Geometric representation of all the points for $cos(z) = 2$, $z in mathbb C$
$begingroup$
So I am having trouble visualizing the solutions to
$$cos(z) = 2, z in mathbb{C}$$
I know that the solution of this (like mentioned of here Solving $cos z = 2$ ), but what points does this equation represent on the complex plane? Does it simply represent all the points which are on $x=2$ and $y=0$?
complex-analysis complex-numbers
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
So I am having trouble visualizing the solutions to
$$cos(z) = 2, z in mathbb{C}$$
I know that the solution of this (like mentioned of here Solving $cos z = 2$ ), but what points does this equation represent on the complex plane? Does it simply represent all the points which are on $x=2$ and $y=0$?
complex-analysis complex-numbers
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
$endgroup$
add a comment |
$begingroup$
So I am having trouble visualizing the solutions to
$$cos(z) = 2, z in mathbb{C}$$
I know that the solution of this (like mentioned of here Solving $cos z = 2$ ), but what points does this equation represent on the complex plane? Does it simply represent all the points which are on $x=2$ and $y=0$?
complex-analysis complex-numbers
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
$endgroup$
So I am having trouble visualizing the solutions to
$$cos(z) = 2, z in mathbb{C}$$
I know that the solution of this (like mentioned of here Solving $cos z = 2$ ), but what points does this equation represent on the complex plane? Does it simply represent all the points which are on $x=2$ and $y=0$?
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
edited Dec 20 '18 at 16:10
Andrei
12.4k21128
12.4k21128
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
asked Dec 20 '18 at 15:52
daljit97daljit97
178111
178111
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $z=x+iy$ then
$$ cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy) = cos(x)cosh(y) - isin(x)sinh(y) $$
Therefore
begin{align} cos(x)cosh(y) &= 2 \ sin(x)sinh(y) &= 0 end{align}
Obviously, no real solution exists, so $yne 0 implies sin x = 0$.
But $cosh (y) > 0, forall y implies cos x > 0 implies cos(x) = 1 implies x = 2npi$, which leaves
$$ cosh(y) = 2 implies y = pm cosh^{-1}(2) = pm ln (2 + sqrt{3}) $$
So the solution is
$$ x = 2npi pm iln(2 + sqrt{3}) $$
For a visualization, think of the solutions as points on the 2 lines $y = pm ln(2+sqrt{3})$, spaced out evenly by $2npi$
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
$endgroup$
$begingroup$
Why is $implies cos(x) = 1$ true? Where does that follow from? Also why there is no real solution?
$endgroup$
– daljit97
Dec 23 '18 at 17:36
1
$begingroup$
If $sin x = 0$ and $cos x > 0$, what does that imply about the value of $cos x$? What's the maximum value of $sin x$ for real $x$?
$endgroup$
– Dylan
Dec 23 '18 at 18:05
$begingroup$
Ok I understand, when $sin x = 0$, $cos x$ can only be 1 or -1. Obvioulsy $|sin x|$ cannot be greater than 1.
$endgroup$
– daljit97
Dec 23 '18 at 18:09
$begingroup$
So the justification behind $sin x =0$ would be that if $sin x sinh y= 0$, either $sin x=0$ or $sinh y =0$. Assuming $sinh y= 0$ implies $y = 2pi n, n in mathbb Z$. Therefore $cosh y = 1$ and $cos x = 2$ which is impossible, therefore $sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this?
$endgroup$
– daljit97
Dec 23 '18 at 18:42
1
$begingroup$
$sinh y = 0$ only has one solution which is $y=0$
$endgroup$
– Dylan
Dec 23 '18 at 18:45
|
show 2 more comments
$begingroup$
Have you thought about considering the polar form of the equation? It might help you think geometrically. $cos z = text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
$endgroup$
2
$begingroup$
If $z in mathbb C$, then $cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(cos x + isin x) $$ $cos z = frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function
$endgroup$
– Dylan
Dec 21 '18 at 18:38
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
$begingroup$
If $z=x+iy$ then
$$ cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy) = cos(x)cosh(y) - isin(x)sinh(y) $$
Therefore
begin{align} cos(x)cosh(y) &= 2 \ sin(x)sinh(y) &= 0 end{align}
Obviously, no real solution exists, so $yne 0 implies sin x = 0$.
But $cosh (y) > 0, forall y implies cos x > 0 implies cos(x) = 1 implies x = 2npi$, which leaves
$$ cosh(y) = 2 implies y = pm cosh^{-1}(2) = pm ln (2 + sqrt{3}) $$
So the solution is
$$ x = 2npi pm iln(2 + sqrt{3}) $$
For a visualization, think of the solutions as points on the 2 lines $y = pm ln(2+sqrt{3})$, spaced out evenly by $2npi$
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
$endgroup$
$begingroup$
Why is $implies cos(x) = 1$ true? Where does that follow from? Also why there is no real solution?
$endgroup$
– daljit97
Dec 23 '18 at 17:36
1
$begingroup$
If $sin x = 0$ and $cos x > 0$, what does that imply about the value of $cos x$? What's the maximum value of $sin x$ for real $x$?
$endgroup$
– Dylan
Dec 23 '18 at 18:05
$begingroup$
Ok I understand, when $sin x = 0$, $cos x$ can only be 1 or -1. Obvioulsy $|sin x|$ cannot be greater than 1.
$endgroup$
– daljit97
Dec 23 '18 at 18:09
$begingroup$
So the justification behind $sin x =0$ would be that if $sin x sinh y= 0$, either $sin x=0$ or $sinh y =0$. Assuming $sinh y= 0$ implies $y = 2pi n, n in mathbb Z$. Therefore $cosh y = 1$ and $cos x = 2$ which is impossible, therefore $sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this?
$endgroup$
– daljit97
Dec 23 '18 at 18:42
1
$begingroup$
$sinh y = 0$ only has one solution which is $y=0$
$endgroup$
– Dylan
Dec 23 '18 at 18:45
|
show 2 more comments
$begingroup$
If $z=x+iy$ then
$$ cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy) = cos(x)cosh(y) - isin(x)sinh(y) $$
Therefore
begin{align} cos(x)cosh(y) &= 2 \ sin(x)sinh(y) &= 0 end{align}
Obviously, no real solution exists, so $yne 0 implies sin x = 0$.
But $cosh (y) > 0, forall y implies cos x > 0 implies cos(x) = 1 implies x = 2npi$, which leaves
$$ cosh(y) = 2 implies y = pm cosh^{-1}(2) = pm ln (2 + sqrt{3}) $$
So the solution is
$$ x = 2npi pm iln(2 + sqrt{3}) $$
For a visualization, think of the solutions as points on the 2 lines $y = pm ln(2+sqrt{3})$, spaced out evenly by $2npi$
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
$endgroup$
$begingroup$
Why is $implies cos(x) = 1$ true? Where does that follow from? Also why there is no real solution?
$endgroup$
– daljit97
Dec 23 '18 at 17:36
1
$begingroup$
If $sin x = 0$ and $cos x > 0$, what does that imply about the value of $cos x$? What's the maximum value of $sin x$ for real $x$?
$endgroup$
– Dylan
Dec 23 '18 at 18:05
$begingroup$
Ok I understand, when $sin x = 0$, $cos x$ can only be 1 or -1. Obvioulsy $|sin x|$ cannot be greater than 1.
$endgroup$
– daljit97
Dec 23 '18 at 18:09
$begingroup$
So the justification behind $sin x =0$ would be that if $sin x sinh y= 0$, either $sin x=0$ or $sinh y =0$. Assuming $sinh y= 0$ implies $y = 2pi n, n in mathbb Z$. Therefore $cosh y = 1$ and $cos x = 2$ which is impossible, therefore $sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this?
$endgroup$
– daljit97
Dec 23 '18 at 18:42
1
$begingroup$
$sinh y = 0$ only has one solution which is $y=0$
$endgroup$
– Dylan
Dec 23 '18 at 18:45
|
show 2 more comments
$begingroup$
If $z=x+iy$ then
$$ cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy) = cos(x)cosh(y) - isin(x)sinh(y) $$
Therefore
begin{align} cos(x)cosh(y) &= 2 \ sin(x)sinh(y) &= 0 end{align}
Obviously, no real solution exists, so $yne 0 implies sin x = 0$.
But $cosh (y) > 0, forall y implies cos x > 0 implies cos(x) = 1 implies x = 2npi$, which leaves
$$ cosh(y) = 2 implies y = pm cosh^{-1}(2) = pm ln (2 + sqrt{3}) $$
So the solution is
$$ x = 2npi pm iln(2 + sqrt{3}) $$
For a visualization, think of the solutions as points on the 2 lines $y = pm ln(2+sqrt{3})$, spaced out evenly by $2npi$
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
$endgroup$
If $z=x+iy$ then
$$ cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy) = cos(x)cosh(y) - isin(x)sinh(y) $$
Therefore
begin{align} cos(x)cosh(y) &= 2 \ sin(x)sinh(y) &= 0 end{align}
Obviously, no real solution exists, so $yne 0 implies sin x = 0$.
But $cosh (y) > 0, forall y implies cos x > 0 implies cos(x) = 1 implies x = 2npi$, which leaves
$$ cosh(y) = 2 implies y = pm cosh^{-1}(2) = pm ln (2 + sqrt{3}) $$
So the solution is
$$ x = 2npi pm iln(2 + sqrt{3}) $$
For a visualization, think of the solutions as points on the 2 lines $y = pm ln(2+sqrt{3})$, spaced out evenly by $2npi$
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
edited Dec 23 '18 at 18:14
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
answered Dec 21 '18 at 18:36
DylanDylan
13.3k31027
13.3k31027
$begingroup$
Why is $implies cos(x) = 1$ true? Where does that follow from? Also why there is no real solution?
$endgroup$
– daljit97
Dec 23 '18 at 17:36
1
$begingroup$
If $sin x = 0$ and $cos x > 0$, what does that imply about the value of $cos x$? What's the maximum value of $sin x$ for real $x$?
$endgroup$
– Dylan
Dec 23 '18 at 18:05
$begingroup$
Ok I understand, when $sin x = 0$, $cos x$ can only be 1 or -1. Obvioulsy $|sin x|$ cannot be greater than 1.
$endgroup$
– daljit97
Dec 23 '18 at 18:09
$begingroup$
So the justification behind $sin x =0$ would be that if $sin x sinh y= 0$, either $sin x=0$ or $sinh y =0$. Assuming $sinh y= 0$ implies $y = 2pi n, n in mathbb Z$. Therefore $cosh y = 1$ and $cos x = 2$ which is impossible, therefore $sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this?
$endgroup$
– daljit97
Dec 23 '18 at 18:42
1
$begingroup$
$sinh y = 0$ only has one solution which is $y=0$
$endgroup$
– Dylan
Dec 23 '18 at 18:45
|
show 2 more comments
$begingroup$
Why is $implies cos(x) = 1$ true? Where does that follow from? Also why there is no real solution?
$endgroup$
– daljit97
Dec 23 '18 at 17:36
1
$begingroup$
If $sin x = 0$ and $cos x > 0$, what does that imply about the value of $cos x$? What's the maximum value of $sin x$ for real $x$?
$endgroup$
– Dylan
Dec 23 '18 at 18:05
$begingroup$
Ok I understand, when $sin x = 0$, $cos x$ can only be 1 or -1. Obvioulsy $|sin x|$ cannot be greater than 1.
$endgroup$
– daljit97
Dec 23 '18 at 18:09
$begingroup$
So the justification behind $sin x =0$ would be that if $sin x sinh y= 0$, either $sin x=0$ or $sinh y =0$. Assuming $sinh y= 0$ implies $y = 2pi n, n in mathbb Z$. Therefore $cosh y = 1$ and $cos x = 2$ which is impossible, therefore $sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this?
$endgroup$
– daljit97
Dec 23 '18 at 18:42
1
$begingroup$
$sinh y = 0$ only has one solution which is $y=0$
$endgroup$
– Dylan
Dec 23 '18 at 18:45
$begingroup$
Why is $implies cos(x) = 1$ true? Where does that follow from? Also why there is no real solution?
$endgroup$
– daljit97
Dec 23 '18 at 17:36
$begingroup$
Why is $implies cos(x) = 1$ true? Where does that follow from? Also why there is no real solution?
$endgroup$
– daljit97
Dec 23 '18 at 17:36
1
1
$begingroup$
If $sin x = 0$ and $cos x > 0$, what does that imply about the value of $cos x$? What's the maximum value of $sin x$ for real $x$?
$endgroup$
– Dylan
Dec 23 '18 at 18:05
$begingroup$
If $sin x = 0$ and $cos x > 0$, what does that imply about the value of $cos x$? What's the maximum value of $sin x$ for real $x$?
$endgroup$
– Dylan
Dec 23 '18 at 18:05
$begingroup$
Ok I understand, when $sin x = 0$, $cos x$ can only be 1 or -1. Obvioulsy $|sin x|$ cannot be greater than 1.
$endgroup$
– daljit97
Dec 23 '18 at 18:09
$begingroup$
Ok I understand, when $sin x = 0$, $cos x$ can only be 1 or -1. Obvioulsy $|sin x|$ cannot be greater than 1.
$endgroup$
– daljit97
Dec 23 '18 at 18:09
$begingroup$
So the justification behind $sin x =0$ would be that if $sin x sinh y= 0$, either $sin x=0$ or $sinh y =0$. Assuming $sinh y= 0$ implies $y = 2pi n, n in mathbb Z$. Therefore $cosh y = 1$ and $cos x = 2$ which is impossible, therefore $sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this?
$endgroup$
– daljit97
Dec 23 '18 at 18:42
$begingroup$
So the justification behind $sin x =0$ would be that if $sin x sinh y= 0$, either $sin x=0$ or $sinh y =0$. Assuming $sinh y= 0$ implies $y = 2pi n, n in mathbb Z$. Therefore $cosh y = 1$ and $cos x = 2$ which is impossible, therefore $sin x =0$. However, since you said that this is obvious is there a "shortcut" to deduce this?
$endgroup$
– daljit97
Dec 23 '18 at 18:42
1
1
$begingroup$
$sinh y = 0$ only has one solution which is $y=0$
$endgroup$
– Dylan
Dec 23 '18 at 18:45
$begingroup$
$sinh y = 0$ only has one solution which is $y=0$
$endgroup$
– Dylan
Dec 23 '18 at 18:45
|
show 2 more comments
$begingroup$
Have you thought about considering the polar form of the equation? It might help you think geometrically. $cos z = text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
$endgroup$
2
$begingroup$
If $z in mathbb C$, then $cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(cos x + isin x) $$ $cos z = frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function
$endgroup$
– Dylan
Dec 21 '18 at 18:38
add a comment |
$begingroup$
Have you thought about considering the polar form of the equation? It might help you think geometrically. $cos z = text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
$endgroup$
2
$begingroup$
If $z in mathbb C$, then $cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(cos x + isin x) $$ $cos z = frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function
$endgroup$
– Dylan
Dec 21 '18 at 18:38
add a comment |
$begingroup$
Have you thought about considering the polar form of the equation? It might help you think geometrically. $cos z = text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
$endgroup$
Have you thought about considering the polar form of the equation? It might help you think geometrically. $cos z = text{Re}(e^{iz})=2$. Then it is important that you consider the complex definition of $ln$ (not just the principal value) in the absence of a branch cut. Then it should be clear of the periodicity of your solutions.
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
answered Dec 20 '18 at 16:56
Joel BiffinJoel Biffin
966
966
2
$begingroup$
If $z in mathbb C$, then $cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(cos x + isin x) $$ $cos z = frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function
$endgroup$
– Dylan
Dec 21 '18 at 18:38
add a comment |
2
$begingroup$
If $z in mathbb C$, then $cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(cos x + isin x) $$ $cos z = frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function
$endgroup$
– Dylan
Dec 21 '18 at 18:38
2
2
$begingroup$
If $z in mathbb C$, then $cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(cos x + isin x) $$ $cos z = frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function
$endgroup$
– Dylan
Dec 21 '18 at 18:38
$begingroup$
If $z in mathbb C$, then $cos z$ is not the real part of $e^{iz}$, rather $$ e^{iz} = e^{i(x+iy)} = e^{-y}(cos x + isin x) $$ $cos z = frac{e^{iz}+e^{-iz}}{2}$ is not a real-valued function
$endgroup$
– Dylan
Dec 21 '18 at 18:38
add a comment |
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