfrac{i}{2k^2} \right) , then what is the value of ( 1 i ) z + Arg ( z ) (1-i)z + \text{Arg}(z) ?

Give your answer to 3 decimal places.

Notation : Arg ( z ) \text{Arg}(z) represents the principal value of the angle of z z .


The answer is 2.382790.

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1 solution

Sri Kanth
Apr 27, 2016

Using Euler's observation , we have that

sin ( w ) = w k = 1 ( 1 w 2 π 2 k 2 ) \displaystyle \sin(w) = w\prod_{k=1}^{\infty} \left(1-\dfrac{w^2}{\pi^2k^2}\right)

Substituting w = ( 1 i ) π 2 w = \dfrac{(1-i)\pi}{2} , and using the exponential formula for complex sine , the required product is

z = sin w w = ( e π 2 + e π 2 2 π ) ( 1 + i ) z = \dfrac{\sin w}{w} = \left(\dfrac{e^\frac{\pi}{2}+e^{-\frac{\pi}{2}}}{2 \pi}\right)\cdot (1+i)

Therefore ( 1 i ) z = 2 cosh ( π 2 ) π (1-i)z = \dfrac{2\cosh(\frac{\pi}2)}{\pi} and Arg ( z ) = π 4 \text{Arg}(z) = \frac{\pi}4 giving the sum 2.382790 2.382790 .

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