Finding values of $\tan^{-1} (2i)$.

Question

I am trying to find all solutions of $\tan^{-1} (2i)$. I don't see anything that I have done wrong, my answer doesn't match the one in the textbook. Here is what I have. (The convention in Brown and Churchill is to use $\log$ for a complex number and $\ln$ for a real number.) \begin{align*} \tan^{-1} (2i) & = \frac{i}{2} \log \frac{i + 2i}{i - 2i} \\ & = \frac{i}{2} \log \frac{3i}{-i} \\ & = \frac{i}{2} \log (-3) \\ & = \frac{i}{2} \left(\ln 3 + (2n + 1) \pi i \right) \\ & = \frac{i}{2} \ln 3 - \frac{2n + 1}{2} \pi \\ & = \frac{i}{2} \ln 3 - \left(n + \frac{1}{2}\right) \pi. \end{align*} The answer in the textbook, however, is: \begin{align*} \frac{i}{2} \ln 3 + \left(n + \frac{1}{2}\right)\pi. \end{align*} This leads me to believe that I have misplaced a sign somewhere, but I cannot see where. Might there just be a typo in the book?

Thanks in advance.

Answer 1

Both are correct.

If $n_1$ is the integer plugged into your solution, plugging $-n_1-1$ into the textbook solution gives the same answer. $$-\left(n_1+\frac{1}{2}\right) = k$$ $$(-n_1-1)+\frac{1}{2} = k$$

Answer 2

Since $\tan(x)=2i$

Using Euler's formula $\tan(x)={\frac {e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}}$ witch must equal $2i$

Manipulating a bit we get $3e^{ix}+e^{-ix}=0$ and if $e^{ix}=y$, $3y+\frac{1}{y}=0$ thus $e^{ix}=y=±i \frac{\sqrt 3}{3}$

This, of course, means that $\ln\left(±i \frac{\sqrt 3}{3}\right)=ix$

Answer 3

$-\left(n_1+\dfrac12\right)$ will be $=n_2+\dfrac12$

$$\iff n_2=-n_1-1$$

As $n_1$ can be any integer, so will be $n_2$ and conversely.