$1 + {alpha}^2 + 2 alpha cos(omega)$ as a square of distance

0

$begingroup$

How to prove that

$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$

starting from

$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$

(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)

complex-analysis complex-numbers

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edited Jan 18 at 15:43

KM101

6,0351525

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Look for the roots of this second degree polynomial in $alpha$.
  $endgroup$
  – Stefan Lafon
  Jan 18 at 15:35
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Just do the 'backward' proof in reverse.
  $endgroup$
  – lightxbulb
  Jan 18 at 15:39
  

  
  
  
  

add a comment | 

0

$begingroup$

How to prove that

$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$

starting from

$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$

(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)

complex-analysis complex-numbers

share|cite|improve this question

edited Jan 18 at 15:43

KM101

6,0351525

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Look for the roots of this second degree polynomial in $alpha$.
  $endgroup$
  – Stefan Lafon
  Jan 18 at 15:35
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Just do the 'backward' proof in reverse.
  $endgroup$
  – lightxbulb
  Jan 18 at 15:39
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

How to prove that

$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$

starting from

$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$

(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)

complex-analysis complex-numbers

share|cite|improve this question

edited Jan 18 at 15:43

KM101

6,0351525

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

$endgroup$

How to prove that

$$ 1 + {alpha}^2 + 2 alpha cos(omega) = |1 - alpha e^{-jomega}|^2$$

starting from

$$ 1 + {alpha}^2 + 2 alpha cos(omega)$$

(That is, no backwards proofs of multiplying out $ |1 - alpha e^{-jomega}|^2$ to see that it matches $1 + {alpha}^2 + 2 alpha cos(omega)$.)

complex-analysis complex-numbers

complex-analysis complex-numbers

share|cite|improve this question

edited Jan 18 at 15:43

KM101

6,0351525

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

share|cite|improve this question

edited Jan 18 at 15:43

KM101

6,0351525

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

share|cite|improve this question

share|cite|improve this question

edited Jan 18 at 15:43

KM101

6,0351525

edited Jan 18 at 15:43

KM101

6,0351525

edited Jan 18 at 15:43

KM101

6,0351525

6,0351525

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

asked Jan 18 at 15:35

DiscreteMathDiscreteMath

414

414

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Look for the roots of this second degree polynomial in $alpha$.
  $endgroup$
  – Stefan Lafon
  Jan 18 at 15:35
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Just do the 'backward' proof in reverse.
  $endgroup$
  – lightxbulb
  Jan 18 at 15:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Look for the roots of this second degree polynomial in $alpha$.
  $endgroup$
  – Stefan Lafon
  Jan 18 at 15:35
  
  
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Just do the 'backward' proof in reverse.
  $endgroup$
  – lightxbulb
  Jan 18 at 15:39
  

  
  
  
  

2

2

$begingroup$
Look for the roots of this second degree polynomial in $alpha$.
$endgroup$
– Stefan Lafon
Jan 18 at 15:35

$begingroup$
Look for the roots of this second degree polynomial in $alpha$.
$endgroup$
– Stefan Lafon
Jan 18 at 15:35

4

4

$begingroup$
Just do the 'backward' proof in reverse.
$endgroup$
– lightxbulb
Jan 18 at 15:39

$begingroup$
Just do the 'backward' proof in reverse.
$endgroup$
– lightxbulb
Jan 18 at 15:39

add a comment | 

2 Answers
2

active

oldest

votes

2

$begingroup$

assuming $alpha$ is a real number (not complex):

$$1-2 alpha cos(omega)+|alpha|^2 $$
$$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
$$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
$$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
$$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
$$|1-alpha e^{jomega}|^2$$

share|cite|improve this answer

edited Jan 18 at 16:31

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

$endgroup$

add a comment | 

0

$begingroup$

Hint

Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.

share|cite|improve this answer

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

$endgroup$

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

assuming $alpha$ is a real number (not complex):

$$1-2 alpha cos(omega)+|alpha|^2 $$
$$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
$$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
$$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
$$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
$$|1-alpha e^{jomega}|^2$$

share|cite|improve this answer

edited Jan 18 at 16:31

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

$endgroup$

add a comment | 

2

$begingroup$

assuming $alpha$ is a real number (not complex):

$$1-2 alpha cos(omega)+|alpha|^2 $$
$$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
$$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
$$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
$$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
$$|1-alpha e^{jomega}|^2$$

share|cite|improve this answer

edited Jan 18 at 16:31

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

$endgroup$

add a comment | 

2

2

2

$begingroup$

assuming $alpha$ is a real number (not complex):

$$1-2 alpha cos(omega)+|alpha|^2 $$
$$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
$$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
$$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
$$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
$$|1-alpha e^{jomega}|^2$$

share|cite|improve this answer

edited Jan 18 at 16:31

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

$endgroup$

assuming $alpha$ is a real number (not complex):

$$1-2 alpha cos(omega)+|alpha|^2 $$
$$1-2 alpha frac{1}{2} (e^{jomega}+e^{-jomega})+|alpha|^2$$
$$1-alpha e^{jomega} - alpha e^{-jomega} + alpha^2 e^{jomega}e^{-jomega}$$
$$(1-alpha e^{jomega}) - alpha e^{-jomega}(1-alpha e^{jomega})$$
$$(1-alpha e^{jomega})(1-alpha e^{-jomega})$$
$$|1-alpha e^{jomega}|^2$$

share|cite|improve this answer

edited Jan 18 at 16:31

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

share|cite|improve this answer

share|cite|improve this answer

edited Jan 18 at 16:31

edited Jan 18 at 16:31

edited Jan 18 at 16:31

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

answered Jan 18 at 16:19

Bill MooreBill Moore

1376

1376

add a comment | 

add a comment | 

0

$begingroup$

Hint

Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.

share|cite|improve this answer

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

$endgroup$

add a comment | 

0

$begingroup$

Hint

Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.

share|cite|improve this answer

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint

Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.

share|cite|improve this answer

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

$endgroup$

Hint

Use $$alpha^2=alpha^2(cos^2omega+sin^2omega)$$and substitute.

share|cite|improve this answer

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

share|cite|improve this answer

share|cite|improve this answer

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

answered Jan 18 at 15:41

Mostafa AyazMostafa Ayaz

15.6k3939

15.6k3939

add a comment | 

add a comment | 

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