In order to define integral of $f(z)$ in a simple we need to first consider a derivative of complex function of a real variable. $w(t)$
$$ w(t) = u(t) + iv(t) $$
Where both $u$ and $v$ are real valued functions and $t \in \R$. So the derivative of $w(t)$ is defined as follow (From rules in real numbers)
$$ w'(t) = u'(t) + iv'(t) $$
Example: $[z_0w(t)]'$
$$ [z_0w(t)] = [z_0u(t) + iz_0v(t)]' = [z_0u(t)]' + i[z_0v(t)]' $$
$$ [z_0u(t)]' = [(x_0+iy_0)u(t)]' = [x_0u'(t)+iy_0u'(t)] = z_0u'(t) $$
$$ [z_0v(t)]' = [(x_0+iy_0)v(t)]' = [x_0v'(t)+iy_0v'(t)] = z_0v'(t) $$
Thus
$$ [z_0w(t)]' = [z_0u(t)]' + i[z_0v(t)]' = z_0u'(t) + iz_0v'(t) = z_0w'(t) $$
<aside> 📢
Note: It is not always the case that same rules from real calculus are true for complex functions. You MUST check before using them.
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The mean value theorem from calculus is no longer applies when working with complex function of real variable. For example, consider the function $w(t) = e^{it}$ in the interval $0\le t\le2\pi$, you can also verify that $w'(t) = ie^{it}$. By mean value theorem we have
$$ \exist_c w'(c) = \frac{w(a) - w(b)}{a-b} $$
And from the fact that $w(0) = w(2\pi) = e^{2\pi}$ then $w(0) - w(2\pi) = 0$, So by mean value theorem we must have:
$$ \exist_c \space ie^{it} = 0 $$
Which is not possible.
<aside> ⚠️
Mean value theorem does not apply to complex function
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