Math Problem Statement

Are the following functions analytic? f(z) - izz

Solution

To determine if the function f(z)=izzˉf(z) = iz\bar{z} is analytic, we need to check whether it satisfies the Cauchy-Riemann equations, which are necessary (but not sufficient) conditions for a function to be analytic.

Step 1: Express f(z)f(z) in terms of z=x+iyz = x + iy

Let z=x+iyz = x + iy, where xx and yy are real variables, and ii is the imaginary unit. The complex conjugate zˉ\bar{z} is given by:

zˉ=xiy\bar{z} = x - iy

Thus, the function f(z)=izzˉf(z) = iz\bar{z} can be written as:

f(z)=i(x+iy)(xiy)=i(x2+y2)f(z) = i(x + iy)(x - iy) = i(x^2 + y^2)

Step 2: Separate f(z)f(z) into its real and imaginary parts

Let f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y), where u(x,y)u(x, y) and v(x,y)v(x, y) are the real and imaginary parts of the function, respectively.

Expanding the expression:

f(z)=i(x2+y2)f(z) = i(x^2 + y^2)

This gives:

u(x,y)=y2+x2,v(x,y)=0u(x, y) = -y^2 + x^2, \quad v(x, y) = 0

Step 3: Apply the Cauchy-Riemann equations

The Cauchy-Riemann equations are:

ux=vyanduy=vx\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

Compute the partial derivatives:

ux=(x2y2)x=2x,uy=(x2y2)y=2y\frac{\partial u}{\partial x} = \frac{\partial (x^2 - y^2)}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = \frac{\partial (x^2 - y^2)}{\partial y} = -2y

vx=0x=0,vy=0y=0\frac{\partial v}{\partial x} = \frac{\partial 0}{\partial x} = 0, \quad \frac{\partial v}{\partial y} = \frac{\partial 0}{\partial y} = 0

Step 4: Check if the Cauchy-Riemann equations hold

  • ux=2x\frac{\partial u}{\partial x} = 2x and vy=0\frac{\partial v}{\partial y} = 0: These are not equal unless x=0x = 0.
  • uy=2y\frac{\partial u}{\partial y} = -2y and vx=0\frac{\partial v}{\partial x} = 0: These are not equal unless y=0y = 0.

Since the Cauchy-Riemann equations do not hold everywhere (they only hold on the x-axis and y-axis individually), the function f(z)=izzˉf(z) = iz\bar{z} is not analytic in any open region of the complex plane.

Would you like more details or have any further questions?

Related Questions:

  1. What are the Cauchy-Riemann equations, and why are they important?
  2. How do you determine if a function is analytic in a specific region?
  3. What is the difference between holomorphic and analytic functions?
  4. Can you explain why a function might satisfy the Cauchy-Riemann equations but still not be analytic?
  5. What is the geometric interpretation of an analytic function in the complex plane?

Tip:

When testing whether a function is analytic, always ensure the Cauchy-Riemann equations hold in an open region, not just at isolated points.

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Math Problem Analysis

Mathematical Concepts

Complex analysis
Analytic functions
Cauchy-Riemann equations

Formulas

Cauchy-Riemann equations

Theorems

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Suitable Grade Level

Advanced undergraduate