Show that $Im(f(re^{i\theta})) = r \sin(\theta)(1+O(r))$ for small $r$

Question

Suppose f is a analytic function such that $f(0) = 1 = f'(0)$ and $/f^{(k)}(0) \in \mathbb{R} \; \forall k$

Show that $v(r,\theta) = Im(f(re^{i\theta})) = r \sin(\theta)(1+O(r))$ for small $r$.

I need to show this, but I don't believe it's true, here's why:

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I know that $f(re^{i\theta}) = \sum a_kr^k(\cos(k\theta)+i\sin(k\theta)) \implies v(r,\theta) = \sum a_kr^k \sin(k\theta)$

where $a_k = \frac{f^{(k)}(0)}{k!}$.

Now, $v(r,\theta) = 1+r\sin(\theta)+\sum_{k\geq 2}a_kr^ksin(k\theta) = r\sin(\theta)\left(1+\frac{1}{r\sin(\theta)}+\sum_{k\geq 2}a_kr^{k-2}\frac{sin(k\theta)}{sin(\theta)}\right)$

Now, $\frac{1}{r\sin(\theta)}$ is not limited by $Cr$ for any constant $C$ for small r, so, how can $\frac{1}{r\sin(\theta)}+\sum_{k\geq 2}a_kr^{k-2}\frac{sin(k\theta)}{sin(\theta)} = O(r)$ ?

I'm sorry, I'm not used to the big Oh notation, regardless I don't know how that statement could be true.

Answer 1

You've made a mistake: The constant term in the expansion of $v(re^{i\theta})$ isn't $1,$ it's $0.$ Thus

$$\text {Im } f(re^{i\theta}) = r\sin \theta + a_2r^2\sin (2\theta) + \cdots.$$

To finish, you may find the inequality $|\sin (n\theta)|\le n|\sin \theta|$ useful (you can prove this by induction).