I got this kind of expression as a value of an infinite product:

$$\prod_{k=1}^{\infty} \left(1-\frac{A}{k^2}+\frac{B}{k^4} \right)$$

It's easy to see how it can be factored into a product of two sines.

$$\frac{1}{\pi^2 \sqrt{B}} \sin \sqrt{a-ib} \sin \sqrt{a+ib}$$



In my case, $4B>A^2$. However, it is obvious both by original expression and numerical computation, that the expression is real valued.

So how do I get rid of $i$ in this expression?

The only idea I have is series expansion. Either expand the sines and multiply the series or move to exponential form and expand the roots.

Is there another, easier way?

  • 2
    $\begingroup$ $\sqrt{a-ib} = \overline{\sqrt{a+ib}}$ probably. Otherwise we just get a factor of $-1$. Now, with $z = x+iy$, we have $$\sin z \sin \overline{z} = \sin z \, \overline{\sin z} = \lvert \sin (x+iy)\rvert^2 = \dotsc = \sin^2 x + \sinh^2 y.$$ $\endgroup$ – Daniel Fischer Feb 23 '16 at 22:49
  • $\begingroup$ Ah, so I have to use the exponential form for $a\pm ib$, then raise it to power $1/2$ and then use this formula? $\endgroup$ – Yuriy S Feb 23 '16 at 22:51
  • $\begingroup$ You don't need to. You can also get the square root in Cartesian form without too much ado. $\endgroup$ – Daniel Fischer Feb 23 '16 at 22:54

I figured it out. Thanks to @DanielFischer for the comment. I still don't know how to proceed without exponential form though.

Let $a,b$ be real positive numbers.

$$a+i b=\sqrt{a^2+b^2} \exp \left(i \arctan \frac{b}{a} \right)$$

$$\sqrt{a+i b}=\sqrt[4]{a^2+b^2} \exp \left(\frac{i}{2} \arctan \frac{b}{a} \right)=$$

I use the principal value of the root.

$$=\sqrt[4]{a^2+b^2}\left( \cos \left(\frac{1}{2} \arctan \frac{b}{a} \right)+i \sin \left(\frac{1}{2} \arctan \frac{b}{a} \right) \right)=$$

I leave out the trigonometric transforms.

$$=\frac{\sqrt[4]{a^2+b^2}}{\sqrt{2}} \left( \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}+i \frac{b}{\sqrt{a^2+b^2} \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}} \right)$$

Letting $b \to -b$ we similarly obtain:

$$\sqrt{a-i b}=\frac{\sqrt[4]{a^2+b^2}}{\sqrt{2}} \left( \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}-i \frac{b}{\sqrt{a^2+b^2} \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}} \right)$$

Now, it is convenient to transform the sine product in the following way:

$$\sin \sqrt{a+i b} \sin \sqrt{a-i b}=\frac{1}{2} \left( \cos (\sqrt{a+i b}-\sqrt{a-i b})-\cos (\sqrt{a+i b}+\sqrt{a-i b}) \right)$$

Let's denote:

$$X=Re (\sqrt{a+i b})=Re (\sqrt{a-i b})=\frac{\sqrt[4]{a^2+b^2}}{\sqrt{2}} \sqrt{1+\frac{a}{\sqrt{a^2+b^2}}}$$

$$Y=|Im (\sqrt{a+i b})|=|Im (\sqrt{a-i b})|=\frac{b}{2 X}$$

Finally we get:

$$\sin \sqrt{a+i b} \sin \sqrt{a-i b}=\frac{1}{2} \left( \cosh \left(\frac{b}{X} \right)-\cos (2X) \right)$$

$$X=\frac{1}{\sqrt{2}} \sqrt{a+\sqrt{a^2+b^2}}$$

Additionally, we obtain:

$$\prod_{k=1}^{\infty} \left(1-\frac{A}{k^2}+\frac{B}{k^4} \right)=\frac{\cosh (\pi \sqrt{2 \sqrt{B}-A})-\cos (\pi \sqrt{2 \sqrt{B}+A})}{2 \pi^2 \sqrt{B}}$$


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