Proving Equivalence of certain Complex numbers

Hello Everyone! Today we are asked to prove that both this expressions are true: $\displaystyle \left | e^{\theta i } \right |=1$ and $\displaystyle \overline{e^{\theta i}}=e^{-\theta i}$. 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 $\displaystyle \sin^2(\theta)+cos^2(\theta)=1$ but was unsure as to how I would be able to turn it into $\displaystyle \sin(\theta)+cos(\theta)=1$ which is what I need to prove the first equality.

As for the second one I did something like this:

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

So that:

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

and by properties of odd and even function:

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

Thus, it should be true to say that :

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

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)

Re: Proving Equivalence of certain Complex numbers

Surely you know that $\displaystyle \displaystyle \begin{align*} \sqrt{ \sin^2{(\theta)} + \cos^2{(\theta)}} \end{align*}$ is NOT $\displaystyle \displaystyle \begin{align*} \sin{(\theta)} + \cos{(\theta)} \end{align*}$...

Re: Proving Equivalence of certain Complex numbers

Quote:

Originally Posted by

**EliteAndoy** $\displaystyle \left | e^{\theta i } \right |=1$ and $\displaystyle \overline{e^{\theta i}}=e^{-\theta i}$. 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 $\displaystyle \sin^2(\theta)+cos^2(\theta)=1$ but was unsure as to how I would be able to turn it into $\displaystyle \sin(\theta)+cos(\theta)=1$ which is what I need to prove the first equality.

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

Quote:

Originally Posted by

**EliteAndoy** $\displaystyle \overline{e^{\theta i}}=e^{-\theta i}$.

As for the second one I did something like this:

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

So that:

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

and by properties of odd and even function:

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

Thus, it should be true to say that :

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

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.

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:

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

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

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

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

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

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

$\displaystyle \sin(\theta)+\cos(\theta)=1$. So, anyone here have any idea as to how I should proceed with the proof? Thanks again to both of you!

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?**

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

__ALL__ of those are __equals__. What don't you get?

Re: Proving Equivalence of certain Complex numbers

Damn it, I forgot about that. Well thanks for verifying that. :D

Re: Proving Equivalence of certain Complex numbers

Quote:

Originally Posted by

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

Watch the language, please.

-Dan

Re: Proving Equivalence of certain Complex numbers

Oh sorry about that hehe.