Why there exists $\theta_0\in [0,\pi]$ such that $\Im m (g(\theta_0))=0$?

Question

Define $g:\mathbb{R}\to \mathbb{C}$ by $$g(\theta)=a e^{-i\theta}+be^{i\theta},\;\theta\in \mathbb{R},$$ with $a,b\in \mathbb{C}$.

Why is there a $\theta_0\in [0,\pi]$ such that $\Im m (g(\theta_0))=0$?

Answer 1

$f(\theta)=\Im m \,g(\theta)$ is a continuous function on $[0,\pi]$ whose values at $0$ and $\pi$ are $\alpha+\beta$ and $-\alpha-\beta$ where $\alpha =\Im m \,a$ and $\beta =\Im m \,b$. Since $0$ lies between these values the result follows by IVP.

Answer 2

Say $a$ is $\rho_a e^{i\theta_a}$ and $b$ is $\rho_b e^{i \theta_b}$.

Then $\theta_0$ is the solution to the equation

$$\rho_a(\sin(\theta_a - \theta_0)) + \rho_b(\sin(\theta_b + \theta_0)) = 0$$

The best way to show this equation has a solution is by defining

$$f(\theta) = \rho_a(\sin(\theta_a - \theta)) + \rho_b(\sin(\theta_b + \theta))$$

Which is a continuous function of $\theta$. Can you use the intermediate value theorem to show this function has a $0$?