How to transform from exponential to trigonometric

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I'm having trouble transforming an expression with exponentials to one with sines or cosines only.

$$u = - b a e^-kz e^i(kz-wt)$$

I know $e^i(kz-wt)$ is $cos(kx-wt)$, but I can't figure out how to go from $e^-kz$ to a trig expression.

Could any enlightened soul please help me out?

trigonometry exponential-function transformation

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edited Jul 18 at 20:39

InterstellarProbe

2,207518

asked Jul 18 at 20:35

isa_cntdr

91

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  math.meta.stackexchange.com/questions/5020/…
  – saulspatz
  Jul 18 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $e^ix=cos x + isin x$
  – saulspatz
  Jul 18 at 20:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to math.SE. Check out e.g. meta.math.stackexchange.com/questions/5020/… for a guide to format math here so it becomes more readable. As a hint for your question: $i^2=-1$.
  – Henrik
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've never really thought about this before, but I guess $e^-x = e^iix = cos(ix) + i sin(ix)$
  – CJD
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @CJC I have to think about this ahah! After all I'm just a puny physics undergraduate
  – Davide Morgante
  Jul 18 at 21:07
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

favorite

I'm having trouble transforming an expression with exponentials to one with sines or cosines only.

$$u = - b a e^-kz e^i(kz-wt)$$

I know $e^i(kz-wt)$ is $cos(kx-wt)$, but I can't figure out how to go from $e^-kz$ to a trig expression.

Could any enlightened soul please help me out?

trigonometry exponential-function transformation

share|cite|improve this question

edited Jul 18 at 20:39

InterstellarProbe

2,207518

asked Jul 18 at 20:35

isa_cntdr

91

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  math.meta.stackexchange.com/questions/5020/…
  – saulspatz
  Jul 18 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $e^ix=cos x + isin x$
  – saulspatz
  Jul 18 at 20:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to math.SE. Check out e.g. meta.math.stackexchange.com/questions/5020/… for a guide to format math here so it becomes more readable. As a hint for your question: $i^2=-1$.
  – Henrik
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've never really thought about this before, but I guess $e^-x = e^iix = cos(ix) + i sin(ix)$
  – CJD
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @CJC I have to think about this ahah! After all I'm just a puny physics undergraduate
  – Davide Morgante
  Jul 18 at 21:07
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I'm having trouble transforming an expression with exponentials to one with sines or cosines only.

$$u = - b a e^-kz e^i(kz-wt)$$

I know $e^i(kz-wt)$ is $cos(kx-wt)$, but I can't figure out how to go from $e^-kz$ to a trig expression.

Could any enlightened soul please help me out?

trigonometry exponential-function transformation

share|cite|improve this question

edited Jul 18 at 20:39

InterstellarProbe

2,207518

asked Jul 18 at 20:35

isa_cntdr

91

I'm having trouble transforming an expression with exponentials to one with sines or cosines only.

$$u = - b a e^-kz e^i(kz-wt)$$

I know $e^i(kz-wt)$ is $cos(kx-wt)$, but I can't figure out how to go from $e^-kz$ to a trig expression.

Could any enlightened soul please help me out?

trigonometry exponential-function transformation

share|cite|improve this question

edited Jul 18 at 20:39

InterstellarProbe

2,207518

asked Jul 18 at 20:35

isa_cntdr

91

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 20:39

InterstellarProbe

2,207518

edited Jul 18 at 20:39

InterstellarProbe

2,207518

edited Jul 18 at 20:39

InterstellarProbe

2,207518

2,207518

asked Jul 18 at 20:35

isa_cntdr

91

asked Jul 18 at 20:35

isa_cntdr

91

asked Jul 18 at 20:35

isa_cntdr

91

91

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  math.meta.stackexchange.com/questions/5020/…
  – saulspatz
  Jul 18 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $e^ix=cos x + isin x$
  – saulspatz
  Jul 18 at 20:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to math.SE. Check out e.g. meta.math.stackexchange.com/questions/5020/… for a guide to format math here so it becomes more readable. As a hint for your question: $i^2=-1$.
  – Henrik
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've never really thought about this before, but I guess $e^-x = e^iix = cos(ix) + i sin(ix)$
  – CJD
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @CJC I have to think about this ahah! After all I'm just a puny physics undergraduate
  – Davide Morgante
  Jul 18 at 21:07
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  math.meta.stackexchange.com/questions/5020/…
  – saulspatz
  Jul 18 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $e^ix=cos x + isin x$
  – saulspatz
  Jul 18 at 20:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Welcome to math.SE. Check out e.g. meta.math.stackexchange.com/questions/5020/… for a guide to format math here so it becomes more readable. As a hint for your question: $i^2=-1$.
  – Henrik
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've never really thought about this before, but I guess $e^-x = e^iix = cos(ix) + i sin(ix)$
  – CJD
  Jul 18 at 20:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @CJC I have to think about this ahah! After all I'm just a puny physics undergraduate
  – Davide Morgante
  Jul 18 at 21:07
  

  
  
  
  

1

1

math.meta.stackexchange.com/questions/5020/…
– saulspatz
Jul 18 at 20:38

math.meta.stackexchange.com/questions/5020/…
– saulspatz
Jul 18 at 20:38

$e^ix=cos x + isin x$
– saulspatz
Jul 18 at 20:39

$e^ix=cos x + isin x$
– saulspatz
Jul 18 at 20:39

Welcome to math.SE. Check out e.g. meta.math.stackexchange.com/questions/5020/… for a guide to format math here so it becomes more readable. As a hint for your question: $i^2=-1$.
– Henrik
Jul 18 at 20:40

Welcome to math.SE. Check out e.g. meta.math.stackexchange.com/questions/5020/… for a guide to format math here so it becomes more readable. As a hint for your question: $i^2=-1$.
– Henrik
Jul 18 at 20:40

I've never really thought about this before, but I guess $e^-x = e^iix = cos(ix) + i sin(ix)$
– CJD
Jul 18 at 20:40

I've never really thought about this before, but I guess $e^-x = e^iix = cos(ix) + i sin(ix)$
– CJD
Jul 18 at 20:40

1

1

@CJC I have to think about this ahah! After all I'm just a puny physics undergraduate
– Davide Morgante
Jul 18 at 21:07

@CJC I have to think about this ahah! After all I'm just a puny physics undergraduate
– Davide Morgante
Jul 18 at 21:07

 | 
show 4 more comments

1 Answer
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In general, $e^pm ix = cos(x) pm isin(x)$. I think you forgot to mention what $z$ is, but I assume (reading that "you know that..." ) that $z = x + iy$ an arbitrary complex number, then:
beginequation
e^-kz = e^-k(x+iy) = e^-kxe^-iky = e^-kx(cos(ky)-isin(ky))
endequation

share|cite|improve this answer

edited Jul 18 at 20:59

answered Jul 18 at 20:51

Giacomo

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1 Answer
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1 Answer
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active

oldest

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up vote
0
down vote

In general, $e^pm ix = cos(x) pm isin(x)$. I think you forgot to mention what $z$ is, but I assume (reading that "you know that..." ) that $z = x + iy$ an arbitrary complex number, then:
beginequation
e^-kz = e^-k(x+iy) = e^-kxe^-iky = e^-kx(cos(ky)-isin(ky))
endequation

share|cite|improve this answer

edited Jul 18 at 20:59

answered Jul 18 at 20:51

Giacomo

62

add a comment | 

up vote
0
down vote

In general, $e^pm ix = cos(x) pm isin(x)$. I think you forgot to mention what $z$ is, but I assume (reading that "you know that..." ) that $z = x + iy$ an arbitrary complex number, then:
beginequation
e^-kz = e^-k(x+iy) = e^-kxe^-iky = e^-kx(cos(ky)-isin(ky))
endequation

share|cite|improve this answer

edited Jul 18 at 20:59

answered Jul 18 at 20:51

Giacomo

62

add a comment | 

up vote
0
down vote

up vote
0
down vote

In general, $e^pm ix = cos(x) pm isin(x)$. I think you forgot to mention what $z$ is, but I assume (reading that "you know that..." ) that $z = x + iy$ an arbitrary complex number, then:
beginequation
e^-kz = e^-k(x+iy) = e^-kxe^-iky = e^-kx(cos(ky)-isin(ky))
endequation

share|cite|improve this answer

edited Jul 18 at 20:59

answered Jul 18 at 20:51

Giacomo

62

In general, $e^pm ix = cos(x) pm isin(x)$. I think you forgot to mention what $z$ is, but I assume (reading that "you know that..." ) that $z = x + iy$ an arbitrary complex number, then:
beginequation
e^-kz = e^-k(x+iy) = e^-kxe^-iky = e^-kx(cos(ky)-isin(ky))
endequation

share|cite|improve this answer

edited Jul 18 at 20:59

answered Jul 18 at 20:51

Giacomo

62

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 20:59

edited Jul 18 at 20:59

edited Jul 18 at 20:59

answered Jul 18 at 20:51

Giacomo

62

answered Jul 18 at 20:51

Giacomo

62

answered Jul 18 at 20:51

Giacomo

62

62

add a comment | 

add a comment | 

 

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