Mixing
Range of real values of sin(z)
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Given $f(z) = sin(z)$ such that $z$ is an element of the complex numbers is the range of the real part of $f(z)$ all the reals?
Is the range of the real part of $f(z)$ all reals given that the imaginary part is zero?
Sorry for the formatting, I’m on mobile.
trigonometry complex-numbers hyperbolic-functions hyperbolic-equations
edited 56 mins ago
Mark Viola
126k1172167
asked 1 hour ago
Matt
183
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Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
favorite
Given $f(z) = sin(z)$ such that $z$ is an element of the complex numbers is the range of the real part of $f(z)$ all the reals?
Is the range of the real part of $f(z)$ all reals given that the imaginary part is zero?
Sorry for the formatting, I’m on mobile.
trigonometry complex-numbers hyperbolic-functions hyperbolic-equations
edited 56 mins ago
Mark Viola
126k1172167
asked 1 hour ago
Matt
183
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Do you know the little Picard Theorem?
– Hirshy
51 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given $f(z) = sin(z)$ such that $z$ is an element of the complex numbers is the range of the real part of $f(z)$ all the reals?
Is the range of the real part of $f(z)$ all reals given that the imaginary part is zero?
Sorry for the formatting, I’m on mobile.
trigonometry complex-numbers hyperbolic-functions hyperbolic-equations
edited 56 mins ago
Mark Viola
126k1172167
asked 1 hour ago
Matt
183
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given $f(z) = sin(z)$ such that $z$ is an element of the complex numbers is the range of the real part of $f(z)$ all the reals?
Is the range of the real part of $f(z)$ all reals given that the imaginary part is zero?
Sorry for the formatting, I’m on mobile.
trigonometry complex-numbers hyperbolic-functions hyperbolic-equations
trigonometry complex-numbers hyperbolic-functions hyperbolic-equations
edited 56 mins ago
Mark Viola
126k1172167
asked 1 hour ago
Matt
183
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 56 mins ago
Mark Viola
126k1172167
asked 1 hour ago
Matt
183
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 56 mins ago
Mark Viola
126k1172167
edited 56 mins ago
Mark Viola
126k1172167
edited 56 mins ago
Mark Viola
126k1172167
126k1172167
asked 1 hour ago
Matt
183
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Matt
183
asked 1 hour ago
Matt
183
183
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Matt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Do you know the little Picard Theorem?
– Hirshy
51 mins ago
add a comment |Â
1
Do you know the little Picard Theorem?
– Hirshy
51 mins ago
1
1
Do you know the little Picard Theorem?
– Hirshy
51 mins ago
Do you know the little Picard Theorem?
– Hirshy
51 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
The equation $sin z=w$ has solution for every complex $w$ (in particular real).
Indeed, if $t=e^iz$, the equation becomes
$$
fract-t^-12i=w
$$
that is
$$
t^2-2iwt-1=0
$$
Therefore $t=iw+sqrt1-w^2$ (or its inverse). In particular $0$ is never a solution and the equation $e^iz=t$ certainly has solution.
answered 49 mins ago
egreg
167k1180189
add a comment |Â
up vote
2
down vote
Building on my comment here is a way to prove it without needing to calculate any solutions:
as $sin:mathbb Crightarrowmathbb C$ is entire and non constant, we can apply the Little Picard Theorem and therefore we have that $sin(mathbb C)=mathbb C$ or $sin(mathbb C)=mathbb Csetminusa$ for one $ainmathbb C$. Let us assume the later. Because $sin$ is an uneven function this would mean that also $-anotinsin(mathbb C)$. This can only be true if $a=0$ (else we would contradict Picard), but we all know that $sin(0)=0$ and therefor $sin(mathbb C)=mathbb C$.
answered 46 mins ago
Hirshy
4,32921336
Nice use of the sledgehammer!
– egreg
42 mins ago
@egreg always fun to use it when (un)necessary ;)
– Hirshy
40 mins ago
add a comment |Â
up vote
0
down vote
$$
sin (i z + fracpi2) = cosh z \
sin (i z - fracpi2) = - cosh z\
$$
So the image of $fracpi2+imathbbR$ takes care of getting $mathbbR_geq 1$. Similarly the image of $frac-pi2+imathbbR$ takes care of getting $mathbbR_leq -1$. The image of $mathbbR$ takes care of getting $[-1,1]$ with $sin$ being viewed as a usual $mathbbR to [-1,1]$.
answered 47 mins ago
AHusain
1,914714
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The equation $sin z=w$ has solution for every complex $w$ (in particular real).
Indeed, if $t=e^iz$, the equation becomes
$$
fract-t^-12i=w
$$
that is
$$
t^2-2iwt-1=0
$$
Therefore $t=iw+sqrt1-w^2$ (or its inverse). In particular $0$ is never a solution and the equation $e^iz=t$ certainly has solution.
answered 49 mins ago
egreg
167k1180189
add a comment |Â
up vote
3
down vote
accepted
The equation $sin z=w$ has solution for every complex $w$ (in particular real).
Indeed, if $t=e^iz$, the equation becomes
$$
fract-t^-12i=w
$$
that is
$$
t^2-2iwt-1=0
$$
Therefore $t=iw+sqrt1-w^2$ (or its inverse). In particular $0$ is never a solution and the equation $e^iz=t$ certainly has solution.
answered 49 mins ago
egreg
167k1180189
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The equation $sin z=w$ has solution for every complex $w$ (in particular real).
Indeed, if $t=e^iz$, the equation becomes
$$
fract-t^-12i=w
$$
that is
$$
t^2-2iwt-1=0
$$
Therefore $t=iw+sqrt1-w^2$ (or its inverse). In particular $0$ is never a solution and the equation $e^iz=t$ certainly has solution.
answered 49 mins ago
egreg
167k1180189
The equation $sin z=w$ has solution for every complex $w$ (in particular real).
Indeed, if $t=e^iz$, the equation becomes
$$
fract-t^-12i=w
$$
that is
$$
t^2-2iwt-1=0
$$
Therefore $t=iw+sqrt1-w^2$ (or its inverse). In particular $0$ is never a solution and the equation $e^iz=t$ certainly has solution.
answered 49 mins ago
egreg
167k1180189
answered 49 mins ago
egreg
167k1180189
answered 49 mins ago
egreg
167k1180189
answered 49 mins ago
egreg
167k1180189
167k1180189
add a comment |Â
add a comment |Â
up vote
2
down vote
Building on my comment here is a way to prove it without needing to calculate any solutions:
as $sin:mathbb Crightarrowmathbb C$ is entire and non constant, we can apply the Little Picard Theorem and therefore we have that $sin(mathbb C)=mathbb C$ or $sin(mathbb C)=mathbb Csetminusa$ for one $ainmathbb C$. Let us assume the later. Because $sin$ is an uneven function this would mean that also $-anotinsin(mathbb C)$. This can only be true if $a=0$ (else we would contradict Picard), but we all know that $sin(0)=0$ and therefor $sin(mathbb C)=mathbb C$.
answered 46 mins ago
Hirshy
4,32921336
Nice use of the sledgehammer!
– egreg
42 mins ago
@egreg always fun to use it when (un)necessary ;)
– Hirshy
40 mins ago
add a comment |Â
up vote
2
down vote
Building on my comment here is a way to prove it without needing to calculate any solutions:
as $sin:mathbb Crightarrowmathbb C$ is entire and non constant, we can apply the Little Picard Theorem and therefore we have that $sin(mathbb C)=mathbb C$ or $sin(mathbb C)=mathbb Csetminusa$ for one $ainmathbb C$. Let us assume the later. Because $sin$ is an uneven function this would mean that also $-anotinsin(mathbb C)$. This can only be true if $a=0$ (else we would contradict Picard), but we all know that $sin(0)=0$ and therefor $sin(mathbb C)=mathbb C$.
answered 46 mins ago
Hirshy
4,32921336
Nice use of the sledgehammer!
– egreg
42 mins ago
@egreg always fun to use it when (un)necessary ;)
– Hirshy
40 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Building on my comment here is a way to prove it without needing to calculate any solutions:
as $sin:mathbb Crightarrowmathbb C$ is entire and non constant, we can apply the Little Picard Theorem and therefore we have that $sin(mathbb C)=mathbb C$ or $sin(mathbb C)=mathbb Csetminusa$ for one $ainmathbb C$. Let us assume the later. Because $sin$ is an uneven function this would mean that also $-anotinsin(mathbb C)$. This can only be true if $a=0$ (else we would contradict Picard), but we all know that $sin(0)=0$ and therefor $sin(mathbb C)=mathbb C$.
answered 46 mins ago
Hirshy
4,32921336
Building on my comment here is a way to prove it without needing to calculate any solutions:
as $sin:mathbb Crightarrowmathbb C$ is entire and non constant, we can apply the Little Picard Theorem and therefore we have that $sin(mathbb C)=mathbb C$ or $sin(mathbb C)=mathbb Csetminusa$ for one $ainmathbb C$. Let us assume the later. Because $sin$ is an uneven function this would mean that also $-anotinsin(mathbb C)$. This can only be true if $a=0$ (else we would contradict Picard), but we all know that $sin(0)=0$ and therefor $sin(mathbb C)=mathbb C$.
answered 46 mins ago
Hirshy
4,32921336
answered 46 mins ago
Hirshy
4,32921336
answered 46 mins ago
Hirshy
4,32921336
answered 46 mins ago
Hirshy
4,32921336
4,32921336
Nice use of the sledgehammer!
– egreg
42 mins ago
@egreg always fun to use it when (un)necessary ;)
– Hirshy
40 mins ago
add a comment |Â
Nice use of the sledgehammer!
– egreg
42 mins ago
@egreg always fun to use it when (un)necessary ;)
– Hirshy
40 mins ago
Nice use of the sledgehammer!
– egreg
42 mins ago
Nice use of the sledgehammer!
– egreg
42 mins ago
@egreg always fun to use it when (un)necessary ;)
– Hirshy
40 mins ago
@egreg always fun to use it when (un)necessary ;)
– Hirshy
40 mins ago
add a comment |Â
up vote
0
down vote
$$
sin (i z + fracpi2) = cosh z \
sin (i z - fracpi2) = - cosh z\
$$
So the image of $fracpi2+imathbbR$ takes care of getting $mathbbR_geq 1$. Similarly the image of $frac-pi2+imathbbR$ takes care of getting $mathbbR_leq -1$. The image of $mathbbR$ takes care of getting $[-1,1]$ with $sin$ being viewed as a usual $mathbbR to [-1,1]$.
answered 47 mins ago
AHusain
1,914714
add a comment |Â
up vote
0
down vote
$$
sin (i z + fracpi2) = cosh z \
sin (i z - fracpi2) = - cosh z\
$$
So the image of $fracpi2+imathbbR$ takes care of getting $mathbbR_geq 1$. Similarly the image of $frac-pi2+imathbbR$ takes care of getting $mathbbR_leq -1$. The image of $mathbbR$ takes care of getting $[-1,1]$ with $sin$ being viewed as a usual $mathbbR to [-1,1]$.
answered 47 mins ago
AHusain
1,914714
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$
sin (i z + fracpi2) = cosh z \
sin (i z - fracpi2) = - cosh z\
$$
So the image of $fracpi2+imathbbR$ takes care of getting $mathbbR_geq 1$. Similarly the image of $frac-pi2+imathbbR$ takes care of getting $mathbbR_leq -1$. The image of $mathbbR$ takes care of getting $[-1,1]$ with $sin$ being viewed as a usual $mathbbR to [-1,1]$.
answered 47 mins ago
AHusain
1,914714
$$
sin (i z + fracpi2) = cosh z \
sin (i z - fracpi2) = - cosh z\
$$
So the image of $fracpi2+imathbbR$ takes care of getting $mathbbR_geq 1$. Similarly the image of $frac-pi2+imathbbR$ takes care of getting $mathbbR_leq -1$. The image of $mathbbR$ takes care of getting $[-1,1]$ with $sin$ being viewed as a usual $mathbbR to [-1,1]$.
answered 47 mins ago
AHusain
1,914714
answered 47 mins ago
AHusain
1,914714
answered 47 mins ago
AHusain
1,914714
answered 47 mins ago
AHusain
1,914714
1,914714
add a comment |Â
add a comment |Â
Matt is a new contributor. Be nice, and check out our Code of Conduct.
Matt is a new contributor. Be nice, and check out our Code of Conduct.
Matt is a new contributor. Be nice, and check out our Code of Conduct.
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1
Do you know the little Picard Theorem?
– Hirshy
51 mins ago