# A step in deriving the entire function is a polynomial is the real part of the entire function is bounded by a polynomial

I am trying to understand ts375_zk26's answer to the question that if the real part of an entire function is bounded by a polynomial, then the function is itself a polynomial.

In the answer given by ts375_zk26, the following statement is claimed:

Let $$f(z)=u+iv=\sum_{k=0}^\infty a_kz^k$$ and $$A(r)=\max _{|z|=r} u(z)$$.
It is well-known that for $$k\ge 1$$ $$a_kr^k=\frac{1}{\pi}\int_0^{2\pi} u(re^{i\theta })e^{-ik\theta }d\theta .$$

However, I have never seen this "well-known" identity before. By applying the Cauchy's integral formula, I get $$a_kr^k=\frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})e^{-ik\theta}d\theta+\frac{1}{2\pi}\int_0^{2\pi}v(re^{i\theta})e^{-ik\theta}id\theta .$$ But I cannot see the reason why $$a_kr^k=\frac{1}{\pi}\int_0^{2\pi} u(re^{i\theta })e^{-ik\theta }d\theta.$$

Cauchy's integral formula gives for $$k \ge 0$$ $$\tag{1} a_k = \frac{1}{2 \pi i} \int_{|z|=r} \frac{f(z)}{z^{k+1}} \, dz = \frac{1}{2 \pi r^k} \int_0^{2 \pi } f(re^{i\theta}) e^{-ik \theta} \, d\theta \, .$$ Cauchy's integral theorem gives for $$k \ge 1$$ $$\tag 2 0 = \frac{1}{2 \pi i} \int_{|z|=r} f(z) z^{k-1} \, dz = \frac{1}{2 \pi r^k} \int_0^{2 \pi} f(re^{i\theta}) e^{ik \theta} \, d\theta \, .$$
Now take the conjugate of $$(2)$$ and add it to $$(1)$$: $$a_k = \frac{1}{2 \pi r^k} \int_0^{2 \pi} 2 \operatorname{Re}f(re^{i\theta}) e^{-ik \theta} \, d\theta$$ which is the desired formula.