Complex exponential to trigonometric simplification

AI Thread Summary
The discussion centers around simplifying the expression (e^(ix) - 1)^2 to show it equals 2 - 2cos(x). Participants initially struggle with the expansion and simplification, particularly with applying trigonometric identities correctly. Key insights involve using Euler's identity and recognizing the magnitude of complex numbers. Ultimately, the correct simplification reveals that the expression does indeed equal 2 - 2cos(x), correcting earlier mistakes in applying identities. The exchange emphasizes the importance of careful application of mathematical identities in complex analysis.
Malgrif
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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
 
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Malgrif said:

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 (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.
 
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.
ixC4zZN.png
 
Malgrif said:
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.
 
SammyS said:
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?
 
Malgrif said:
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.
 
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?
 
Starting with |(ejx - 1)2|, show us what you did.
 
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
Malgrif said:
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
Malgrif said:
using sin^2(x)=1-cos^2(x)

= 2cos^2(x) - 2cos(x)
 
  • #11
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
Mark44 said:
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!
 
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