ecuation

[nestor95]

how can I solve |e^ix - 1| = 2 (x E (0,2pi)) ? what kind of ecuation is it?

 

[Steven G]

e^i = e^(i*1) = cos(1) + i sin(1)

Just substitute that back into your equation and see what you can do with it. Please post back.

 

[topsquark]

how can I solve |e^ix - 1| = 2 (x E (0,2pi)) ? what kind of ecuation is it?

By definition, [math]|z| = \sqrt{ \overline{z} z }[/math], where [math]\overline{z}[/math] is the complex conjugate of z. So we have
[math]\left | e^{-ix} - 1 \right | = \sqrt{ (e^{-iz} - 1)(e^{iz} - 1) } = \sqrt{ e^{-ix} \cdot e^{ix} - (e^{ix} + e^{-ix} ) + 1 } = \sqrt{2 - (e^{ix} + e^{-ix} ) }[/math]
Knowing that [math]cos(x) = \dfrac{ e^{ix} + e^{-ix} }{2}[/math] can you finish?

-Dan

 

[topsquark]

@nestor: You're fast. I didn't even have enough time to fix the typo! (It was minor, you probably didn't notice it.)

-Dan

 

[nestor95]

what I get is this: [MATH]\sqrt{Cos[x]^2 + Sin[x]^2} -1 = 2[/MATH]

 

[topsquark]

what I get is this: [MATH]\sqrt{Cos[x]^2 + Sin[x]^2} -1 = 2[/MATH]

Where did the sine function come from?

[math]\sqrt{ 2 - (e^{ix} + e^{-ix} ) } = \sqrt{ 2 - 2 ~ cos(x) } = \sqrt{2(1 - cos(x))}[/math]
Now, you could probably leave it there but, if you like, you can now do a half-angle substitution:
[math]sin^2 \left ( \dfrac{x}{2} \right ) = 2(1 - cos(x))[/math], so we get
[math]\sqrt{2(1 - cos(x))} = \sqrt{ 4 ~ sin^2 \left ( \dfrac{x}{2} \right ) }= \pm 2 ~ sin \left ( \dfrac{x}{2} \right )[/math]
A word about the signs. The OP dealt with an absolute value so the final answer is required to be non-negative. So we really have
[math]= \left \| \pm 2 ~ sin \left ( \dfrac{x}{2} \right ) \right \| = 2 ~ \left \| sin \left ( \dfrac{x}{2} \right ) \right \| [/math]. (Watch the intervals where [math]sin \left ( \dfrac{x}{2} \right )[/math] is negative.) This would normally take care of itself but we played with the trig function using the half-angle formula and anyway you always need to check.

-Dan

 

[pka]

how can I solve |e^ix - 1| = 2 (x E (0,2pi)) ? what kind of ecuation is it?

I think we should follow Jomo's approach. Taking note that \(|z|^2=\Re ^2(z) +\Im ^2(z)\)
We get \((\cos(x)-1)^2+\sin^2(x)=4\).
Can you proceed?

 

[nestor95]

Yes, much better, thanks!