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

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:

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.

• You seem to be correct, this would be the case only if $f(0)=0$. – DinosaurEgg Oct 2 '18 at 20:52
• What does $/f^{(k)}(0)$ mean? – zhw. Oct 4 '18 at 15:48
• @zhw. It means the k-th derivative of f on 0. I've found out today that the I was asked to prove a false statement because of a typo of my teacher. He meant to use a lower case 'o', not a upper case one. – Matheus barros castro Oct 16 '18 at 20:31

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).