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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 $z=e^{\frac{1}{2}}$.

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It should be $2 \Theta = 0$, not $2 \Theta = 1$. What's the imaginary part of $1$? – Antonio Vargas Nov 5 '12 at 23:26

Caution: if $z = r e^{i\theta}$, $1/z = (1/r) e^{-i\theta}$, but if $-\pi < \theta \le \pi$ you have $-\pi \le -\theta < \pi$, not $-\pi < \theta \le \pi$. So $\text{Log} (1/z) = - \text{Log}(z)$ if $-\pi < \theta < \pi$, but if $\theta = \pi$, $\text{Log}(1/z) = -\log r + \pi i = - \text{Log}(z) + 2 \pi i$.

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