Proving Equivalence of certain Complex numbers

• Apr 28th 2013, 02:44 AM
EliteAndoy
Proving Equivalence of certain Complex numbers
Hello Everyone! Today we are asked to prove that both this expressions are true: $\left | e^{\theta i } \right |=1$ and $\overline{e^{\theta i}}=e^{-\theta i}$. To be quite honest, I am not very sure as to how I can prove these, but my attempt for the first one, I was able to prove that $\sin^2(\theta)+cos^2(\theta)=1$ but was unsure as to how I would be able to turn it into $\sin(\theta)+cos(\theta)=1$ which is what I need to prove the first equality.

As for the second one I did something like this:

$e^{\theta i}=\cos(\theta)+i\sin(\theta)$

So that:

$\overline{e^{\theta i}}=\cos(\theta)-i\sin(\theta)$

and by properties of odd and even function:

$\overline{e^{\theta i}}=\cos(-\theta)+i\sin(-\theta)$

Thus, it should be true to say that :

$\overline{e^{\theta i}}=e^{-\theta i}$

The proof above, for me, is quite sloppy, but I think I could live with that. But for the first one, I really have no idea as to how to prove it. Anyone got some ideas as to how I should prove both equivalence? Thanks everyone in advance!(Wink)
• Apr 28th 2013, 03:01 AM
Prove It
Re: Proving Equivalence of certain Complex numbers
Surely you know that \displaystyle \begin{align*} \sqrt{ \sin^2{(\theta)} + \cos^2{(\theta)}} \end{align*} is NOT \displaystyle \begin{align*} \sin{(\theta)} + \cos{(\theta)} \end{align*}...
• Apr 28th 2013, 03:30 AM
Plato
Re: Proving Equivalence of certain Complex numbers
Quote:

Originally Posted by EliteAndoy
$\left | e^{\theta i } \right |=1$ and $\overline{e^{\theta i}}=e^{-\theta i}$. To be quite honest, I am not very sure as to how I can prove these, but my attempt for the first one, I was able to prove that $\sin^2(\theta)+cos^2(\theta)=1$ but was unsure as to how I would be able to turn it into $\sin(\theta)+cos(\theta)=1$ which is what I need to prove the first equality.

You must know that $\sqrt{a^2+b^2}\ne a+b~!$ and $\sqrt{1}=1$

Quote:

Originally Posted by EliteAndoy
$\overline{e^{\theta i}}=e^{-\theta i}$.
As for the second one I did something like this:
$e^{\theta i}=\cos(\theta)+i\sin(\theta)$
So that:
$\overline{e^{\theta i}}=\cos(\theta)-i\sin(\theta)$
and by properties of odd and even function:
$\overline{e^{\theta i}}=\cos(-\theta)+i\sin(-\theta)$
Thus, it should be true to say that :
$\overline{e^{\theta i}}=e^{-\theta i}$

Yours above is the standard proof for this.
You might add that this works because cosine is an even function and sine is an odd function.
• Apr 28th 2013, 08:44 AM
EliteAndoy
Re: Proving Equivalence of certain Complex numbers
Yeah, that's absolutely true,that's why I was unable to prove the equality. Here's how my little failed proof goes:
$\left |e^{i \theta} \right |= \sin^2(\theta)+\cos^2(\theta)$

From the equation above, we can see that the modulus is 1 so that:

$\sqrt{\sin^2(\theta)+\cos^2(\theta)}=1$

For the equation above to be true, the radicand should equal to 1, so that

$\sin^2(\theta))+\cos^2(\theta)=1$

From here, I tried to play with a lot of trig identities which all I was able to do is prove that this equation is true. But yeah, like Plato and Prove it said, this is not equal to
$\sin(\theta)+\cos(\theta)=1$. So, anyone here have any idea as to how I should proceed with the proof? Thanks again to both of you!
• Apr 28th 2013, 09:04 AM
Plato
Re: Proving Equivalence of certain Complex numbers
Quote:

Originally Posted by EliteAndoy
So, anyone here have any idea as to how I should proceed with the proof? Thanks again to both of you!

What in the world are you on about?

$\left|e^{i\theta}\right|=\left|\cos(\theta)+i\sin( \theta)\right|=\sqrt{\cos^2(\theta)+\sin^2(\theta) }=\sqrt1=1$

ALL of those are equals. What don't you get?
• Apr 28th 2013, 09:41 AM
EliteAndoy
Re: Proving Equivalence of certain Complex numbers
Damn it, I forgot about that. Well thanks for verifying that. :D
• Apr 28th 2013, 12:31 PM
topsquark
Re: Proving Equivalence of certain Complex numbers
Quote:

Originally Posted by EliteAndoy
(Swear) it, I forgot about that. Well thanks for verifying that. :D