Suppose that the Cauchy-Riemann equations hold for $u$ and $v$ in their entire domain. Let $h, k \in \R \setminus \left\{ {0}\right\}$, and put $t = h + i k \in \C$. Let $\left({x, y}\right) \in \R^2$ be a point in the domain of $u$ and $v$. Put: :$a = \dfrac{\partial u}{\partial x} \left({x, y}\right) = \dfrac{\partial v}{\partial y} \left({x, y}\right)$ and: :$b = - \dfrac{\partial u}{\partial y} \left({x, y}\right) = \dfrac{\partial v}{\partial x} \left({x, y}\right)$ From the Alternative Differentiability Condition, it follows that: :$u \left({x + h, y}\right) = u \left({x, y}\right) + h \left({a + \epsilon_{u x} \left({h}\right) }\right)$ :$u \left({x, y + k}\right) = u \left({x, y}\right) + k \left({-b + \epsilon_{u y} \left({k}\right) }\right)$ :$v \left({x + h, y}\right) = v \left({x, y}\right) + h \left({b + \epsilon_{v x} \left({k}\right) }\right)$ :$v \left({x, y + k}\right) = v \left({x, y}\right) + k \left({a + \epsilon_{v y} \left({h}\right) }\right)$ where $\epsilon_{u x}, \epsilon_{u y}, \epsilon_{v x}, \epsilon_{v y}$ are continuous real functions that converge to $0$ as $h$ and $k$ tend to $0$. With $z = x + i y$, it follows that: {{begin-eqn}} {{eqn | l = f \left({z + t}\right)       | r = u \left({x + h, y + k}\right) + i v \left({x + h, y + k}\right) }} {{eqn | r = u \left({x, y}\right) + iv \left({x, y}\right) + h \left({a + i b + \epsilon_{u x} \left({h}\right) + i \epsilon_{v x} \left({h}\right)}\right) + k \left({i a - b + \epsilon_{u y} \left({k}\right) + i \epsilon_{v y} \left({k}\right)}\right) }} {{eqn | r = f \left({z}\right) + t \left({a + i b + \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)}\right) }} {{end-eqn}} With the assumption that the partial derivatives are continuous, we have $a + i b$,, which is a function of h, converges to itself by taking h -> 0. {{explain|What does that actually mean: "constant with limit sign"? Can the above statement be explained and proved with a link?}} Thus: {{begin-eqn}} {{eqn | l = \epsilon \left({t}\right)       | r = \dfrac h t \epsilon_{u x} \left({h}\right) + \dfrac h t i \epsilon_{v x} \left({h}\right) - \dfrac k t i \epsilon_{u y} \left({k}\right) + \dfrac k t \epsilon_{v y} \left({k}\right)       | c =  }} {{eqn | ll= \implies       | l = \left\vert{\epsilon \left({t}\right) }\right\vert       | o = \le       | r = \left\vert{\dfrac h t \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t \epsilon_{v y} \left({k}\right) }\right\vert       | c = Triangle Inequality for Complex Numbers }} {{eqn | r = \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert       | c = Complex Modulus of Product of Complex Numbers }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to 0} \left\vert{\epsilon \left({t}\right) }\right\vert       | o = \le       | r = \lim_{t \mathop \to 0} \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{u x} \left({h}\right) }\right\vert + \left\vert{\dfrac h t}\right\vert \left\vert{ \epsilon_{v x} \left({h}\right) }\right\vert       | c = Combined Sum Rule for Limits of Functions }} {{eqn | o =        | ro= +       | r = \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{u y} \left({k}\right) }\right\vert + \left\vert{\dfrac k t}\right\vert \left\vert{ \epsilon_{v y} \left({k}\right) }\right\vert }} {{eqn | o = \le       | r = \lim_{t \mathop \to 0} \left\vert{ \epsilon_{ux} \left({h}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vx} \left({h}\right) }\right\vert       | c = as $\left\vert{\dfrac h t}\right\vert \le 1$ and $\left\vert{\dfrac k t}\right\vert \le 1$ by Modulus Larger than Real Part and Imaginary Part }} {{eqn | o =        | ro= +       | r = \lim_{t \mathop \to 0} \left\vert{ \epsilon_{uy} \left({k}\right) }\right\vert + \lim_{t \mathop \to 0} \left\vert{ \epsilon_{vy} \left({k}\right) }\right\vert }} {{eqn | r = 0 }} {{end-eqn}} {{explain|It is not immediately apparent why $\left\vert{\dfrac h t}\right\vert \le 1$ and $\left\vert{\dfrac k t}\right\vert \le 1$ by Modulus Larger than Real Part and Imaginary Part.}} This shows that: : $\displaystyle \lim_{t \mathop \to 0} \epsilon \left({t}\right) = 0$ From Continuity of Composite Mapping, it follows that $\epsilon \left({t}\right)$ is continuous. Then the Alternative Differentiability Condition shows that: :$f' \left({z}\right) = a + i b$ {{qed|lemma}}