# Real & Imaginary parts of a finite product

1. Mar 5, 2006

### benorin

So I'm trying to work-out the real and imaginary parts of a finite product, put

$$P_n = \prod_{k=1}^{n} \left( x_k + iy_k\right)$$

where the x's and y's are real numbers like you would expect.

2. Mar 5, 2006

### matt grime

Well, what is the general formula for expanding that product? And when are the terms going to be real? I know I'm just restating your question, but if you can answer each of those question you can do this. So where have you got to?

3. Mar 5, 2006

### shmoe

If you can write them in polar form it may be much neater. Do you know anything specific about the x's and y's? Of course a general version will be possible, just icky.

4. Mar 5, 2006

### benorin

NOTE: I will edit this post in some time, but I have got the pattern all mixed-up in my head right now and must then come back to this later.

I cut my post short on account of something more important, allow me to continue: I have considered the first several expansions of $$P_n$$ for small n, I have noticed that the each term in the expansion contains exactly n factors, and that, alternately, the terms of the real and imaginary parts in the expansion of the product contain only an even or an odd number of x's or y's according as n even or n odd. The last sentence is complicated for me to write (and is almost certianly wrong,) but, perhaps, some examples will make up for that

$$P_1 = \prod_{k=1}^{1} \left( x_k + iy_k\right)= x_1 + iy_1$$

the real part: $$\Re P_1 = x_1$$ contains only terms having an odd number of x's and an even number of y's (I speak not of the oddness nor the evenness of the subscripts, but rather that there is one (odd) x and zero (even) y's

vice-versa for the imaginary part

$$P_2 = \prod_{k=1}^{2} \left( x_k + iy_k\right)= x_1x_2-y_1y_2 + i(x_1y_2+x_2y_1)$$

the real part: $$\Re P_2 = x_1x_2-y_1y_2$$ contains only terms having an even number of x's and an even number of y's

5. Mar 5, 2006

### benorin

I intend to try to relate the pattern I fail to describe above to the symmetric polynomials, and get some sort of combinatorial closed form for the real and imaginary parts.

6. Mar 7, 2006

### benorin

Try polar form

Try polar form? Let $$z_k=r_{k}e^{i\theta_k},$$ where $$r_k:=|z_k| \mbox{ and } \theta_k:=\mbox{Arg } z_{k}$$ so that

$$P_n = \prod_{k=1}^{n} z_k = \prod_{k=1}^{n} r_{k}e^{i\theta_k} = \left(\prod_{k=1}^{n} r_{k} \right) e^{i\sum_{j=1}^{n}\theta_j}$$

let us put $$r:=\prod_{k=1}^{n} r_{k}\mbox{ and } \theta:=\sum_{j=1}^{n}\theta_j$$ so that we have

$$P_n =re^{i\theta} = r(\cos{\theta}+i\sin{\theta})$$

and hence $$\Re{P_n} = r\cos{\theta}\mbox{ and } \Im{P_n} = r\sin{\theta}$$

there is some considerations to be had presently, firstly (and without trouble,) we have

$$r = \prod_{k=1}^{n} r_{k} = \prod_{k=1}^{n} |z_{k}| = \prod_{k=1}^{n} \sqrt{x_{k}^{2}+y_{k}^{2}} = \sqrt{ \prod_{k=1}^{n} \left( x_{k}^{2}+y_{k}^{2}\right) }$$

and here is the trouble,

$$\theta = \sum_{j=1}^{n}\theta_j = \sum_{j=1}^{n}\mbox{Arg } z_{j} \neq \mbox{Arg } \left( \prod_{j=1}^{n} z_{j}\right)$$

where I have put $$\neq$$ since we have that pesky brach-cut structure to worry about, since

$$\mbox{Arg } \left( z_{1}z_{2}\right) = \mbox{Arg } z_1 + \mbox{Arg } z_2 + 2\pi i \cdot\mbox{floor}\left(\frac{\pi - \mbox{Arg } z_1 - \mbox{Arg } z_2 }{2\pi}\right)$$

how does this generalize? Above, floor() is the floor function (a.k.a. the greatest integer function.)

Last edited: Mar 7, 2006
7. Mar 7, 2006

### benorin

Heck with it, close enough is: there exists an integer k such that

$$\theta = \sum_{j=1}^{n}\mbox{Arg } z_{j} = \mbox{Arg } \left( \prod_{j=1}^{n} z_{j}\right) +k\pi i$$

How do I translate these back into x's and y's ?

Last edited: Mar 7, 2006
8. Mar 7, 2006

### shmoe

There's no reason to be concerned with the Arg of the product. When you take the cos and sin to find the real and imaginary parts this is irrelevant.

The downside here is that in terms of x and y you have the cos or sin of a sum of arctan's. Whether this is more or less desirable than a hideous sum is up to you. Either way is not going to be too nice with a general batch of z's.

Of course you've probably noticed how much cleaner it is if you started in polar form, these are much more natural coordinates for complex multiplication.

9. Mar 7, 2006

### benorin

Re & Im parts of infinite products

I am also considering finding the Re & Im parts of infinite products, (as they interest me.) For example, we know that for all complex $z:=x+iy$ the following hold:

$$\sin {z} = z\prod_{k=1}^{\infty} \left( 1-\frac{z^2}{k^2\pi ^2}\right)$$

$$\mbox{sinh} {z} = z\prod_{k=1}^{\infty} \left( 1+\frac{z^2}{k^2\pi ^2}\right)$$

$$\cos {z} = \prod_{k=1}^{\infty} \left( 1-\frac{4z^2}{(2k-1)^2\pi ^2}\right)$$

$$\mbox{cosh} {z} = \prod_{k=1}^{\infty} \left( 1+\frac{4z^2}{(2k-1)^2\pi ^2}\right)$$

and from

$$\sin (x+iy) = \sin (x)\mbox{cosh}(y)+i\cos (x)\mbox{sinh}(y)$$

we have, for example

$$\Re \sin {(x+iy)} = \Re \left[ (x+iy)\prod_{k=1}^{\infty} \left( 1-\frac{(x+iy)^2}{k^2\pi ^2}\right) \right] = x\prod_{k=1}^{\infty} \left[ \left( 1-\frac{x^2}{k^2\pi ^2}\right) \left( 1+\frac{4y^2}{(2k-1)^2\pi ^2}\right)\right] = \sin (x)\mbox{cosh}(y)$$

the Im part is similar...

What is the formula for the Re & Im parts of an arbitrary convergent infinite product in terms of an infinite product?

Last edited: Mar 7, 2006