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Are $ cos z$ and $ sin z$ bounded functions with $z$ a complex number?

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0

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Really I guess since , $cos ^2 z+sin^2 z=1$ for every complex number $z$ then they behave like that in $mathbbR$ , Right ?

real-analysis complex-analysis

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asked Mar 15 at 16:43

zeraoulia rafikzeraoulia rafik

2,44211032

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $cos^2z+sin^2z=1$ doesn't mean they're bounded when complex
  $endgroup$
  – J. W. Tanner
  Mar 15 at 16:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  in My guess sin z and cos z should lie in (-1,1)
  $endgroup$
  – zeraoulia rafik
  Mar 15 at 16:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  not necessarily the case if $z$ is complex
  $endgroup$
  – J. W. Tanner
  Mar 15 at 16:49
  

  
  
  
  

add a comment | 

0

$begingroup$

Really I guess since , $cos ^2 z+sin^2 z=1$ for every complex number $z$ then they behave like that in $mathbbR$ , Right ?

real-analysis complex-analysis

share|cite|improve this question

asked Mar 15 at 16:43

zeraoulia rafikzeraoulia rafik

2,44211032

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $cos^2z+sin^2z=1$ doesn't mean they're bounded when complex
  $endgroup$
  – J. W. Tanner
  Mar 15 at 16:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  in My guess sin z and cos z should lie in (-1,1)
  $endgroup$
  – zeraoulia rafik
  Mar 15 at 16:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  not necessarily the case if $z$ is complex
  $endgroup$
  – J. W. Tanner
  Mar 15 at 16:49
  

  
  
  
  

add a comment | 

$begingroup$

Really I guess since , $cos ^2 z+sin^2 z=1$ for every complex number $z$ then they behave like that in $mathbbR$ , Right ?

real-analysis complex-analysis

share|cite|improve this question

asked Mar 15 at 16:43

zeraoulia rafikzeraoulia rafik

2,44211032

$endgroup$

share|cite|improve this question

asked Mar 15 at 16:43

zeraoulia rafikzeraoulia rafik

2,44211032

share|cite|improve this question

asked Mar 15 at 16:43

zeraoulia rafikzeraoulia rafik

2,44211032

asked Mar 15 at 16:43

zeraoulia rafikzeraoulia rafik

2,44211032

asked Mar 15 at 16:43

zeraoulia rafikzeraoulia rafik

2,44211032

3 Answers
3

active

oldest

votes

3

$begingroup$

The formulas for the complex cosine and sine can be derived from Euler's formula
$$e^iz = cos(z) + i sin(z)
$$
By substituting $-z$, and using that $cos$ is even and $sin$ is odd, one obtains
$$e^-iz = cos(z) - i sin(z)
$$
Thinking of this as a system of two equations in two unknowns, solving gives
$$cos(z) = frac12 (e^iz + e^-iz)
$$
and
$$sin(z) = frac12i (e^iz - e^-iz)
$$
From this you can see that $cos(z)$ is unbounded along the imaginary axis, because
$$cos(iy) = frac12 (e^-y + e^y)
$$
and similarly for the $sin$ function.

share|cite|improve this answer

answered Mar 15 at 16:54

Lee MosherLee Mosher

51.1k33889

$endgroup$

add a comment | 

4

$begingroup$

Another way to look at this is from the perspective of Liouville's theorem: If $f(z)$ is a holomorphic function, defined on all of $Bbb C$, and $|f(z)|$ is bounded, then $f(z)$ is constant.

Since $cos z$ and $sin z$ are nonconstant, holomorphic, and defined for every $zin Bbb C$ (the series defining them converge absolutely and uniformly on arbitrarily large compact subsets of $Bbb C$), they are unbounded by Liouville's theorem.

share|cite|improve this answer

edited Mar 15 at 19:11

answered Mar 15 at 17:04

Alex OrtizAlex Ortiz

11k21441

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add a comment | 

2

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Using Picard’s little theorem and the knowledge that $sin z$ and $cos z$ are entire (i.e. holomorphic for all $zinmathbbC$, we can also conclude that $sin z$ and $cos z$ are unbounded. Picard’s little theorem states that an entire function $f:mathbb Ctomathbb C$ takes on every value in $mathbb C$ with at most one exception.

share|cite|improve this answer

answered Mar 15 at 17:02

csch2csch2

6031314

$endgroup$

add a comment | 

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3

$begingroup$

The formulas for the complex cosine and sine can be derived from Euler's formula
$$e^iz = cos(z) + i sin(z)
$$
By substituting $-z$, and using that $cos$ is even and $sin$ is odd, one obtains
$$e^-iz = cos(z) - i sin(z)
$$
Thinking of this as a system of two equations in two unknowns, solving gives
$$cos(z) = frac12 (e^iz + e^-iz)
$$
and
$$sin(z) = frac12i (e^iz - e^-iz)
$$
From this you can see that $cos(z)$ is unbounded along the imaginary axis, because
$$cos(iy) = frac12 (e^-y + e^y)
$$
and similarly for the $sin$ function.

share|cite|improve this answer

answered Mar 15 at 16:54

Lee MosherLee Mosher

51.1k33889

$endgroup$

add a comment | 

3

$begingroup$

The formulas for the complex cosine and sine can be derived from Euler's formula
$$e^iz = cos(z) + i sin(z)
$$
By substituting $-z$, and using that $cos$ is even and $sin$ is odd, one obtains
$$e^-iz = cos(z) - i sin(z)
$$
Thinking of this as a system of two equations in two unknowns, solving gives
$$cos(z) = frac12 (e^iz + e^-iz)
$$
and
$$sin(z) = frac12i (e^iz - e^-iz)
$$
From this you can see that $cos(z)$ is unbounded along the imaginary axis, because
$$cos(iy) = frac12 (e^-y + e^y)
$$
and similarly for the $sin$ function.

share|cite|improve this answer

answered Mar 15 at 16:54

Lee MosherLee Mosher

51.1k33889

$endgroup$

add a comment | 

$begingroup$

The formulas for the complex cosine and sine can be derived from Euler's formula
$$e^iz = cos(z) + i sin(z)
$$
By substituting $-z$, and using that $cos$ is even and $sin$ is odd, one obtains
$$e^-iz = cos(z) - i sin(z)
$$
Thinking of this as a system of two equations in two unknowns, solving gives
$$cos(z) = frac12 (e^iz + e^-iz)
$$
and
$$sin(z) = frac12i (e^iz - e^-iz)
$$
From this you can see that $cos(z)$ is unbounded along the imaginary axis, because
$$cos(iy) = frac12 (e^-y + e^y)
$$
and similarly for the $sin$ function.

share|cite|improve this answer

answered Mar 15 at 16:54

Lee MosherLee Mosher

51.1k33889

$endgroup$

share|cite|improve this answer

answered Mar 15 at 16:54

Lee MosherLee Mosher

51.1k33889

answered Mar 15 at 16:54

Lee MosherLee Mosher

51.1k33889

answered Mar 15 at 16:54

Lee MosherLee Mosher

51.1k33889

4

$begingroup$

Another way to look at this is from the perspective of Liouville's theorem: If $f(z)$ is a holomorphic function, defined on all of $Bbb C$, and $|f(z)|$ is bounded, then $f(z)$ is constant.

Since $cos z$ and $sin z$ are nonconstant, holomorphic, and defined for every $zin Bbb C$ (the series defining them converge absolutely and uniformly on arbitrarily large compact subsets of $Bbb C$), they are unbounded by Liouville's theorem.

share|cite|improve this answer

edited Mar 15 at 19:11

answered Mar 15 at 17:04

Alex OrtizAlex Ortiz

11k21441

$endgroup$

add a comment | 

4

$begingroup$

Another way to look at this is from the perspective of Liouville's theorem: If $f(z)$ is a holomorphic function, defined on all of $Bbb C$, and $|f(z)|$ is bounded, then $f(z)$ is constant.

Since $cos z$ and $sin z$ are nonconstant, holomorphic, and defined for every $zin Bbb C$ (the series defining them converge absolutely and uniformly on arbitrarily large compact subsets of $Bbb C$), they are unbounded by Liouville's theorem.

share|cite|improve this answer

edited Mar 15 at 19:11

answered Mar 15 at 17:04

Alex OrtizAlex Ortiz

11k21441

$endgroup$

add a comment | 

$begingroup$

Another way to look at this is from the perspective of Liouville's theorem: If $f(z)$ is a holomorphic function, defined on all of $Bbb C$, and $|f(z)|$ is bounded, then $f(z)$ is constant.

Since $cos z$ and $sin z$ are nonconstant, holomorphic, and defined for every $zin Bbb C$ (the series defining them converge absolutely and uniformly on arbitrarily large compact subsets of $Bbb C$), they are unbounded by Liouville's theorem.

share|cite|improve this answer

edited Mar 15 at 19:11

answered Mar 15 at 17:04

Alex OrtizAlex Ortiz

11k21441

$endgroup$

share|cite|improve this answer

edited Mar 15 at 19:11

answered Mar 15 at 17:04

Alex OrtizAlex Ortiz

11k21441

answered Mar 15 at 17:04

Alex OrtizAlex Ortiz

11k21441

answered Mar 15 at 17:04

Alex OrtizAlex Ortiz

11k21441

2

$begingroup$

Using Picard’s little theorem and the knowledge that $sin z$ and $cos z$ are entire (i.e. holomorphic for all $zinmathbbC$, we can also conclude that $sin z$ and $cos z$ are unbounded. Picard’s little theorem states that an entire function $f:mathbb Ctomathbb C$ takes on every value in $mathbb C$ with at most one exception.

share|cite|improve this answer

answered Mar 15 at 17:02

csch2csch2

6031314

$endgroup$

add a comment | 

2

$begingroup$

Using Picard’s little theorem and the knowledge that $sin z$ and $cos z$ are entire (i.e. holomorphic for all $zinmathbbC$, we can also conclude that $sin z$ and $cos z$ are unbounded. Picard’s little theorem states that an entire function $f:mathbb Ctomathbb C$ takes on every value in $mathbb C$ with at most one exception.

share|cite|improve this answer

answered Mar 15 at 17:02

csch2csch2

6031314

$endgroup$

add a comment | 

$begingroup$

Using Picard’s little theorem and the knowledge that $sin z$ and $cos z$ are entire (i.e. holomorphic for all $zinmathbbC$, we can also conclude that $sin z$ and $cos z$ are unbounded. Picard’s little theorem states that an entire function $f:mathbb Ctomathbb C$ takes on every value in $mathbb C$ with at most one exception.

share|cite|improve this answer

answered Mar 15 at 17:02

csch2csch2

6031314

$endgroup$

share|cite|improve this answer

answered Mar 15 at 17:02

csch2csch2

6031314

answered Mar 15 at 17:02

csch2csch2

6031314

answered Mar 15 at 17:02

csch2csch2

6031314