Complex exponential to trigonometric simplification

1. Apr 15, 2013

Malgrif

1. The problem statement, all variables and given/known data
Given (e^(ix) - 1)^2 , show that it is equal to 2-2cosx

2. Relevant equations
e^ix = cosx + isinx

3. The attempt at a solution
After subbing in Euler identity and expanding I get:

cos(x)^2+sin(x)^2-2cosx-2jsinx+2jcosxsinx + 1

after using the addtion formulas I get

cos(2x)+jsin(2x)-2cosx-2jsinx+1

seems like i'm missing a fundamental trig ID... Please help, and let me know if my approach is correct.

Thanks

2. Apr 15, 2013

SammyS

Staff Emeritus
Please state the problem, word for word as it was given to you.

It's obvious that (eix-1)2 ≠ 2 - 2cos(x) .

Also, please stick with one symbol for the imaginary unit: either i or j . Don't switch back and forth.

3. Apr 15, 2013

Malgrif

Sorry EE student here.

Hm, maybe I'm missing the bigger picture then. This is part of my digital systems homework, regarding filter design. The question I posed is of my own formulation, as that is where I am stuck with the with problem. If the question is nonsensical i must be missing a cancellation. I attached a picture so you can hopefully help me figure it out.

4. Apr 15, 2013

SammyS

Staff Emeritus
That makes things clearer.

It does look like $\displaystyle \ \left|\left(e^{\,j\,\theta}-1 \right)^2\right|=2-2\cos(\theta) \,, \$ where j is the imaginary unit.

5. Apr 15, 2013

Malgrif

Ok. Could you go through the steps?

6. Apr 15, 2013

SammyS

Staff Emeritus

One thing that may help: (a + bj)(a - bj) = a2 + b2 = |a + bj|2 .

The other is the Euler identity, so that $\displaystyle \ e^{\,x\,j}-1=(\cos(x)-1)+j\sin(x)\ .$

See what you can do with that.

7. Apr 15, 2013

Malgrif

Well, I got that far on my own. The final simplification is the issue.

so after expansion I get:

cos(x)^2-2cosx+1-sin(x)^2

I use Pythagorean identity to get:

2cos(x)^2 - 1 + 2cosx + 1

so 2cos(x)^2 + 2cosx

What's next? Or where's the error?

8. Apr 15, 2013

Staff: Mentor

Starting with |(ejx - 1)2|, show us what you did.

9. Apr 15, 2013

Malgrif

I did...

So from the start:

|(e^jx - 1)^2| = | (cosx +jsinx -1)^2|

using |a+jb|^2=a^2+b^2

we get

(cosx - 1)^2 + (jsinx)^2

= cos^2(x)-2cosx +1 - sin^2(x)

using sin^2(x)=1-cos^2(x)

= 2cos^2(x) - 2cos(x)

10. Apr 15, 2013

Staff: Mentor

The last term above has the wrong sign. In the formula you used, b is the coefficient of j, and doesn't include it. You should get (cosx - 1)2 + sin2x

11. Apr 15, 2013

Malgrif

Oh I found my mistake. I applied the identity incorrectly.

So it is just (cosx -1)^2 + (sin^2(x))

which will come to 2 - 2cos(x)

Thanks Sammy!

12. Apr 15, 2013

Malgrif

Yup just saw it myself! Thank yoU!