Solving a complex equation.

Question

Let $\gamma : [0, 2\pi] \to \mathbb C$ be defined as $\gamma(t)=\frac{\sqrt{3}}{2}+\frac{i}{2}+e^{it}$. Then the image of $\gamma$ is a circle centered at $\frac{\sqrt{3}}{2}+\frac{i}{2}$ and of radius $1$. This circle passes through $0$, and I want to find $t_0 \in [0, 2\pi]$ such that $\gamma(t_0)=0$. So, I want to solve the equation $\gamma(t)=\frac{\sqrt{3}}{2}+\frac{i}{2}+e^{it}=0$, and I get $t = \frac{1} {i} \text{Log} \left( \frac{-\sqrt{3}}{2}-\frac{i}{2}\right)= \text{Arg} \left( \frac{-\sqrt{3}}{2}-\frac{i}{2}\right)-i$, but this $t$ is not real, so it's not in $[0, 2\pi]$. What am I doing wrong?

Answer 1

$$\gamma(t)=0\iff e^{\pi i/6} +e^{it}=0\iff e^{it}=-e^{\pi/6}=e^{7\pi i/6}\iff t=\frac{7\pi}6$$

keeping in mind that we have chosen a specific value for the angle and etc.

Answer 2

$\gamma(t_0) = 0$ if $\displaystyle e^{it} = -\frac{\sqrt{3}}{2} - \frac{i}{2} = e^{(7\pi /6 + 2k\pi)i}$ for $k \in \mathbb{Z}$. Hence ...