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

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