Verifying CR Equations for $z\cos z$

In summary, when verifying the Cauchy-Riemann equations for $z\cos z$, there was a sign error in the calculation of $f(z)$, which resulted in incorrect values for $u(x,y)$ and $v(x,y)$. After correcting the sign error, the correct values for $u(x,y)$ and $v(x,y)$ are $x\cos x\cosh y + y\sin x\sinh y$ and $y\cos x\cosh y - x\sin x\sinh y$, respectively.
  • #1
Dustinsfl
2,281
5
$z\cos z$

Let $z = x + yi$.
Then $f(z) = (x + yi)\cos (x + yi)$.
By the addition rule for cosine and the identities $\cos yi = \cosh y$ and $-i\sin yi = \sinh y\Leftrightarrow \sin yi = i\sinh y$, we have that $\cos (x + yi) = \cos x\cosh y + i\sin x\sinh y$.
So
$$
f(z) = z\cos z = x\cos x\cosh y - y\sin x\sinh y + i(x\sin x\sinh y + y\cos x\cosh y).
$$
Then
$$
u(x,y) = x\cos x\cosh y - y\sin x\sinh y\quad\text{and}\quad
v(x,y) = y\cos x\cosh y + x\sin x\sinh y,
$$

I am trying to verify the CR equations but there is a negative sign difference. There has to be an error in my algebra but I can't find it. What is wrong with the above?
 
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  • #2
dwsmith said:
$z\cos z$

Let $z = x + yi$.
Then $f(z) = (x + yi)\cos (x + yi)$.
By the addition rule for cosine and the identities $\cos yi = \cosh y$ and $-i\sin yi = \sinh y\Leftrightarrow \sin yi = i\sinh y$, we have that $\cos (x + yi) = \cos x\cosh y\,{\color{red}-}\,i\sin x\sinh y$.
So
$$
f(z) = z\cos z = x\cos x\cosh y\,{\color{red}+}\, y\sin x\sinh y + i({\color{red}-}\,x\sin x\sinh y + y\cos x\cosh y).
$$
Then
$$
u(x,y) = x\cos x\cosh y\,{\color{red}+}\, y\sin x\sinh y\quad\text{and}\quad
v(x,y) = y\cos x\cosh y\,{\color{red}-}\, x\sin x\sinh y,
$$

I am trying to verify the CR equations but there is a negative sign difference. There has to be an error in my algebra but I can't find it. What is wrong with the above?

You had a sign error. See all my changes in red.
 
  • #3
Chris L T521 said:
You had a sign error. See all my changes in red.

Bad admin for missing the solved tag in the title for before you posted.:confused:
 
  • #4
dwsmith said:
Bad admin for missing the solved tag in the title for before you posted.:confused:

I clicked on the link from the home page, and it didn't show a solved tag. :P
 
  • #5


After reviewing your calculations, I can see that the error lies in the second term of each function, $u(x,y)$ and $v(x,y)$.
The correct expressions should be:
$$
u(x,y) = x\cos x\cosh y + y\sin x\sinh y\quad\text{and}\quad
v(x,y) = y\cos x\cosh y - x\sin x\sinh y.
$$
You may have mistakenly switched the signs in front of the terms involving $\sin$ and $\sinh$.
After making this correction, the CR equations are satisfied:
$$
u_x = \cos x\cosh y - x\sin x\cosh y = \cos(x+yi) = v_y,
$$
$$
u_y = x\cos x\sinh y + \sin x\sinh y = \sin(x+yi) = -v_x.
$$
Therefore, the equations are verified and the error has been corrected.
 

FAQ: Verifying CR Equations for $z\cos z$

How do we verify CR equations for $z\cos z$?

The Cauchy-Riemann (CR) equations for a complex function are given by:
$u_x = v_y$ and $u_y = -v_x$, where $u$ and $v$ represent the real and imaginary parts of the function, respectively. To verify these equations for $z\cos z$, we first find the real and imaginary parts of the function:
$z\cos z = (x + iy)(\cos x \cosh y - i\sin x \sinh y) = x\cos x \cosh y - y\sin x \sinh y + i(x\sin x \sinh y + y\cos x \cosh y)$.
Then, we take the partial derivatives of $u$ and $v$ with respect to $x$ and $y$ and equate them to the corresponding partial derivatives of $u$ and $v$.

What is the domain of $z\cos z$?

The domain of a complex function is the set of all complex numbers for which the function is defined. In the case of $z\cos z$, the function is defined for all complex numbers, i.e. the domain is the entire complex plane.

How do we prove that $z\cos z$ is analytic?

A complex function is said to be analytic if it is differentiable at every point in its domain. To prove that $z\cos z$ is analytic, we need to show that the limit of the difference quotient exists at every point in its domain. Since we have verified the CR equations for $z\cos z$, we can conclude that the function is differentiable at every point in its domain and therefore, it is analytic.

What are the singularities of $z\cos z$?

A singularity of a complex function is a point in its domain where the function is not defined or is not analytic. For $z\cos z$, the function is defined for all complex numbers, so there are no points in its domain where it is not defined. However, since the function is analytic for all complex numbers, there are no singularities.

Can we express $z\cos z$ in terms of $x$ and $y$?

Yes, we can express $z\cos z$ in terms of $x$ and $y$ by expanding the function and separating the real and imaginary parts, as shown in the answer to the first question. This will give us the expression:
$z\cos z = x\cos x \cosh y - y\sin x \sinh y + i(x\sin x \sinh y + y\cos x \cosh y)$.

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