Why does $\arg (f \circ \gamma)' (t_0) = \arg (f'(z_0)) + \arg (\gamma' (t_0))\ $?

Question

Can anybody help me understanding as to why $\arg (f \circ \gamma)' (t_0) = \arg (f'(z_0)) + \arg (\gamma' (t_0)) $ holds?

I know that $\arg (z_1z_2) \neq \arg (z_1) + \arg (z_2)$ unless $-\pi \lt \arg (z_ 1) + \arg (z_2) \leq \pi.$

How do I make sure that $\arg (f'(z_0)) + \arg (\gamma'(t_0)) \in (-\pi, \pi]\ $? Any help in this regard will be highly appreciated.

Thanks in advance.

Answer 1

Use the formula for the derivative of the composition of two functions:

$$(f \circ \gamma)' (t_0) = f'(z_0)\gamma' (t_0)$$

then, take the argument of both sides:

$$\arg (f \circ \gamma)' (t_0) = \arg (f'(z_0)) + \arg (\gamma' (t_0))$$