1. ## cosine inverse

Solve the following equation:

$\displaystyle \cos z = 3 - 2i$.

My book says that

$\displaystyle \cos^{-1}z= -i \log[z+i(1-z^2)^{\frac{1}{2}}]$. It also says that $\displaystyle (1-z^2)^{\frac{1}{2}}$ is a double valued fuction. How do I use this formula or is there a better way to do this? I don't see how to get the double value it is talking about.

2. Originally Posted by xboxlive89128
Solve the following equation:

$\displaystyle \cos z = 3 - 2i$.

My book says that

$\displaystyle \cos^{-1}z= -i \log[z+i(1-z^2)^{\frac{1}{2}}]$. It also says that $\displaystyle (1-z^2)^{\frac{1}{2}}$ is a double valued fuction. How do I use this formula or is there a better way to do this? I don't see how to get the double value it is talking about.

let

$\displaystyle x = \cos ^{-1} z$ x here present cos inverse of z we want to find x value without using cos function , x is a complex number

so

$\displaystyle \cos x = z$

$\displaystyle \frac{e^{ix}+e^{-ix}}{2} = z$

$\displaystyle e^{ix} + e^{-ix} =2z$

$\displaystyle e^{2ix} - 2z e^{ix} +1 = 0$

let $\displaystyle v=e^{ix}$

$\displaystyle v^2 -2z v +1 =0$

$\displaystyle v=\frac{-(-2z)\mp \sqrt{4z^2 - 4}}{2}$

$\displaystyle e^{ix} = \frac{2z \mp 2\sqrt{z^2-1}}{2}$

$\displaystyle e^{ix} = z\mp \sqrt{z^2-1}$

$\displaystyle ix = \ln (z + \sqrt{z^2 -1 } )$

$\displaystyle x = -i \ln (z+\sqrt{z^2-1})$

$\displaystyle x = -i \ln (z+\sqrt{-1(1-z^2)}$

$\displaystyle x = -i \ln (z +\sqrt{-1}(\sqrt{1-z^2}))$

$\displaystyle x = -i \ln (z+i \sqrt{1-z^2})=\cos ^{-1} z$ End

3. Originally Posted by xboxlive89128
I don't see how to get the double value it is talking about.
Because the square root is multi-valued and defined as:

$\displaystyle \sqrt{z}=r^{1/2}e^{i/2(\theta+2k\pi)},\quad k=0,1$

You've seen that formula right? And also, the log function is infinitely valued:

$\displaystyle \log(z)=\ln r+i(\theta+2k\pi),\quad k=0,\pm 1, \pm 2,\cdots$

Therefore, $\displaystyle cos^{-1}(z)$ is infinitely-valued.