Cauchy Riemann Equation

Let \(f:U\to \mathbb{C}\) function which is differentiable and \(U\) is open in \(\mathbb{C}\). Suppose \(f'(z_0) \) exists where \(z_0=a+ib\in \)  \(U\subset \mathbb{C}\). \(f(z)=u+iv\) where \(u:U\to \mathbb{R}\) and \(v:U\to \mathbb{R}\). First take \(h=t\in \mathbb{R}\).$$f'(z_0)  = \lim_{t\to 0}\frac{f(a+t+ib)-f(a+ib)}{t}$$Breaking it we get $$ \lim_{t\to 0} \frac{u(a+t,b)-u(a,b)}{t}+i\lim_{t\to 0}\frac{v(a+t,b)-v(a,b)}{t}= \left.\frac{\partial u}{\partial x}\right|_{z_0}+i\left. \frac{\partial v}{\partial x}\right|_{z_0} \label{cd1}$$

Now take \(h=it\), \(t\in \mathbb{R}\)

$$f'(z_0) = \lim_{t\to 0}\frac{f(a+ib+it)-f(a+ib)}{it}$$

Breaking it down we get $$ \lim_{t\to 0} \frac{u(a,b+t)-u(a,b)}{it}+i\lim_{t\to 0}\frac{v(a,b+t)-v(a,b)}{it}= \left. \frac{\partial v}{\partial y}\right|_{z_0} -i\left. \frac{\partial u}{\partial y}\right|_{z_0}\label{cd2}$$

Equating the two equations we get \(f\) is complex differentiable at \(z_0\) and $$\boxed{\left.\frac{\partial u}{\partial x}\right|_{z_0}=\left.\frac{\partial v}{\partial y}\right|_{z_0}\qquad \left.\frac{\partial v}{\partial x}\right|_{z_0}=-\left.\frac{\partial u}{\partial y}\right|_{z_0}}$$These two equations are called Cauchy Riemann Equation