Solve the following equation for $z\in \mathbb{C}$: $\text{Log}(z)-\text{Log}\left(\frac{1}{z}\right)=1$ where $\text{Log}{z}=\ln{r}+i\Theta$ for $-\pi<\Theta<\pi$ and $z\neq 0$. This is what I have so far:
\begin{align*} \text{Log}(z)-\text{Log}\left(\frac{1}{z}\right)&=1\\ \text{Log}{\frac{z}{\frac{1}{z}}}&=1\\ \text{Log}{(z^2)}&=1\\ 2\text{Log}{z}&=1\\ 2\left[\ln{r}+i\Theta\right]&=1\\ 2\ln{r}+i2\Theta&=1+i0\\ \end{align*}
From here we are able to break up the equation into its reals and imaginary components. The real equation is $2\ln{r}=1\rightarrow\ln{r}=\frac{1}{2}\rightarrow r=e^{\frac{1}{2}}$. And the imaginary parts can be broken up into $2\Theta=0\rightarrow \Theta=0$.
Thus, the solution to is $z=e^{\frac{1}{2}}$.