Complex exponential to trigonometric simplification

[Malgrif]

Homework Statement

Given (e^(ix) - 1)^2 , show that it is equal to 2-2cosx

Homework Equations

e^ix = cosx + isinx

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

 

[SammyS]

Homework Statement

Given (e^(ix) - 1)^2 , show that it is equal to 2-2cosx

Homework Equations

e^ix = cosx + isinx

...

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

It's obvious that (e^ix-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.

 

[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.

 

[SammyS]

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.

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.

 

[Malgrif]

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.

Ok. Could you go through the steps?

 

[SammyS]

Ok. Could you go through the steps?

Let's help you go through the steps.

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.

 

[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?

 

[Mark44]

Starting with |(e^jx - 1)²|, show us what you did.

 

[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)

 

[Mark44]

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)

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 +[/color] sin²x

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

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

 

[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!

 

[Malgrif]

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 +[/color] sin²x

Yup just saw it myself! Thank yoU!