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Euler's definition of complex analysis [closed]

Complex analysis simultaneous mappingcontour integrals complex analysis 2Complex Analysis Weierstrass TestLimit in complex analysisComplex Analysis - Calculating ResiduesImage drawing complex analysisProve Euler's reflection formula with complex analysisanalysis complexComplex Analysis Question - Cosine and quadratic combinedSingular points in Complex analysis

-1

1

$begingroup$

Can someone help? Need to find the answer of
$$frace^iz + e^iz2 .$$

complex-analysis

share|cite|improve this question

edited Mar 22 at 16:08

Ertxiem

671112

asked Mar 22 at 15:17

JibiJibi

42

$endgroup$

closed as unclear what you're asking by Lord Shark the Unknown, Chickenmancer, Lee Mosher, tatan, Sil Mar 24 at 18:36

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your question? What do you mean by the answer of an expression?
  $endgroup$
  – Brian
  Mar 22 at 15:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is it $e^iz$?
  $endgroup$
  – Lord Shark the Unknown
  Mar 22 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  please reformat, we're not sure what the questions is. As it looks now, $(e^iz + e^iz)/2 = (2e^iz)/2 = e^iz$
  $endgroup$
  – NazimJ
  Mar 22 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, $cos z=(e^iz+e^pmb-iz)/2$
  $endgroup$
  – J. W. Tanner
  Mar 22 at 15:21
  
  

  
  
  
  

add a comment | 

-1

1

$begingroup$

Can someone help? Need to find the answer of
$$frace^iz + e^iz2 .$$

complex-analysis

share|cite|improve this question

edited Mar 22 at 16:08

Ertxiem

671112

asked Mar 22 at 15:17

JibiJibi

42

$endgroup$

closed as unclear what you're asking by Lord Shark the Unknown, Chickenmancer, Lee Mosher, tatan, Sil Mar 24 at 18:36

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is your question? What do you mean by the answer of an expression?
  $endgroup$
  – Brian
  Mar 22 at 15:18
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is it $e^iz$?
  $endgroup$
  – Lord Shark the Unknown
  Mar 22 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  please reformat, we're not sure what the questions is. As it looks now, $(e^iz + e^iz)/2 = (2e^iz)/2 = e^iz$
  $endgroup$
  – NazimJ
  Mar 22 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, $cos z=(e^iz+e^pmb-iz)/2$
  $endgroup$
  – J. W. Tanner
  Mar 22 at 15:21
  
  

  
  
  
  

add a comment | 

$begingroup$

Can someone help? Need to find the answer of
$$frace^iz + e^iz2 .$$

complex-analysis

share|cite|improve this question

edited Mar 22 at 16:08

Ertxiem

671112

asked Mar 22 at 15:17

JibiJibi

42

$endgroup$

share|cite|improve this question

edited Mar 22 at 16:08

Ertxiem

671112

asked Mar 22 at 15:17

JibiJibi

42

share|cite|improve this question

edited Mar 22 at 16:08

Ertxiem

671112

asked Mar 22 at 15:17

JibiJibi

42

edited Mar 22 at 16:08

Ertxiem

671112

edited Mar 22 at 16:08

Ertxiem

671112

asked Mar 22 at 15:17

JibiJibi

42

asked Mar 22 at 15:17

JibiJibi

42

2 Answers
2

active

oldest

votes

1

$begingroup$

The answer is simply $e^iz$ (though I suspect a typo)
This is because there are two copies of the term in the numerator, which cancels with the 2 in the denominator.
Then, simplification can be done using $e^iz=cos(z)+isin(z)$ which is Euler's formula.

However, if you meant $(e^iz+e^-iz)/2$ the answer is $cos z$

Video

share|cite|improve this answer

edited Mar 22 at 19:09

Ertxiem

671112

answered Mar 22 at 15:29

amanaman

34210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then what is (e^z - e^-z)/2
  $endgroup$
  – Jibi
  Mar 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is sinh (z)
  $endgroup$
  – aman
  Mar 22 at 15:54
  

  
  
  
  

add a comment | 

1

$begingroup$

$$beginaligneddfrace^iz+e^-iz2&=cos z\dfrace^iz-e^-iz2i&=sin zendaligned$$

Therefore, we have $isin z=1/2cdot left[e^iz-e^-izright]$.

---

$$dfrace^z-e^-z2=sinh z$$

share|cite|improve this answer

edited Mar 22 at 15:49

answered Mar 22 at 15:40

Paras KhoslaParas Khosla

2,883523

$endgroup$

add a comment | 

1

$begingroup$

The answer is simply $e^iz$ (though I suspect a typo)
This is because there are two copies of the term in the numerator, which cancels with the 2 in the denominator.
Then, simplification can be done using $e^iz=cos(z)+isin(z)$ which is Euler's formula.

However, if you meant $(e^iz+e^-iz)/2$ the answer is $cos z$

Video

share|cite|improve this answer

edited Mar 22 at 19:09

Ertxiem

671112

answered Mar 22 at 15:29

amanaman

34210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then what is (e^z - e^-z)/2
  $endgroup$
  – Jibi
  Mar 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is sinh (z)
  $endgroup$
  – aman
  Mar 22 at 15:54
  

  
  
  
  

add a comment | 

1

$begingroup$

The answer is simply $e^iz$ (though I suspect a typo)
This is because there are two copies of the term in the numerator, which cancels with the 2 in the denominator.
Then, simplification can be done using $e^iz=cos(z)+isin(z)$ which is Euler's formula.

However, if you meant $(e^iz+e^-iz)/2$ the answer is $cos z$

Video

share|cite|improve this answer

edited Mar 22 at 19:09

Ertxiem

671112

answered Mar 22 at 15:29

amanaman

34210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then what is (e^z - e^-z)/2
  $endgroup$
  – Jibi
  Mar 22 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is sinh (z)
  $endgroup$
  – aman
  Mar 22 at 15:54
  

  
  
  
  

add a comment | 

$begingroup$

The answer is simply $e^iz$ (though I suspect a typo)
This is because there are two copies of the term in the numerator, which cancels with the 2 in the denominator.
Then, simplification can be done using $e^iz=cos(z)+isin(z)$ which is Euler's formula.

However, if you meant $(e^iz+e^-iz)/2$ the answer is $cos z$

Video

share|cite|improve this answer

edited Mar 22 at 19:09

Ertxiem

671112

answered Mar 22 at 15:29

amanaman

34210

$endgroup$

share|cite|improve this answer

edited Mar 22 at 19:09

Ertxiem

671112

answered Mar 22 at 15:29

amanaman

34210

edited Mar 22 at 19:09

Ertxiem

671112

edited Mar 22 at 19:09

Ertxiem

671112

answered Mar 22 at 15:29

amanaman

34210

answered Mar 22 at 15:29

amanaman

34210

1

$begingroup$

$$beginaligneddfrace^iz+e^-iz2&=cos z\dfrace^iz-e^-iz2i&=sin zendaligned$$

Therefore, we have $isin z=1/2cdot left[e^iz-e^-izright]$.

---

$$dfrace^z-e^-z2=sinh z$$

share|cite|improve this answer

edited Mar 22 at 15:49

answered Mar 22 at 15:40

Paras KhoslaParas Khosla

2,883523

$endgroup$

add a comment | 

1

$begingroup$

$$beginaligneddfrace^iz+e^-iz2&=cos z\dfrace^iz-e^-iz2i&=sin zendaligned$$

Therefore, we have $isin z=1/2cdot left[e^iz-e^-izright]$.

---

$$dfrace^z-e^-z2=sinh z$$

share|cite|improve this answer

edited Mar 22 at 15:49

answered Mar 22 at 15:40

Paras KhoslaParas Khosla

2,883523

$endgroup$

add a comment | 

$begingroup$

$$beginaligneddfrace^iz+e^-iz2&=cos z\dfrace^iz-e^-iz2i&=sin zendaligned$$

Therefore, we have $isin z=1/2cdot left[e^iz-e^-izright]$.

---

$$dfrace^z-e^-z2=sinh z$$

share|cite|improve this answer

edited Mar 22 at 15:49

answered Mar 22 at 15:40

Paras KhoslaParas Khosla

2,883523

$endgroup$

share|cite|improve this answer

edited Mar 22 at 15:49

answered Mar 22 at 15:40

Paras KhoslaParas Khosla

2,883523

answered Mar 22 at 15:40

Paras KhoslaParas Khosla

2,883523

answered Mar 22 at 15:40

Paras KhoslaParas Khosla

2,883523