How do I write arccos 4 as a + ib (complex number)?

ssadi · #1 $Old$ November 27th, 2008, 09:06 AM

This a question from complex number chapter of fp3
Find the following in the form a+ib, a,b member of real number set:
(a) arccos 4 (b) arcsin 2 (c) arcsin i
I tried to these sums by: x=arccos4, so cosx=4, so e^(ix)-e^-(ix)=8
The answers:
(a) My answer:+/- i arcosh 4
Book's answer: 2m*pi +/- i arcosh 4
(b) My answer: +/- arcosh 2(i am not sure how i get that)
Book's answer: (4m+1)*pi/2 +/- i arcosh 2
(c) My answer: i ln ((-2+/- sqrt 5)/2))
Book's answer: n*pi + i arsinh [(-1)^n]

So I gathered that I am missing something major here. Please anyone tell me the correct method to do these sums, thus enabling me to try it out myself. Just tell me the method, please

This is my first sums of this kind:woo:
P.s. i=imaginary.

Mathstud28 · #2 $Old$ November 27th, 2008, 10:28 AM

Quote:

Originally Posted by ssadi $View Post$

This a question from complex number chapter of fp3
Find the following in the form a+ib, a,b member of real number set:
(a) arccos 4 (b) arcsin 2 (c) arcsin i
I tried to these sums by: x=arccos4, so cosx=4, so e^(ix)-e^-(ix)=8
The answers:
(a) My answer:+/- i arcosh 4
Book's answer: 2m*pi +/- i arcosh 4
(b) My answer: +/- arcosh 2(i am not sure how i get that)
Book's answer: (4m+1)*pi/2 +/- i arcosh 2
(c) My answer: i ln ((-2+/- sqrt 5)/2))
Book's answer: n*pi + i arsinh [(-1)^n]

So I gathered that I am missing something major here. Please anyone tell me the correct method to do these sums, thus enabling me to try it out myself. Just tell me the method, please

This is my first sums of this kind:woo:
P.s. i=imaginary.

The first thing you might want to check is in your first step it should be $e^{ix}{\color{red}+}e^{-ix}=8$

Or what you might want to try is solving $y=\frac{e^{ix}+e^{-ix}}{2}$ for $y$ by letting $e^{ix}=\varphi$ . But be sure to remember that $e^{ix}$ is a many valued (many-to-one) function so there isn't going to be just one number as the answer.

ssadi · #3 $Old$ November 27th, 2008, 10:30 AM

Quote:

Originally Posted by Mathstud28 $View Post$

The first thing you might want to check is in your first step it should be $e^{ix}{\color{red}+}e^{-ix}=8$

Or what you might want to try is solving $y=\frac{e^{ix}+e^{-ix}}{2}$ for $y$ by letting $e^{ix}=\varphi$ . But be sure to remember that $e^{ix}$ is a many valued (many-to-one) function so there isn't going to be just one number as the answer.

oh i typed wrong here, in the sum i used e^ix + e^-ix :embarassed:

ssadi · #4 $Old$ November 27th, 2008, 10:31 AM

Quote:

Originally Posted by Mathstud28 $View Post$

The first thing you might want to check is in your first step it should be $e^{ix}{\color{red}+}e^{-ix}=8$

Or what you might want to try is solving $y=\frac{e^{ix}+e^{-ix}}{2}$ for $y$ by letting $e^{ix}=\varphi$ . But be sure to remember that $e^{ix}$ is a many valued (many-to-one) function so there isn't going to be just one number as the answer.

and from where the m and n comes into book's answer? I am really nonplussed.

ssadi · #5 $Old$ November 27th, 2008, 10:32 AM

Quote:

Originally Posted by Mathstud28 $View Post$

The first thing you might want to check is in your first step it should be $e^{ix}{\color{red}+}e^{-ix}=8$

Or what you might want to try is solving $y=\frac{e^{ix}+e^{-ix}}{2}$ for $y$ by letting $e^{ix}=\varphi$ . But be sure to remember that $e^{ix}$ is a many valued (many-to-one) function so there isn't going to be just one number as the answer.

Isn't there a general formula or something i can use to do this type of sum?

Mathstud28 · #6 $Old$ November 27th, 2008, 10:47 AM

Quote:

Originally Posted by ssadi $View Post$

and from where the m and n comes into book's answer? I am really nonplussed.

That is because as I said, the imaginary exponential function is multi-valued, $e^{ix}=\cos(x)+i\sin(x)\implies{e^{i(2\pi{n}+x)}=\cos(x)+i\sin(x)}=e^{ix}$

Quote:

Originally Posted by ssadi $View Post$

oh i typed wrong here, in the sum i used e^ix + e^-ix :embarassed:

Dont worry! We all make typos!

Quote:

Originally Posted by ssadi $View Post$

Isn't there a general formula or something i can use to do this type of sum?

Try my suggestion and see if it helps.

There is no need to have three seperate posts

ssadi · #7 $Old$ November 28th, 2008, 05:40 AM

Quote:

Originally Posted by Mathstud28 $View Post$

That is because as I said, the imaginary exponential function is multi-valued, $e^{ix}=\cos(x)+i\sin(x)\implies{e^{i(2\pi{n}+x)}=\cos(x)+i\sin(x)}=e^{ix}$

Dont worry! We all make typos!

Try my suggestion and see if it helps.

There is no need to have three seperate posts

My ideas were coming randomly, hence the array of posts

I am not actually familiar with multi-valuation of imaginary exponential funtion, but I think i can solve the equation in the trigonometric form. But I am not sure how sin x=0 can lead to (4m+1)*pi/2 as the value of x, especially when 3/2 pi is a solution of x. Can you give me a site which addresses how to get the general formula for cos x=0, sinx =0 , sin x=c etc. I will be perfectly happy then

shawsend · #8 $Old$ November 28th, 2008, 06:56 AM

The key to understanding these is understanding the multi-valued logarithm function. Until you understand that, you're going to have problems throughout Complex Analysis. Now $\sin^{-1} z=-i\log\bigg[iz+\sqrt{1-z^2}\bigg]$

That formula is a big deal because both logarithm and the root object are multivalued:

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

$\log(z)=\ln(r)+i(\Theta_z+2n\pi); n\in\mathbb{Z}$

Then:

$\begin{aligned}\sin^{-1} 2&=-i\log\bigg[2i+\sqrt{3}e^{i/2(\pi+2k\pi)}\bigg]\\&=-i\log\bigg[2i\pm i\sqrt{3}\bigg]\end{aligned}$

This then bifurcates into two infinite sets of values:

$\sin^{-1} 2=-i\bigg[\ln|2+\sqrt{3}|+i(\Theta_a+2n\pi)\bigg]$

$\sin^{-1} 2=-i\bigg[\ln|2-\sqrt{3}|+i(\Theta_b+2n\pi)\bigg]$

Where in this case $\Theta_a=\Theta_b=Arg(2\pm\sqrt{3})$ and $n\in\mathbb{Z}$