Let $z = x+iy$ denote the input into a complex-valued function $f$, and let $f(z) = u(z) + iv(z)$ denote the output. Letting $\delta$ denote a small change in the real direction (positive or negative), then the derivative of $f$ at $z_0 = x_0 + iy_0$ must satisfy \begin{eqnarray*} f'(z_0) & = & \lim_{\delta \to 0} \frac{u(x_0+\delta + iy_0) +iv(x_0+\delta + iy_0)- u(x_0+iy_0) - iv(x_0+iy_0)}{\delta} \\ & = & \lim_{\delta \to 0} \frac{u(x_0+\delta + iy_0) - u(x_0+iy_0) }{\delta} + \lim_{\delta \to 0} i \frac{v(x_0+\delta + iy_0)- v(x_0+iy_0)}{\delta} \\ & = & \left.\frac{\partial u}{\partial x}\right|_{x_0 + i y_0} + i \left.\frac{\partial v}{\partial x}\right|_{x_0+iy_0} \\ & & {\rm or\ more \ simply}\\ \frac{df}{dz} & = & \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}. \end{eqnarray*}

But we need to get the same result if we approach $z_0 = x_0 + iy_0$ from the imaginary direction.

\begin{eqnarray*} f'(z_0) & = & \lim_{\delta \to 0} \frac{u(x_0+\ iy_0 + i\delta) +iv(x_0 + iy_0)- u(x_0+iy_0+ i\delta) - iv(x_0+iy_0)}{i\delta} \\ & = & \lim_{\delta \to 0} \frac{u(x_0+ iy_0 + i\delta) - u(x_0+iy_0) }{i\delta} + \lim_{\delta \to 0} i \frac{v(x_0 + iy_0+ i\delta)- v(x_0+iy_0)}{i\delta} \\ & = & \frac{1}{i}\left.\frac{\partial u}{\partial y}\right|_{x_0 + i y_0} + \left.\frac{\partial v}{\partial y}\right|_{x_0+iy_0} \\ & & {\rm or\ more \ simply}\\ \frac{df}{dz} & = & -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}. \end{eqnarray*}

This leads us to what are known as the Cauchy-Riemann equations, first stated by d'Alembert in 1752, long before Cauchy or Riemann were born. Cauchy observed these conditions as a desirable property of a complex-valued function of a complex variable, but it was Riemann in his doctoral thesis of 1851 who built the study of complex analysis around the requirement of differentiability.

In order for the derivative of $f$ to exist at a point $z$, it is necessary that $$ \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y} $$ which implies that $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad {\rm and} \quad \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} . $$ More simply, this condition can be stated as $$ i\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}. $$

This is a necessary condition. It is not quite sufficient because we must guarantee that we get the same limit no matter from which direction we approach $z_0 = x_0 + iy_0$. I will spare you the details, but it is enough that the partial derivatives of $u$ and $v$ exist and are continuous in some open set containing $z_0$. (Recall that a set is open if every element of the set is the center of an open disc---which could be extremely small but must have positive radius---contained within the set.)

So what functions are analytic? We start with $z^n$ for a positive integer $n$, $$ f(z) = z^n = (x+iy)^n = \sum_{j=0}^n \frac{n!}{ j!\,(n-j)!} x^j i^{n-j} y^{n-j}. $$ The two partial derivatives we need to examine are \begin{eqnarray*} i\frac{\partial f}{\partial x} & = & i \sum_{j=0}^n \frac{n!}{ j!\,(n-j)!} \, j \, x^{j-1} i^{n-j} y^{n-j}\\ & = & \sum_{j=1}^n \frac{n!}{ (j-1)!\,(n-j)!}\, x^{j-1} i^{n-j+1} y^{n-j} \\ & = & \sum_{j=0}^{n-1} \frac{n!}{ j!\,(n-j-1)!}\, x^j i^{n-j} y^{n-j-1}, \end{eqnarray*} where we have replaced $j$ by $j+1$ in the last line, and \begin{eqnarray*} \frac{\partial f}{\partial y} & = & \sum_{j=0}^n \frac{n!}{ j!\,(n-j)!}\, x^j\, i^{n-j} (n-j) y^{n-j-1}\\ & = & \sum_{j=0}^{n-1} \frac{n!}{ j!\,(n-j-1)!}\, x^j\, i^{n-j} y^{n-j-1}. \end{eqnarray*}

Since the partial derivatives are clearly continuous, the function defined by $f(z) = z^n$ is differentiable. Furthermore, since the partial derivatives satisfied the required equation, we can find the derivative by taking the partial derivative with respect to $x$: \begin{eqnarray*} \frac{df}{dz}\ = \ \frac{\partial f}{\partial x} & = & \sum_{j=0}^{n-1} \frac{n!}{ j!\,(n-j-1)!} \,x^j\, i^{n-j-1} y^{n-j-1}\\ & = & n \sum_{j=0}^{n-1} \frac{(n-1)!}{ j!\,(n-j-1)!} \,x^j\, i^{n-j-1} y^{n-j-1} \\ & = & n(x+iy)^{n-1}\ =\ nz^{n-1}. \end{eqnarray*} Whew. We would have been in trouble if it had been anything else.