I am facing hard in complex analysis and so can't understand why the following steps happened? If $z,z+\delta z$ are two points in complex plane and to differentiate a complex function $f(z)$ we do,
$$\frac{df}{dz}=\lim_{\delta z \to 0} \frac{f(z+\delta z)-f(z)}{\delta z}$$
for  $\delta z=\epsilon e^{i\alpha}$. So 
$$\frac{df}{dz}=e^{-i\alpha}\left(\frac{\partial u}{\partial x}\cos (\alpha) +i\frac{\partial v}{\partial y}\sin (\alpha)\right)+\left(\frac{\partial u}{\partial y}\sin (\alpha) +i\frac{\partial v}{\partial x}\cos (\alpha)\right)$$ 
I am unable to understand the last step, if anyone helps it is apreciable
 A: If $\delta z=\epsilon e^{i\alpha}$, then $\delta z=\epsilon \cos(\alpha)+i\epsilon \sin(\alpha)$.  Writing $f(z)=u(x,y)+iv(x,y)$, we have
$$\begin{align}
f'(z)&=\lim_{\delta z\to 0}\frac{f(z+\delta z)-f(z)}{\delta z}\\\\
&=\lim_{\epsilon \to 0}\left(\frac{u(x+\epsilon \cos(\alpha),y+\epsilon \sin(\alpha))-u(x,y)}{\epsilon e^{i\alpha}}+i\frac{v(x+\epsilon \cos(\alpha),y+\epsilon \sin(\alpha))-v(x,y)}{\epsilon e^{i\alpha}}\right)\\\\
&=e^{-i\alpha}\cos(\alpha)\left(\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x}\right)+e^{-i\alpha}\sin(\alpha)\left(\frac{\partial u(x,y)}{\partial y}+i\frac{\partial v(x,y)}{\partial y}\right) \tag 1
\end{align}$$
If $f(z)$ is differentiable, then the result in $(1)$must be independent of $\alpha$.  If $\alpha=0$, we obtain
$$f'(z)=\frac{\partial u(x,y)}{\partial x}+i\frac{\partial v(x,y)}{\partial x} \tag 2$$
while if $\alpha =\pi/2$, we obtain
$$f'(z)=\frac{\partial v(x,y)}{\partial y}-i\frac{\partial u(x,y)}{\partial y} \tag 3$$
Equating the real and imaginary parts of $(2)$ and $(3)$ yields the Cauchy-Riemann Equations
$$\frac{\partial u(x,y)}{\partial x}=\frac{\partial v(x,y)}{\partial y}$$
and
$$\frac{\partial u(x,y)}{\partial y}=-\frac{\partial v(x,y)}{\partial x}$$
A: Fix $\alpha$, and let $\epsilon\rightarrow0$. Then we want the limit
$$ \lim_{\epsilon\rightarrow0} \frac{f(z+\delta z)-f(z)}{\delta z}$$
Using L'Hospital's rule (differentiating with respect to $\epsilon$) gives
$$ \lim_{\epsilon\rightarrow0}\frac{\frac{df}{d\epsilon}}{e^{i\alpha}}$$
And now just note that 
$$ \frac{df}{d\epsilon} = f_x\frac{dx}{d\epsilon} + f_y\frac{dy}{d\epsilon}$$
where
\begin{align}
x &= \epsilon\cos\alpha\\
y &= \epsilon\sin\alpha
\end{align}
