Recently, I was looking at the generalized solution for a sum of the geometric series and/or sequence where $r = p+qi$ where $p$ and $q$ are real numbers. I started with the binomial theorem which gave me $$\sum_{n=0}^\infty\sum_{k=0}^n\binom{n}{k}p^{n-k}q^ki^k$$
I was having difficulty with the imaginary numbers, so I decided to treat it as a vector sum, isolating the imaginary and real coordinates from each other. As such, I broke down the original sum into 4 separate sums. I made these sums of infinite sequences for every 4th term, then combined the real with the real and imaginary with imaginary
real
$$\sum_{n=0}^{\infty}\sum_{j=0}^{3}\sum_{k=0}^{n}\binom{4n+j}{4k}p^{4n+j-4k}q^{4k}-\sum_{n=0}^{\infty}\sum_{j=0}^{3}\sum_{k=0}^{n}\binom{4n+j}{4k+2}p^{4n+j-4k-2}q^{4k+2}$$
imaginary
$$\sum_{n=0}^\infty \sum_{j=0}^3 \sum_{k=0}^n \binom{4n+j}{4k+1}p^{4n+j-4k-1}q^{4k+1} - \sum_{n=0}^\infty \sum_{j=0}^3 \sum_{k=0}^n \binom{4n+j}{4k+3}p^{4n+j-4k-3}q^{4k+3}$$
I also tried making finite summations, but I'm not sure how well those turned out. I've been experimenting with the $\bmod(x,4)$ on Desmos to keep the terms on track, but my problem is that I am having trouble with how the sequences start on different terms and are clustered in groups of 4.
My question is how do I write in summation form the 4 sequences that isolate the various power values of $i$ in the geometric sequence given term n of the original series?
Also, I am aware of $\frac{1}{1-z}$ and $(1-z^{n+1})/(1-z)$ I am solely interested in the proper notation for the 4 finite summations that add up to the finite sum $$\sum_{n=0}^m \sum_{k=0}^n \binom{n}{k}p^{n-k}q^k i^k$$ where $m$ is any positive integer in terms of $i^0$, $i^1$, $i^2$, and $i^3$.
Edit: In the finite summation, I expanded the terms a bit to show why I'm having a bit of trouble. Here are terms $a_0$ through $a_7$
$$a_0 = \sum_{k=0}^{0}\binom{0}{4k+0}p^{0-(4k+0)}q^{4k+0}i^{0}$$
$$a_1 = \sum_{k=0}^{0}\binom{1}{4k+0}p^{1-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{0}\binom{1}{4k+1}p^{1-(4k+1)}q^{4k+1}i^{1}$$
$$a_2 = \sum_{k=0}^{0}\binom{2}{4k+0}p^{2-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{0}\binom{2}{4k+1}p^{2-(4k+1)}q^{4k+1}i^{1}+ \sum_{k=0}^{0}\binom{2}{4k+2}p^{2-(4k+2)}q^{4k+2}i^{2}$$
$$a_3 = \sum_{k=0}^{0}\binom{3}{4k+0}p^{3-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{0}\binom{3}{4k+1}p^{3-(4k+1)}q^{4k+1}i^{1}+ \sum_{k=0}^{0}\binom{3}{4k+2}p^{3-(4k+2)}q^{4k+2}i^{2}+ \sum_{k=0}^{0}\binom{3}{4k+3}p^{3-(4k+3)}q^{4k+3}i^{3}$$
$$a_4 = \sum_{k=0}^{1}\binom{4}{4k+0}p^{4-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{0}\binom{4}{4k+1}p^{4-(4k+1)}q^{4k+1}i^{1}+ \sum_{k=0}^{0}\binom{4}{4k+2}p^{4-(4k+2)}q^{4k+2}i^{2}+ \sum_{k=0}^{0}\binom{4}{4k+3}p^{4-(4k+3)}q^{4k+3}i^{3}$$
$$a_5 = \sum_{k=0}^{1}\binom{5}{4k+0}p^{5-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{1}\binom{5}{4k+1}p^{5-(4k+1)}q^{4k+1}i^{1}+ \sum_{k=0}^{0}\binom{5}{4k+2}p^{5-(4k+2)}q^{4k+2}i^{2}+ \sum_{k=0}^{0}\binom{5}{4k+3}p^{5-(4k+3)}q^{4k+3}i^{3}$$
$$a_6 = \sum_{k=0}^{1}\binom{6}{4k+0}p^{6-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{1}\binom{6}{4k+1}p^{6-(4k+1)}q^{4k+1}i^{1}+ \sum_{k=0}^{1}\binom{6}{4k+2}p^{6-(4k+2)}q^{4k+2}i^{2}+ \sum_{k=0}^{0}\binom{6}{4k+3}p^{6-(4k+3)}q^{4k+3}i^{3}$$
$$a_7 = \sum_{k=0}^{1}\binom{7}{4k+0}p^{7-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{1}\binom{7}{4k+1}p^{7-(4k+1)}q^{4k+1}i^{1}+ \sum_{k=0}^{1}\binom{7}{4k+2}p^{7-(4k+2)}q^{4k+2}i^{2}+ \sum_{k=0}^{1}\binom{7}{4k+3}p^{7-(4k+3)}q^{4k+3}i^{3}$$
As you can see, the k maximum for the summation representing each power of i increases by 1 every 4 terms. When mod(n,4) = 0, the k maximum for the i^0 term increases by 1. When mod(n,4) = 1, the k maximum for the i^1 term increases by 1. When mod(n,4) = 2, the k maximum for the i^2 term increases by 1. When mod(n,4) = 3, the k maximum for the i^3 term increases by 1. These $k_{maximum values}$ can be found with the following functions. $$k_{{max}_{i^0}}(n)=\frac{n-0-(n-0)\bmod4}{4}$$ $$k_{{max}_{i^1}}(n)=\frac{n-1-(n-1)\bmod4}{4}$$ $$k_{{max}_{i^2}}(n)=\frac{n-2-(n-2)\bmod4}{4}$$ $$k_{{max}_{i^3}}(n)=\frac{n-3-(n-3)\bmod4}{4}$$
So, I have an $a_n$ value at where $n\geq3$ $$ \sum_{k=0}^{\frac{n-0-(n-0)\bmod4}{4}}\binom{n}{4k+0}p^{n-(4k+0)}q^{4k+0}i^{0}+ \sum_{k=0}^{\frac{n-1-(n-1)\bmod4}{4}}\binom{n}{4k+1}p^{n-(4k+1)}q^{4k+1}i^{1}+ \sum_{k=0}^{\frac{n-2-(n-2)\bmod4}{4}}\binom{n}{4k+2}p^{n-(4k+2)}q^{4k+2}i^{2}+ \sum_{k=0}^{\frac{n-3-(n-3)\bmod4}{4}}\binom{n}{4k+3}p^{n-(4k+3)}q^{4k+3}i^{3} $$ or $$ (\sum_{k=0}^{\frac{n-0-(n-0)\bmod4}{4}}\binom{n}{4k+0}p^{n-(4k+0)}q^{4k+0}- \sum_{k=0}^{\frac{n-2-(n-2)\bmod4}{4}}\binom{n}{4k+2}p^{n-(4k+2)}q^{4k+2}) + (\sum_{k=0}^{\frac{n-1-(n-1)\bmod4}{4}}\binom{n}{4k+1}p^{n-(4k+1)}q^{4k+1}- \sum_{k=0}^{\frac{n-3-(n-3)\bmod4}{4}}\binom{n}{4k+3}p^{n-(4k+3)}q^{4k+3})i $$ To reiterate, I am looking for(if possible)
- A simplified formula for $S_n$, preferably similar to the geometric finite sum $\frac{1-z^{n+1}}{1-z}$ with the real terms separate from the imaginary terms such that it is conducive to plotting a point $$(S_{n_x},S_{n_y})$$
- If that is impossible, then the simplified condensed version of $ (\sum_{m=0}^{n}(\sum_{k=0}^{\frac{m-0-(m-0)\bmod4}{4}}\binom{m}{4k+0}p^{m-(4k+0)}q^{4k+0}- \sum_{k=0}^{\frac{m-2-(m-2)\bmod4}{4}}\binom{m}{4k+2}p^{m-(4k+2)}q^{4k+2}) , \sum_{m=0}^{n}(\sum_{k=0}^{\frac{m-1-(m-1)\bmod4}{4}}\binom{m}{4k+1}p^{m-(4k+1)}q^{4k+1}- \sum_{k=0}^{\frac{m-3-(m-3)\bmod4}{4}}\binom{m}{4k+3}p^{m-(4k+3)}q^{4k+3})) $
So, it turns out I was overthinking this. With de Moivre's formula, $$\left(r\cos\theta+ir\sin\theta\right)^{n}=\left(r^{n}\cos\left(n\theta\right)+ir^{n}\sin\left(n\theta\right)\right)$$ we can convert $p+qi$ to polar coordinates and proceed from there. First, we find the $r$ and $\theta$ values of $p+qi$. $$r=\sqrt{p^{2}+q^{2}}$$ $$\theta=\left[\cos^{-1}\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right),\sin^{-1}\left(\frac{q}{\sqrt{p^{2}+q^{2}}}\right)\right]$$ $$p+qi=\sqrt{p^{2}+q^{2}}\cos\left(\cos^{-1}\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right)\right)+\sqrt{p^{2}+q^{2}}\sin\left(\sin^{-1}\left(\frac{q}{\sqrt{p^{2}+q^{2}}}\right)\right)\cdot i$$
So for the finite geometric sum, the formula is $S_n=\frac{a_1(1-r^n)}{1-r}$ The $r$ value here is not to be confused with $r=\sqrt{p^2+q^2}$, but instead represents the complex number $p+qi$. To avoid confusion, I will rewrite the formula from this point on as $$S_n=\frac{1-\left(p+qi\right)^{n}}{1-\left(p+qi\right)}$$ leaving $a_1$ as $1$ for the purposes of simplicity. $$\frac{1-\left(p+qi\right)^{n}}{1-\left(p+qi\right)}=\frac{1-\left(r\cos\theta+ir\sin\theta\right)^{n}}{1-\left(r\cos\theta+ir\sin\theta\right)}=\frac{1-\left(r^{n}\cos\left(n\theta\right)+ir^{n}\sin\left(n\theta\right)\right)}{1-\left(r\cos\theta+ir\sin\theta\right)}$$ $$=\frac{\left(1-r^{n}\cos\left(n\theta\right)\right)-ir^{n}\sin\left(n\theta\right)}{\left(1-r\cos\theta\right)-ir\sin\theta}$$ To rationalize the denominator, we multiply by $\frac{\left(1-r\cos\theta\right)+ir\sin\theta}{\left(1-r\cos\theta\right)+ir\sin\theta}$ $$=\frac{\left(\left(1-r^{n}\cos\left(n\theta\right)\right)-ir^{n}\sin\left(n\theta\right)\right)\left(\left(1-r\cos\theta\right)+ir\sin\theta\right)}{\left(1-r\cos\theta\right)^{2}+r^{2}\sin^{2}\theta}$$ $$=\frac{1-r\cos\left(θ\right)-r^{n}\cos\left(nθ\right)+r^{n+1}\cos\left(nθ\right)\cos\left(θ\right)+r^{n+1}\sin\left(nθ\right)\sin\left(θ\right)}{1-2r\cos\left(θ\right)+r^{2}\cos^{2}\left(θ\right)+r^{2}\sin^{2}\left(θ\right)}+\frac{r\sin\left(θ\right)-r^{n+1}\cos\left(nθ\right)\sin\left(θ\right)-r^{n}\sin\left(nθ\right)+r^{n+1}\sin\left(nθ\right)\cos\left(θ\right)}{1-2r\cos\left(θ\right)+r^{2}\cos^{2}\left(θ\right)+r^{2}\sin^{2}\left(θ\right)}i$$ The only terms that can't be immediately converted back to $p$ and $q$ form in this expression are the $cos(n\theta)$ and $sin(n\theta)$ terms. Anyway, substituting in $p=r*cos(\theta)$, $q=r*sin(\theta)$ and $r=\sqrt{p^2+q^2}$, we get $$S_n=\frac{1-p-\sqrt{p^{2}+q^{2}}^{n}\cos\left(nθ\right)+\sqrt{p^{2}+q^{2}}^{n}\cos\left(nθ\right)p+\sqrt{p^{2}+q^{2}}^{n}\sin\left(nθ\right)q}{1-2p+p^{2}+q^{2}}+\frac{q-\sqrt{p^{2}+q^{2}}^{n}\cos\left(nθ\right)q-\sqrt{p^{2}+q^{2}}^{n}\sin\left(nθ\right)+\sqrt{p^{2}+q^{2}}^{n}\sin\left(nθ\right)p}{1-2p+p^{2}+q^{2}}i$$ $$=\frac{\left(1-\sqrt{p^{2}+q^{2}}^{n}\cos\left(nθ\right)\right)\left(1-p\right)+\left(\sqrt{p^{2}+q^{2}}^{n}\sin\left(nθ\right)\right)q}{\left(1-p\right)^{2}+q^{2}}+\frac{\left(1-\sqrt{p^{2}+q^{2}}^{n}\cos\left(nθ\right)\right)q-\left(\sqrt{p^{2}+q^{2}}^{n}\sin\left(nθ\right)\right)\left(1-p\right)}{\left(1-p\right)^{2}+q^{2}}i$$ $$S_n=\frac{\left(1-\sqrt{p^{2}+q^{2}}^{n}\cos\left(n\cos^{-1}\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right)\right)\right)\left(1-p\right)+\left(\sqrt{p^{2}+q^{2}}^{n}\sin\left(n\sin^{-1}\left(\frac{q}{\sqrt{p^{2}+q^{2}}}\right)\right)\right)q}{\left(1-p\right)^{2}+q^{2}}+\frac{\left(1-\sqrt{p^{2}+q^{2}}^{n}\cos\left(n\cos^{-1}\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right)\right)\right)q-\left(\sqrt{p^{2}+q^{2}}^{n}\sin\left(n\sin^{-1}\left(\frac{q}{\sqrt{p^{2}+q^{2}}}\right)\right)\right)\left(1-p\right)}{\left(1-p\right)^{2}+q^{2}}i$$ We can actually convert the $cos(n\theta)$ and $sin(n\theta)$ terms into a summation. $$sin(n\theta)=\sum_{k=0}^{2k+1\le n}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k+1\right)\cos^{n-2k-1}\left(\theta\right)\sin^{2k+1}\left(\theta\right)$$ $$cos(n\theta)=\sum_{k=0}^{2k\le n}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k\right)\cos^{n-2k}\left(\theta\right)\sin^{2k}\left(\theta\right)$$ The $k_{max}$ values here are calculated as such:
So, if we substitute p and q into these sums we get $$cos(n\theta)=\sum_{k=0}^{2k\le n}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k\right)\cos^{n-2k}\left(\cos^{-1}\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right)\right)\sin^{2k}\left(\sin^{-1}\left(\frac{q}{\sqrt{p^{2}+q^{2}}}\right)\right)$$ $$=\sum_{k=0}^{2k\le n}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k\right)\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right)^{n-2k}\left(\frac{q}{\sqrt{p^{2}+q^{2}}}\right)^{2k}$$ $$=\sum_{k=0}^{⌊n/2⌋ }\left(-1\right)^{k}\operatorname{nCr}\left(n,2k\right)\left(\frac{p}{\sqrt{p^{2}+q^{2}}}\right)^{n-2k}\left(\frac{q}{\sqrt{p^{2}+q^{2}}}\right)^{2k}$$
$$=\sum_{k=0}^{⌊n/2⌋}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k\right)\frac{p^{n-2k}q^{2k}}{\sqrt{p^{2}+q^{2}}^{n}}$$
$$\cos\left(n\theta\right)=\frac{1}{\sqrt{p^{2}+q^{2}}^{n}}\sum_{k=0}^{⌊n/2⌋}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k\right)p^{n-2k}q^{2k}$$ and $$sin(n\theta)=\sum_{k=0}^{⌊(n-1)/2⌋}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k+1\right)\cos^{n-2k-1}\left(\theta\right)\sin^{2k+1}\left(\theta\right)$$ $$=\sum_{k=0}^{⌊(n-1)/2⌋}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k+1\right)\frac{p}{\sqrt{p^{2}+q^{2}}}^{n-2k-1}\frac{q}{\sqrt{p^{2}+q^{2}}}^{2k+1}$$ $$=\sum_{k=0}^{⌊(n-1)/2⌋}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k+1\right)\frac{p^{n-2k-1}q^{2k+1}}{\sqrt{p^{2}+q^{2}}^{n}}$$ $$sin(n\theta)=\frac{1}{\sqrt{p^{2}+q^{2}}^{n}}\sum_{k=0}^{⌊(n-1)/2⌋}\left(-1\right)^{k}\operatorname{nCr}\left(n,2k+1\right)p^{n-2k-1}q^{2k+1}$$ We can see that if we plug these values for $sin(n\theta)$ and $cos(n\theta)$ into the equation $$$$ $S_n=\frac{\left(1-\sqrt{p^{2}+q^{2}}^{n}\cos\left(nθ\right)\right)\left(1-p\right)+\left(\sqrt{p^{2}+q^{2}}^{n}\sin\left(nθ\right)\right)q}{\left(1-p\right)^{2}+q^{2}}+\frac{\left(1-\sqrt{p^{2}+q^{2}}^{n}\cos\left(nθ\right)\right)q-\left(\sqrt{p^{2}+q^{2}}^{n}\sin\left(nθ\right)\right)\left(1-p\right)}{\left(1-p\right)^{2}+q^{2}}i$ , we will get
$S_n=\frac{(1-\sum_{k=0}^{⌊n/2⌋}(-1)^{k}\binom{n}{2k}p^{n-2k}q^{2k})(1-p)+(\sum_{k=0}^{⌊(n-1)/2⌋}(-1)^{k}\binom{n}{2k+1}p^{n-2k-1}q^{2k+1})q} {(1-p)^2+q^2}+\frac{(1-\sum_{k=0}^{⌊n/2⌋}(-1)^{k}\binom{n}{2k}p^{n-2k}q^{2k})q-(\sum_{k=0}^{⌊(n-1)/2⌋}(-1)^{k}\binom{n}{2k+1}p^{n-2k-1}q^{2k+1})(1-p)}{(1-p)^{2}+q^{2}}i$ $$$$In conclusion, we have two forms for the summation of $\sum_{k=0}^{n}\left(p+qi\right)^{k}$. The real values can be written as either
while the imaginary values can be written as