ecuation

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.
 
how can I solve |e^ix - 1| = 2 (x E (0,2pi)) ? what kind of ecuation is it?
By definition, z=zz|z| = \sqrt{ \overline{z} z }, where z\overline{z} is the complex conjugate of z. So we have
eix1=(eiz1)(eiz1)=eixeix(eix+eix)+1=2(eix+eix)\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} ) }
Knowing that cos(x)=eix+eix2cos(x) = \dfrac{ e^{ix} + e^{-ix} }{2} can you finish?

-Dan
 
@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
 
what I get is this: [MATH]\sqrt{Cos[x]^2 + Sin[x]^2} -1 = 2[/MATH]
Where did the sine function come from?

2(eix+eix)=22 cos(x)=2(1cos(x))\sqrt{ 2 - (e^{ix} + e^{-ix} ) } = \sqrt{ 2 - 2 ~ cos(x) } = \sqrt{2(1 - cos(x))}
Now, you could probably leave it there but, if you like, you can now do a half-angle substitution:
sin2(x2)=2(1cos(x))sin^2 \left ( \dfrac{x}{2} \right ) = 2(1 - cos(x)), so we get
2(1cos(x))=4 sin2(x2)=±2 sin(x2)\sqrt{2(1 - cos(x))} = \sqrt{ 4 ~ sin^2 \left ( \dfrac{x}{2} \right ) }= \pm 2 ~ sin \left ( \dfrac{x}{2} \right )
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
=±2 sin(x2)=2 sin(x2)= \left \| \pm 2 ~ sin \left ( \dfrac{x}{2} \right ) \right \| = 2 ~ \left \| sin \left ( \dfrac{x}{2} \right ) \right \| . (Watch the intervals where sin(x2)sin \left ( \dfrac{x}{2} \right ) 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
 
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 z2=2(z)+2(z)|z|^2=\Re ^2(z) +\Im ^2(z)
We get (cos(x)1)2+sin2(x)=4(\cos(x)-1)^2+\sin^2(x)=4.
Can you proceed?
 
Top