# Arithmetic progression with complex common difference?

Suppose we have the following sequence:

$$\{0,i,2i,3i,4i,5i\}$$

Can we call this sequence an arithmetic progression with first term $0$ and common difference of $i$ ?

Clarification: Here, $i$ is referring to the imaginary unit, i.e., $i=\sqrt{-1}$

In general, I want to know if the common difference of an AP can be any complex value and not just real value.

Thanks!

• I don't see why not. In fact, those elementary formulae (the sum of $n$ terms, etc) can be applied. – ajotatxe May 26 '15 at 17:59
• @ajotatxe, I thought so too but lately I saw a few websites where they are suggesting that the common difference is restricted to reals, which is the reason for me asking this question. – tom_cruise May 26 '15 at 18:01
• @tom_cruise If you have a specific result in mind about arithmetic progressions, that might impose further restrictions. Else you can go as far as to have a group or even just a monoid. – AlexR May 26 '15 at 18:04
• An arithmetic sequence can be thought of as a set of (equally-spaced) points along a line in $\ \mathbf{R}^2 \ \$; the common difference between terms is related to the "slope" of the line. One can perfectly well define a line in the complex plane in this fashion, except that the integer parameter corresponding to each point is not plotted on such a "graph". – colormegone May 26 '15 at 18:04
• @AlexR, in my case, the number of terms is odd, so can't we create w.l.o.g the following AP: $\{a+j\delta\}_{j=-k}^{j=k}$ where $n=2k+1$ is the number of terms and $a,\delta$ are constant values with $\delta$ being the common difference. My question is: Is there necessarily any restriction on the domain of $\delta$ ? – tom_cruise May 26 '15 at 18:14

## 1 Answer

You can define an arithmetic progression in any monoid $(M,+)$. It is then defined by a starting element $a\in M$ and an increment $b\in M$ and the recursion $$a_0 = a\\ a_{n+1} = a_n + b$$

There is no reason to restrict to reals $(\mathbb R,+)$ or complex numbers $(\mathbb C, +)$. For some results about arithmetic progressions, you might want $M$ to be an (abelian) group or even a field (both is true for the two settings mentioned here).

For a complex finite arithmetic progression $\{z,z+w, \ldots, z+nw\}$ to have a real sum, you must actually force $$\Im \sum_{k=0}^n (z+kw) = \Im \left((n+1)z + \frac{n(n+1)}2w\right) = (n+1)\Im z + \frac{n(n+1)}2 \Im w \stackrel!=0$$ In other words you can freely pick the real parts of $z$ and $w$, but the imaginary parts must be related by $$\Im w = - \frac2n \Im z$$ for some $n\in\mathbb N$ wich will double as the number of terms minus one (since we sum from $k=0$ to $n$, wich has $n+1$ summands)