# Complex Pascal triangle

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The complex Pascal triangle is the expansion of Pascal's triangle to the complex numbers and opens up the entire spectrum of the complex numbers to the binomial triangles. It establishes the complete complex binomial distribution, the partial complex binomial distribution and the imaginary binomial distribution.

## Definition: complex Pascal triangle

The entry in the ${\displaystyle n-}$th row and ${\displaystyle k-}$th column of the complex Pascal triangle is denoted by:

${\displaystyle \displaystyle\binom{n}{k}_{1+i}}$

We set:

${\displaystyle \begin{array}{lll} \displaystyle\binom{0}{0}_{1+i} := 1+i\text{,} & \hspace{10pt} \displaystyle\binom{n}{0}_{1+i} := 1+i\text{,} & \hspace{10pt} \displaystyle\binom{n}{n}_{1+i} := 1+i \\ \end{array}}$

for all ${\displaystyle n \geq 1}$. Further, the ${\displaystyle 1+i}$ complex triangle fulfills the addition rule:

${\displaystyle \displaystyle\binom{n + 1}{k + 1}_{1+i} = \displaystyle\binom{n}{k}_{1+i} + \displaystyle\binom{n}{k + 1}_{1+i} }$

for every ${\displaystyle k}$, ${\displaystyle n \geq 2}$. With the above binomial coefficients the triangle can be visualized as follows:

${\displaystyle \begin{array}{ccccccccc} \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & \\ & \displaystyle\binom{3}{0}_{1+i} & & \displaystyle\binom{3}{1}_{1+i} & & \displaystyle\binom{3}{2}_{1+i} & & \displaystyle\binom{3}{3}_{1+i} &\\ & & & & & & & &\\ & & \displaystyle\binom{2}{0}_{1+i} & & \displaystyle\binom{2}{1}_{1+i} & & \displaystyle\binom{2}{2}_{1+i} & & \\ & & & & & & & &\\ & & & \displaystyle\binom{1}{0}_{1+i} & & \displaystyle\binom{1}{1}_{1+i} & & & \\ & & & & & & & & \\ & & & & \displaystyle\binom{0}{0}_{1+i} & & & & \\ \end{array}}$

which continues for ${\displaystyle n = 4}$, ${\displaystyle 5}$, ${\displaystyle \dots}$. The equivalent entries to the binomial coefficients are as follows:

${\displaystyle \begin{array}{cccccccccccc} \dots & & \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & & & \\ & 1+i & & 4+4i & & 6+6i & & 4+4i & & 1+i & \\ & & & & & & & & & & \\ & & 1+i& & 3+3i & & 3+3i & & 1+i & & \\ & & & & & & & & & & \\ & & & 1+i & & 2+2i & & 1+i & & & \\ & & & & & & & & & & \\ & & & & 1+i & & 1+i & & & & \\ & & & & & & & & & & \\ & & & & & 1+i & & & & & \\ \end{array}}$

## Consequence

From the complex (binomial) Pascal triangle we can conclude two things:

• firstly, that we can construct ${\displaystyle p-}$nomial triangles with entries out of the complex numbers;
• and secondly, that the structure ${\displaystyle -}$ and the patterns of the entries, with the same underlying starting sequence of numbers for the separated real ${\displaystyle -}$ and imaginary part, does not change.

## Application

Although the complex Pascal triangle seems appearing out of nowhere, we can show that it has a real application. We set:

${\displaystyle \begin{array}{ll} z_{n} & = \displaystyle\sum\limits_{k = 0}^{n}~\displaystyle\binom{n}{k} \, x^{n - k} a^{k} +i\displaystyle\sum\limits_{k = 0}^{n}~\displaystyle\binom{n}{k} \, y^{n - k} b^{k}\\ & \\ & = (x + a)^{n} +i[(y+b)^{n}] \\ \end{array}}$

for ${\displaystyle \mathbb{N} \ni n \geq 1}$, where ${\displaystyle z_{n} \in \mathbb{C}}$ and ${\displaystyle a}$, ${\displaystyle b}$, ${\displaystyle x}$, ${\displaystyle y \in \mathbb{R}}$. This yields that both, the real ${\displaystyle -}$ and the imaginary part, are represented by the same binomial; and since ${\displaystyle \operatorname{Re}(z_{n})}$ and ${\displaystyle \operatorname{Im}(z_{n})}$ are already separated, we know that either part is binomial distributed for every ${\displaystyle n}$, which in return can be represented by the complex Pascal triangle.

## Definition: partial complex Pascal triangle

The entry in the ${\displaystyle n-}$th row and ${\displaystyle k-}$th column of the partial complex Pascal triangle is denoted by:

${\displaystyle \displaystyle\binom{n}{k}_{1|i}}$

We set:

${\displaystyle \begin{array}{lll} \displaystyle\binom{0}{0}_{1|i} := 0\text{,} & \hspace{10pt} \displaystyle\binom{n}{0}_{1|i} := 1\text{,} & \hspace{10pt} \displaystyle\binom{n}{n}_{1|i} := i \\ \end{array}}$

for all ${\displaystyle n \geq 1}$. Further, the ${\displaystyle 1|i}$ complex triangle fulfills the addition rule:

${\displaystyle \displaystyle\binom{n + 1}{k + 1}_{1|i} = \displaystyle\binom{n}{k}_{1|i} + \displaystyle\binom{n}{k + 1}_{1|i} }$

for every ${\displaystyle k}$, ${\displaystyle n \geq 2}$. With the above binomial coefficients the triangle can be visualized as follows:

${\displaystyle \begin{array}{ccccccccc} \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & \\ & \displaystyle\binom{3}{0}_{1|i} & & \displaystyle\binom{3}{1}_{1|i} & & \displaystyle\binom{3}{2}_{1|i} & & \displaystyle\binom{3}{3}_{1|i} &\\ & & & & & & & &\\ & & \displaystyle\binom{2}{0}_{1|i} & & \displaystyle\binom{2}{1}_{1|i} & & \displaystyle\binom{2}{2}_{1|i} & & \\ & & & & & & & &\\ & & & \displaystyle\binom{1}{0}_{1|i} & & \displaystyle\binom{1}{1}_{1|i} & & & \\ & & & & & & & & \\ & & & & \displaystyle\binom{0}{0}_{1|i} & & & & \\ \end{array}}$

which continues for ${\displaystyle n = 4}$, ${\displaystyle 5}$, ${\displaystyle \dots}$. The equivalent entries to the binomial coefficients are as follows:

${\displaystyle \begin{array}{cccccccccccc} \dots & & \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & & & \\ & 1 & & 3+i & & 3+3i & & 1+3i & & \hspace{10pt}i & \\ & & & & & & & & & & \\ & & 1& & 2+i & & 1+2i & & i & & \\ & & & & & & & & & & \\ & & & 1 & & 1+i & & i & & & \\ & & & & & & & & & & \\ & & & & 1 & & i & & & & \\ & & & & & & & & & & \\ & & & & & 0 & & & & & \\ \end{array}}$

As a consequence we have a partial complex binomial distribution.

## Definition: imaginary Pascal triangle

The entry in the ${\displaystyle n-}$th row and ${\displaystyle k-}$th column of the imaginary Pascal triangle is denoted by:

${\displaystyle \displaystyle\binom{n}{k}_{i}}$

We set:

${\displaystyle \begin{array}{lll} \displaystyle\binom{0}{0}_{i} := i\text{,} & \hspace{10pt} \displaystyle\binom{n}{0}_{i} := i\text{,} & \hspace{10pt} \displaystyle\binom{n}{n}_{i} := i \\ \end{array}}$

for all ${\displaystyle n \geq 1}$. Further, the ${\displaystyle 1|i}$ complex triangle fulfills the addition rule:

${\displaystyle \displaystyle\binom{n + 1}{k + 1}_{i} = \displaystyle\binom{n}{k}_{i} + \displaystyle\binom{n}{k + 1}_{i} }$

for every ${\displaystyle k}$, ${\displaystyle n \geq 2}$. With the above binomial coefficients the triangle can be visualized as follows:

${\displaystyle \begin{array}{ccccccccc} \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & \\ & \displaystyle\binom{3}{0}_{i} & & \displaystyle\binom{3}{1}_{i} & & \displaystyle\binom{3}{2}_{1|i} & & \displaystyle\binom{3}{3}_{i} &\\ & & & & & & & &\\ & & \displaystyle\binom{2}{0}_{i} & & \displaystyle\binom{2}{1}_{i} & & \displaystyle\binom{2}{2}_{i} & & \\ & & & & & & & &\\ & & & \displaystyle\binom{1}{0}_{i} & & \displaystyle\binom{1}{1}_{i} & & & \\ & & & & & & & & \\ & & & & \displaystyle\binom{0}{0}_{i} & & & & \\ \end{array}}$

which continues for ${\displaystyle n = 4}$, ${\displaystyle 5}$, ${\displaystyle \dots}$. The equivalent entries to the binomial coefficients are as follows:

${\displaystyle \begin{array}{cccccccccccc} \dots & & \dots & & \dots & & \dots & & \dots & & \dots\\ & & & & & & & & & & \\ & i & & 4i & & 6i & & 4i & & i & \\ & & & & & & & & & & \\ & & i& & 3i & & 3i & & i & & \\ & & & & & & & & & & \\ & & & i & & 2i & & i & & & \\ & & & & & & & & & & \\ & & & & i & & i & & & & \\ & & & & & & & & & & \\ & & & & & i & & & & & \\ \end{array}}$

As a consequence we have a purely imaginary distribution.

## Corollary

As a result of the theorem, establishing the operations on ${\displaystyle p-}$nomial triangles, we have:

${\displaystyle \binom{n}{k} \oplus \binom{n}{k}_{i} = \binom{n}{k}_{1+i}}$

where ${\displaystyle \binom{n}{k}}$ denotes Pascal's ${\displaystyle -}$, ${\displaystyle \binom{n}{k}_{i}}$ denotes the imaginary Pascal ${\displaystyle -}$ and ${\displaystyle \binom{n}{k}_{1+i}}$ denotes the complex Pascal triangle.