Find points where $f(z)$ is holomophic and compute $f'(z)$ [closed]
$begingroup$
If we have a complex function
$f(x+iy) = cos(x)cos(y)-isin(x)sin(y)$ for $x,y ∈ {rm I!R}$
Find the set of points at which f:$mathbb{C}$ $Rightarrow$ $mathbb{C}$ are holomorphic and compute $f '(z)$.
I understand how to differentiate the function however i'm unsure as to go about proving at which points the function is holomorphic. And how to show it.
complex-analysis analysis holomorphic-functions
edited Dec 2 '18 at 16:33
Larry
2,27231028
asked Dec 2 '18 at 16:27
L GL G
248
$endgroup$
closed as unclear what you're asking by Did, Cesareo, Rebellos, MisterRiemann, Lee David Chung Lin Dec 4 '18 at 21:15
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.
add a comment |
$begingroup$
If we have a complex function
$f(x+iy) = cos(x)cos(y)-isin(x)sin(y)$ for $x,y ∈ {rm I!R}$
Find the set of points at which f:$mathbb{C}$ $Rightarrow$ $mathbb{C}$ are holomorphic and compute $f '(z)$.
I understand how to differentiate the function however i'm unsure as to go about proving at which points the function is holomorphic. And how to show it.
complex-analysis analysis holomorphic-functions
edited Dec 2 '18 at 16:33
Larry
2,27231028
asked Dec 2 '18 at 16:27
L GL G
248
$endgroup$
closed as unclear what you're asking by Did, Cesareo, Rebellos, MisterRiemann, Lee David Chung Lin Dec 4 '18 at 21:15
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.
add a comment |
$begingroup$
If we have a complex function
$f(x+iy) = cos(x)cos(y)-isin(x)sin(y)$ for $x,y ∈ {rm I!R}$
Find the set of points at which f:$mathbb{C}$ $Rightarrow$ $mathbb{C}$ are holomorphic and compute $f '(z)$.
I understand how to differentiate the function however i'm unsure as to go about proving at which points the function is holomorphic. And how to show it.
complex-analysis analysis holomorphic-functions
edited Dec 2 '18 at 16:33
Larry
2,27231028
asked Dec 2 '18 at 16:27
L GL G
248
$endgroup$
If we have a complex function
$f(x+iy) = cos(x)cos(y)-isin(x)sin(y)$ for $x,y ∈ {rm I!R}$
Find the set of points at which f:$mathbb{C}$ $Rightarrow$ $mathbb{C}$ are holomorphic and compute $f '(z)$.
I understand how to differentiate the function however i'm unsure as to go about proving at which points the function is holomorphic. And how to show it.
complex-analysis analysis holomorphic-functions
complex-analysis analysis holomorphic-functions
edited Dec 2 '18 at 16:33
Larry
2,27231028
asked Dec 2 '18 at 16:27
L GL G
248
edited Dec 2 '18 at 16:33
Larry
2,27231028
asked Dec 2 '18 at 16:27
L GL G
248
edited Dec 2 '18 at 16:33
Larry
2,27231028
edited Dec 2 '18 at 16:33
Larry
2,27231028
edited Dec 2 '18 at 16:33
Larry
2,27231028
2,27231028
asked Dec 2 '18 at 16:27
L GL G
248
asked Dec 2 '18 at 16:27
L GL G
248
asked Dec 2 '18 at 16:27
L GL G
248
248
closed as unclear what you're asking by Did, Cesareo, Rebellos, MisterRiemann, Lee David Chung Lin Dec 4 '18 at 21:15
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.
closed as unclear what you're asking by Did, Cesareo, Rebellos, MisterRiemann, Lee David Chung Lin Dec 4 '18 at 21:15
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.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Holomorphic functions $f:Omegato mathbb{C}$ satisfy a pair of partial differential equations called the Cauchy-Riemann Equations. If we write $f(x,y)=u(x,y)+iv(x,y)$ then the Cauchy-Riemann Equations read
$$ begin{cases}
u_x=v_y\
u_y=-v_x.
end{cases}$$
Compute the set of $(x,y)in mathbb{C}$ so that $f$ satisfies these equations. Then, to differentiate, observe that
$$ f'(z)=lim_{hto 0}frac{f(z+h)-f(z)}{h}.$$
However, because $h$ can tend to $0$ from whichever direction we wish, $f'(z)=u_x+iv_x$. In particular, the complex derivative is equal to either of the $x$ or $y$ partial derivatives.
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
yeah this makes a lot of sense I've figured out the Cauchy-Reimann equations and found that cos(x)sin(y)=-cos(x)sin(y) which can't be true so since they don't satisfy the Cauchy Reimann equations the function isn't holomorphic for any values of x,y
$endgroup$
– L G
Dec 2 '18 at 16:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Holomorphic functions $f:Omegato mathbb{C}$ satisfy a pair of partial differential equations called the Cauchy-Riemann Equations. If we write $f(x,y)=u(x,y)+iv(x,y)$ then the Cauchy-Riemann Equations read
$$ begin{cases}
u_x=v_y\
u_y=-v_x.
end{cases}$$
Compute the set of $(x,y)in mathbb{C}$ so that $f$ satisfies these equations. Then, to differentiate, observe that
$$ f'(z)=lim_{hto 0}frac{f(z+h)-f(z)}{h}.$$
However, because $h$ can tend to $0$ from whichever direction we wish, $f'(z)=u_x+iv_x$. In particular, the complex derivative is equal to either of the $x$ or $y$ partial derivatives.
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
yeah this makes a lot of sense I've figured out the Cauchy-Reimann equations and found that cos(x)sin(y)=-cos(x)sin(y) which can't be true so since they don't satisfy the Cauchy Reimann equations the function isn't holomorphic for any values of x,y
$endgroup$
– L G
Dec 2 '18 at 16:56
add a comment |
$begingroup$
Holomorphic functions $f:Omegato mathbb{C}$ satisfy a pair of partial differential equations called the Cauchy-Riemann Equations. If we write $f(x,y)=u(x,y)+iv(x,y)$ then the Cauchy-Riemann Equations read
$$ begin{cases}
u_x=v_y\
u_y=-v_x.
end{cases}$$
Compute the set of $(x,y)in mathbb{C}$ so that $f$ satisfies these equations. Then, to differentiate, observe that
$$ f'(z)=lim_{hto 0}frac{f(z+h)-f(z)}{h}.$$
However, because $h$ can tend to $0$ from whichever direction we wish, $f'(z)=u_x+iv_x$. In particular, the complex derivative is equal to either of the $x$ or $y$ partial derivatives.
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
yeah this makes a lot of sense I've figured out the Cauchy-Reimann equations and found that cos(x)sin(y)=-cos(x)sin(y) which can't be true so since they don't satisfy the Cauchy Reimann equations the function isn't holomorphic for any values of x,y
$endgroup$
– L G
Dec 2 '18 at 16:56
add a comment |
$begingroup$
Holomorphic functions $f:Omegato mathbb{C}$ satisfy a pair of partial differential equations called the Cauchy-Riemann Equations. If we write $f(x,y)=u(x,y)+iv(x,y)$ then the Cauchy-Riemann Equations read
$$ begin{cases}
u_x=v_y\
u_y=-v_x.
end{cases}$$
Compute the set of $(x,y)in mathbb{C}$ so that $f$ satisfies these equations. Then, to differentiate, observe that
$$ f'(z)=lim_{hto 0}frac{f(z+h)-f(z)}{h}.$$
However, because $h$ can tend to $0$ from whichever direction we wish, $f'(z)=u_x+iv_x$. In particular, the complex derivative is equal to either of the $x$ or $y$ partial derivatives.
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
Holomorphic functions $f:Omegato mathbb{C}$ satisfy a pair of partial differential equations called the Cauchy-Riemann Equations. If we write $f(x,y)=u(x,y)+iv(x,y)$ then the Cauchy-Riemann Equations read
$$ begin{cases}
u_x=v_y\
u_y=-v_x.
end{cases}$$
Compute the set of $(x,y)in mathbb{C}$ so that $f$ satisfies these equations. Then, to differentiate, observe that
$$ f'(z)=lim_{hto 0}frac{f(z+h)-f(z)}{h}.$$
However, because $h$ can tend to $0$ from whichever direction we wish, $f'(z)=u_x+iv_x$. In particular, the complex derivative is equal to either of the $x$ or $y$ partial derivatives.
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Dec 2 '18 at 16:38
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
9,91741640
$begingroup$
yeah this makes a lot of sense I've figured out the Cauchy-Reimann equations and found that cos(x)sin(y)=-cos(x)sin(y) which can't be true so since they don't satisfy the Cauchy Reimann equations the function isn't holomorphic for any values of x,y
$endgroup$
– L G
Dec 2 '18 at 16:56
add a comment |
$begingroup$
yeah this makes a lot of sense I've figured out the Cauchy-Reimann equations and found that cos(x)sin(y)=-cos(x)sin(y) which can't be true so since they don't satisfy the Cauchy Reimann equations the function isn't holomorphic for any values of x,y
$endgroup$
– L G
Dec 2 '18 at 16:56
$begingroup$
yeah this makes a lot of sense I've figured out the Cauchy-Reimann equations and found that cos(x)sin(y)=-cos(x)sin(y) which can't be true so since they don't satisfy the Cauchy Reimann equations the function isn't holomorphic for any values of x,y
$endgroup$
– L G
Dec 2 '18 at 16:56
$begingroup$
yeah this makes a lot of sense I've figured out the Cauchy-Reimann equations and found that cos(x)sin(y)=-cos(x)sin(y) which can't be true so since they don't satisfy the Cauchy Reimann equations the function isn't holomorphic for any values of x,y
$endgroup$
– L G
Dec 2 '18 at 16:56
add a comment |
