Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
add comment

1 Answer 1

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 $\theta = \pi$, $\text{Log}(1/z) = -\log r + \pi i = - \text{Log}(z) + 2 \pi i$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.