MHB Plus or minus question for the complex log

[cbarker1]

Dear Everybody, I am having troubles figuring out why the plus or minus sign in this problem. The question is:

Solve the equation $\sin\left({z}\right)=2$ for $z$ by using $\arcsin\left({z}\right)$

The work for this problem is the following:
$\sin\left({z}\right)=2$
$z=\arcsin\left({2}\right)$
$=-i\log\left({2i+\sqrt{1-{2}^{2}}}\right)$
$=-i\log\left({2i+i\sqrt{3}}\right)$
$=-i\log\left({i\left(2+\sqrt{3}\right)}\right)$
$=-i\left[\log\left({i}\right)+\log\left({2+\sqrt{3}}\right)\right]$
$=-i\left[\log\left({i}\right)+\ln\left({2+\sqrt{3}}\right)\right]$
$=-i\left[\ln\left({1}\right)+i\frac{\pi}{2}+i2n\pi+\ln\left({2+\sqrt{3}}\right)\right]$
$=-i\left[\ln\left({2+\sqrt{3}}\right)+i\pi(2n+\frac{1}{2})\right]$
$=\pi(2n+\frac{1}{2})-i\ln\left({2+\sqrt{3}}\right)$ where $n\in\Bbb{Z}$

And the answer in the book:
$=\pi(2n+\frac{1}{2})\pm i\ln\left({2+\sqrt{3}}\right)$ where
$n\in\Bbb{Z}$

Where did I make a mistake?

Thanks,
Cbarker1

 

[Euge]

Hi Cbarker1,

You had $\sqrt{1-2^2} = i\sqrt{3}$ but isn't it $\pm i\sqrt{3}$ as there are two branches of the square root? If we consider $-i\sqrt{3}$ then we would have an $\ln(2-\sqrt{3})$ term which is the same as $-\ln(2 + \sqrt{3})$ because $(2 - \sqrt{3})(2 + \sqrt{3}) = 1$.