Let $\Omega$ be a domain in $\mathbb{C}$, and let $z_0 \in \Omega$. Let $f$ be analytic on $\Omega$.
Let $z=z_0+re^{i\theta}$ for $r$ small.
Assume that $f(z_0) \neq 0$ and $f'(z_0) \neq 0$.
I want to show $|f(z)| = |f(z_0)|\big(1+\lambda r \cos(\theta+\phi)+O(r^2)\big)$ where $\displaystyle \frac{f'(z_0)}{f(z_0)}=\lambda e^{i\phi}, \lambda > 0.$
Now:
$f(z)=f(z_0)+f'(z_0)(z-z_0)+O((z-z_0)^2) = $ $ f(z_0)+f'(z_0)(re^{i\theta})+O(r^2) =$
$|f(z_0)|\big | 1+\lambda r e^{i(\theta+\phi)} + O(r^2) \big | =$ $|f(z_0)|\big | 1+\lambda r (\cos(\theta+\phi)+i\sin(\theta + \phi)) + O(r^2) \big |$
It is not clear to me how I finish this off...
Let us for simplicity write $\psi = \theta + \phi$. When dealing with absolute moduli, it is often convenient to square them. Here, that yields
$$\begin{align} \lvert 1 + \lambda re^{i\psi}\rvert^2 &= \lvert (1+\lambda r\cos\psi) + i\lambda r\sin\psi\rvert^2\\ &= (1+\lambda r\cos\psi)^2 + \lambda^2 r^2\sin^2\psi\\ &= 1 + 2\lambda r\cos\psi + \lambda^2r^2. \end{align}$$
Taking the square root, using $\sqrt{1+x} = 1 + \frac{x}{2} + O(x^2)$ yields
$$\lvert 1 +\lambda re^{i\psi}\rvert = 1 + \lambda r\cos\psi + \frac12\lambda^2r^2 + O(\lambda^2r^2) = 1 + \lambda r\cos\psi + O(r^2)$$
as desired.