Imaginary Part of Integer Power of Complex Number is Harmonic

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Theorem

Let $z \in \C$ be a complex number.

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $z^n$ denote $z$ raised to the $n$th power.


Then the imaginary part $\map \Im {z^n}$ of $z^n$ is a harmonic polynomial.


Proof

Let $z = x + i y$.

Then:

\(\ds z^n\) \(=\) \(\ds \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ even} } } \paren {-1}^{j / 2} \dbinom n j x^{n - j} y^j} + i \paren {\sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j x^{n - j} y^j}\) Arbitrary Power of Complex Number
\(\ds \leadsto \ \ \) \(\ds \map \Im {z^n}\) \(=\) \(\ds \sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j x^{n - j} y^j\) Definition of Imaginary Part

Then we have:

\(\ds \map {\dfrac \partial {\partial x} } {\map \Im {z^n} }\) \(=\) \(\ds \sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j \paren {n - j} x^{n - j - 1} y^j\) Definition of Partial Derivative
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac {\partial^2} {\partial x^2} } {\map \Im {z^n} }\) \(=\) \(\ds \sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j \paren {n - j} \paren {n - j - 1} x^{n - j - 2} y^j\) Definition of Partial Derivative
\(\ds \) \(=\) \(\ds \sum_{\substack {0 \mathop \le j \mathop \le n - 2 \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j \paren {n - j} \paren {n - j - 1} x^{n - j - 2} y^j\) as $n - j = 0$ when $j = n$ and $n - j - 1 = 0$ when $j = n - 1$, so terms vanish

and:

\(\ds \map {\dfrac \partial {\partial y} } {\map \Im {z^n} }\) \(=\) \(\ds \sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j j x^{n - j} y^{j - 1}\) Definition of Partial Derivative
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac {\partial^2} {\partial y^2} } {\map \Im {z^n} }\) \(=\) \(\ds \sum_{\substack {0 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j j \paren {j - 1} x^{n - j} y^{j - 2}\) Definition of Partial Derivative
\(\ds \) \(=\) \(\ds \sum_{\substack {2 \mathop \le j \mathop \le n \\ \text {$j$ odd} } } \paren {-1}^{\paren {j - 1} / 2} \dbinom n j j \paren {j - 1} x^{n - j} y^{j - 2}\) as terms vanish when $j = 0$ and $j = 1$
\(\ds \) \(=\) \(\ds \sum_{\substack {0 \mathop \le j \mathop \le n - 2 \\ \text {$j$ odd} } } \paren {-1}^{\paren {j + 1} / 2} \dbinom n {j + 2} \paren {j + 2} \paren {j + 1} x^{n - j - 2} y^j\) Translation of Index Variable of Summation


It remains to be shown that for $j \in \set {0, 1, \ldots, n - 2}$ that:

$\paren {-1}^{\paren {j - 1} / 2} \dbinom n j \paren {n - j} \paren {n - j - 1} = -\paren {-1}^{\paren {j + 1} / 2} \dbinom n {j + 2} \paren {j + 2} \paren {j + 1}$


First we have that:

\(\ds \paren {-1}^{\paren {j + 1} / 2}\) \(=\) \(\ds \paren {-1}^{\paren {j - 1} / 2 + 1}\)
\(\ds \) \(=\) \(\ds -\paren {-1}^{\paren {j - 1} / 2}\)


Hence it remains to be shown that:

$\dbinom n j \paren {n - j} \paren {n - j - 1} = \dbinom n {j + 2} \paren {j + 2} \paren {j + 1}$


We have:

\(\ds \dbinom n j \paren {n - j} \paren {n - j - 1}\) \(=\) \(\ds \dbinom n {n - j} \paren {n - j} \paren {n - j - 1}\) Symmetry Rule for Binomial Coefficients
\(\ds \) \(=\) \(\ds n \dbinom {n - 1} {n - j - 1} \paren {n - j - 1}\) Factors of Binomial Coefficient
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dbinom n j \paren {n - j} \paren {n - j - 1}\) \(=\) \(\ds n \paren {n - 1} \dbinom {n - 2} {n - j - 2}\) Factors of Binomial Coefficient


and:

\(\ds \dbinom n {j + 2} \paren {j + 2} \paren {j + 1}\) \(=\) \(\ds n \dbinom {n - 1} {j + 1} \paren {j + 1}\) Factors of Binomial Coefficient
\(\ds \) \(=\) \(\ds n \paren {n - 1} \dbinom {n - 2} j\) Factors of Binomial Coefficient
\(\ds \) \(=\) \(\ds n \paren {n - 1} \dbinom {n - 2} {n - j - 2}\) Symmetry Rule for Binomial Coefficients
\(\ds \) \(=\) \(\ds \dbinom n j \paren {n - j} \paren {n - j - 1}\) from $(1)$

Hence the result.

$\blacksquare$


Sources

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