If $u : \mathbb{C} \to \mathbb{R}$ satisfies $$u(x + iy) =\frac{1}{4}[u(x + a + iy) + u(x − a + iy) + u(x + i(y + a)) + u(x + i(y − a))]\tag{$*$}$$
for all $a$ then determine whether $u$ is harmonic, where (a) $a\in \mathbb{R}$, (b) $a\in \mathbb{C}$.
Attempt: $(*)$ is true iff $$u(x + a + iy) + u(x − a + iy) + u(x + i(y + a)) + u(x + i(y − a))-4u(x+iy)=0\\ $$ In the latest equation we have four terms $u(x\pm a+iy)-u(x+iy),u(x+i(y\pm a)-u(x+iy)$, so for $a\not=0$ we divide by $a$ and take the limit as $a\to 0$ to obtain $$2u_x+2u_y=0\tag{$**$}$$ I am not sure what to do at this point. Taking more partial derivatives doesn't get me any further. Is there a nicer way to go about this? I don't think this technique applies if $a\in \mathbb{C}$.
Any real-linear function $\phi(x,y):=px+qy$ trivially satisfies $(*)$. Therefore it is impossible to deduce $(**)$.
If you assume $u\in C^2$ to begin with then you can argue as follows (I shall treat the case $a\in{\Bbb R}$, which is sufficient): For given $(x,y)$ one has $$u(x+a,y)-u(x,y)=u_x(x,y) a+{1\over2}u_{xx}(x,y)a^2 +o(a^2)\qquad(a\to0)$$ and three more equations of this kind. Adding these four equations up we obtain, using $(*)$: $$\eqalign{0&=\bigl(u(x+a,y)+u(x,y+a)+u(x-a,y)+u(x,y-a)\bigr)-4u(x,y)\cr &=\bigl(u_{xx}(x,y)+u_{yy}(x,y)\bigr) a^2+o(a^2)\qquad(a\to0)\ .\cr}$$ This implies $$\Delta u(x,y)=0\ ,$$ and as $(x,y)$ was arbitrary it follows that $u$ is harmonic.