Harmonic polynomials property The textbook i am currently reading states that:
If $a_1^2+ \cdots + a_n^2 = 0$ and $a_i \in \mathbb{C}$ then the polynomial
$$ f(x) = (a_1x_1 + \cdots a_nx_n)^m $$
is a harmonic polynomial and homogeneous of degree $m$.
I just just can seem to figure out why this is true?? Well, i tried computating for n = 2 and m= 2. And i could see that it was true. But how would i see it for arbitrary n and m??.
 A: While many people regard dots ("$\cdots$") as easier to read than sigma notation, and while many textbooks pose problems involving "$\cdots$" instead of sigma notation, sigma notation is often the simplest way of performing calculations mechanically, without cleverness or effort, when the number of terms in a sum is not fixed in advance.  In the absence of better approaches you should always default to using sigma notation.
Write $S(x) = \sum_{j=1}^n a_j x_j$.  Then $f(x) = S(x)^m$ and by the chain rule the partial derivatives of $f$ will involve the partial derivatives of $S$.  So let's compute those.
For any $1 \leq k \leq n$
$$
\frac{\partial S}{\partial x_k} (x) = \frac{\partial}{\partial x_k} \sum_{j=1}^n a_j x_j = \sum_{j=1}^n a_j \frac{\partial x_j}{\partial x_k} = a_k,
$$
because $\frac{\partial}{\partial x_k}$ is a linear operator (i.e. it can be brought inside the sum indicated by sigma notation, and have the scalars $a_j$ pulled out of it) and because $\frac{\partial x_j}{\partial x_k}$ is $0$ when $j \neq k$ and is $1$ when $j = k$.
It follows that
$$
\frac{\partial f}{\partial x_k} (x) = m S(x)^{m-1} \frac{\partial S}{\partial x_k} (x) = m S(x)^{m-1} a_k, \qquad 1 \leq k \leq n,
$$
by application of the chain rule and the use of the fact deduced above.
Applying $\frac{\partial}{\partial x_k}$ to the last equation and again using what we calculated about $\frac{\partial S}{\partial x_k}$ we deduce
$$
\frac{\partial^2 f}{\partial x_k^2} (x) = m a_k (m-1) S(x)^{m-2} \frac{\partial S}{\partial x_k} (x) = m a_k (m-1) S(x)^{m-2} a_k = a_k^2 m(m-1) S(x)^{m-2}.
$$
Thus
$$
\sum_{k=1}^n \frac{\partial^2 f}{\partial x_k^2} (x) = \sum_{k=1}^n a_k^2 m(m-1) S(x)^{m-2}  = m(m-1) S(x)^{m-2} \sum_{k=1}^n a_k^2
$$
and if the sum at the extreme right is zero then $f$ will be harmonic.
As for proving that $f$ is a homogeneous polynomial, this may depend on how you have defined "homogeneous polynomial."  If a polynomial in multiple variables is homogeneous if it is a sum of terms of the same total degree, then a similar use of sigma notation can establish the multinomial theorem (see https://en.wikipedia.org/wiki/Multinomial_theorem) which will show explicitly that $f$ is homogeneous of degree $m$.
Alternatively, should you happen to know that a polynomial $f(x)$ is homogeneous of degree $m$ if and only if $f(\lambda x) = \lambda^m f(x)$ holds for all $x$ and all scalars $\lambda$ (this is sometimes taken to be the definition of "homogeneous," but sometimes not), this provides another way of approaching the problem without fully expanding the product.  The calculation that helps you here is that $S(\lambda x) = \lambda S(x)$ (try proving it using sigma notation) so that $f(\lambda x) = \lambda^m S(x)^m = \lambda^m f(x)$.
To reiterate the key point: try sigma notation, not $\cdots$, when asked to prove anything about a sum whose number of terms is not known in advance.
