Use the Cauchy-Riemann equations to show that f(z)=z― is not analytic.

Question

Liyana Mansell · Accepted Answer

Step 1
A function is said to be analytic when Cauchy-Riemann equations are satisfied.
The Cauchy-Riemann equations are satisfied when

u_{x} = v_{y} and v_{x} = - u_{y}

.
The given function is

f (z) = \overset{―}{z}

Rewrite the given function as follows.

f (z) = \overset{―}{z}

=x-iy
Here u(x,y)=x and v(x,y)=-y.
Step 2
Evaluate

u_{x}

as follows.

u_{x} = \frac{\partial}{\partial x} (x)

=1
Thus,

u_{x} = 1

.
Evaluate

u_{y}

as follows.

u_{y} = \frac{\partial}{\partial y} (x)

=0
Thus,

u_{y} = 0

.
Step 3
Evaluate

v_{x}

as follows.

v_{x} = \frac{\partial}{\partial x} (- y)

=0
Thus,

v_{x} = 0

.
Evaluate

v_{y}

as follows.

v_{y} = \frac{\partial}{\partial y} (- y)

=-1
Thus,

v_{y} = - 1

.
Clearly,

u_{x} \neq v_{y}

. So, the function did not satisfy Cauchy-Riemann equations.
Therefore, the given function is not analytic.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

alenahelenash · Accepted Answer

According to the Cauchy-Riemann equations, for a function

f (z)

to be analytic, the partial derivatives of its real and imaginary parts must satisfy the following conditions:

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}

and

\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}

, where

u

represents the real part of

f (z)

and

v

represents its imaginary part.
For

f (z) = \overset{―}{z}

, we have

u (x, y) = x

and

v (x, y) = - y

. Let's compute the partial derivatives:

\frac{\partial u}{\partial x} = 1

,

\frac{\partial u}{\partial y} = 0

,

\frac{\partial v}{\partial x} = 0

,

\frac{\partial v}{\partial y} = - 1

.
We can see that the first condition

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}

is satisfied, but the second condition

\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}

is not satisfied.
Since the Cauchy-Riemann equations are not satisfied for

f (z) = \overset{―}{z}

, we conclude that

f (z) = \overset{―}{z}

is not analytic.

user_27qwe · Accepted Answer

Step 1:To show that f(z)=z― is not analytic, we can utilize the Cauchy-Riemann equations. The Cauchy-Riemann equations state that if a complex function f(z)=u(x,y)+iv(x,y) is analytic, where u(x,y) and v(x,y) are real-valued functions of the real variables x and y, then the partial derivatives of u and v satisfy the following conditions:∂u∂x=∂v∂y(1) and ∂u∂y=−∂v∂x(2)Let&#039;s apply these equations to the function f(z)=z―. Here, z― represents the complex conjugate of z, which can be written as z―=x−iy.We can express f(z) in terms of its real and imaginary parts as follows:f(z)=z―=x−iy=u(x,y)+iv(x,y)From this representation, we can identify u(x,y)=x and v(x,y)=−y.Step 2:Now, let&#039;s calculate the partial derivatives of u and v:∂u∂x=1(3)∂v∂y=−1(4)∂u∂y=0(5)∂v∂x=0(6)Comparing equations (3) and (4), we can see that ∂u∂x≠∂v∂y. Similarly, comparing equations (5) and (6), we find that ∂u∂y≠−∂v∂x.Since the Cauchy-Riemann equations are not satisfied for f(z)=z―, we conclude that f(z) is not analytic.

karton · Accepted Answer

Result:f(z)=z― is not analyticSolution:Let&#039;s first express the complex variable z in terms of its real and imaginary parts: z=x+iy, where x and y are real numbers. Now, we can rewrite f(z) as:f(z)=z―=x+iy―Taking the complex conjugate of x+iy yields:f(z)=z―=x−iyNow, we can apply the Cauchy-Riemann equations to f(z) and check if they are satisfied.The Cauchy-Riemann equations are as follows:∂u∂x=∂v∂y(1)∂u∂y=−∂v∂x(2)where u and v are the real and imaginary parts of the function f(z), respectively.For f(z)=z―=x−iy, we can identify u(x,y)=x and v(x,y)=−y. Let&#039;s calculate the partial derivatives:∂u∂x=1(3)∂v∂y=−1(4)∂u∂y=0(5)∂v∂x=0(6)Comparing equations (3) and (4), we see that they are not equal, which means that the Cauchy-Riemann equations are not satisfied for f(z)=z―. Therefore, f(z)=z― is not an analytic function.

Use the Cauchy-Riemann equations to show that f(z)=bar(z) is not analytic.

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