Math Problem Statement

Using Cauchy Riemann equations, check whether complex function 𝑓(𝑧) = ((π‘₯ + 1)3 βˆ’ 3(π‘₯ + 1)𝑦2 βˆ’ 3π‘₯) + 𝑖(𝑦3 + 3𝑦 βˆ’ 3(π‘₯ + 1)2𝑦) is analytic in the complex plane or not.

Solution

To determine whether the complex function f(z)=((x+1)3βˆ’3(x+1)y2βˆ’3x)+i(y3+3yβˆ’3(x+1)2y)f(z) = ((x + 1)^3 - 3(x + 1)y^2 - 3x) + i(y^3 + 3y - 3(x + 1)^2y) is analytic, we need to use the Cauchy-Riemann equations. These equations provide necessary conditions for a function to be analytic (i.e., differentiable) at a point in the complex plane.

Step 1: Express the Function in Terms of u(x,y)u(x, y) and v(x,y)v(x, y)

Given f(z)=u(x,y)+iv(x,y)f(z) = u(x, y) + iv(x, y), where z=x+iyz = x + iy:

  • u(x,y)=(x+1)3βˆ’3(x+1)y2βˆ’3xu(x, y) = (x + 1)^3 - 3(x + 1)y^2 - 3x
  • v(x,y)=y3+3yβˆ’3(x+1)2yv(x, y) = y^3 + 3y - 3(x + 1)^2y

Step 2: Compute the Partial Derivatives

We need to compute the partial derivatives of u(x,y)u(x, y) and v(x,y)v(x, y) with respect to xx and yy:

Partial Derivatives of u(x,y)u(x, y):

ux=βˆ‚βˆ‚x[(x+1)3βˆ’3(x+1)y2βˆ’3x]u_x = \frac{\partial}{\partial x} \left[(x + 1)^3 - 3(x + 1)y^2 - 3x\right] uy=βˆ‚βˆ‚y[(x+1)3βˆ’3(x+1)y2βˆ’3x]u_y = \frac{\partial}{\partial y} \left[(x + 1)^3 - 3(x + 1)y^2 - 3x\right]

Calculating these:

ux=3(x+1)2βˆ’3y2βˆ’3u_x = 3(x + 1)^2 - 3y^2 - 3 uy=βˆ’6(x+1)yu_y = -6(x + 1)y

Partial Derivatives of v(x,y)v(x, y):

vx=βˆ‚βˆ‚x[y3+3yβˆ’3(x+1)2y]v_x = \frac{\partial}{\partial x} \left[y^3 + 3y - 3(x + 1)^2y\right] vy=βˆ‚βˆ‚y[y3+3yβˆ’3(x+1)2y]v_y = \frac{\partial}{\partial y} \left[y^3 + 3y - 3(x + 1)^2y\right]

Calculating these:

vx=βˆ’6(x+1)yv_x = -6(x + 1)y vy=3y2+3βˆ’3(x+1)2v_y = 3y^2 + 3 - 3(x + 1)^2

Step 3: Apply the Cauchy-Riemann Equations

The Cauchy-Riemann equations state that for f(z)f(z) to be analytic, the following must hold:

  1. ux=vyu_x = v_y
  2. uy=βˆ’vxu_y = -v_x

Substitute the partial derivatives we calculated:

  1. ux=3(x+1)2βˆ’3y2βˆ’3u_x = 3(x + 1)^2 - 3y^2 - 3 and vy=3y2+3βˆ’3(x+1)2v_y = 3y^2 + 3 - 3(x + 1)^2

    Here, uxu_x and vyv_y are not equal unless specific conditions are met (i.e., 3(x+1)2βˆ’3y2βˆ’3=3y2+3βˆ’3(x+1)23(x + 1)^2 - 3y^2 - 3 = 3y^2 + 3 - 3(x + 1)^2), which simplifies to 2(3(x+1)2βˆ’3y2)=02(3(x + 1)^2 - 3y^2) = 0. This implies (x+1)2=y2(x + 1)^2 = y^2.

  2. uy=βˆ’6(x+1)yu_y = -6(x + 1)y and βˆ’vx=6(x+1)y-v_x = 6(x + 1)y

    Here, uy=βˆ’vxu_y = -v_x holds true.

Step 4: Conclusion

The function f(z)f(z) satisfies the Cauchy-Riemann equations only when (x+1)2=y2(x + 1)^2 = y^2, which defines a set of curves in the complex plane. Outside of this set, the function does not satisfy the Cauchy-Riemann equations.

Therefore, the function f(z)f(z) is not analytic in the entire complex plane. It is only analytic on the specific curves where (x+1)2=y2(x + 1)^2 = y^2.

Would you like more details on any part of this solution, or do you have any questions?

Relative Questions:

  1. How can we determine the analyticity of a function using the Cauchy-Riemann equations?
  2. What does it mean for a function to be analytic in the complex plane?
  3. Can a function satisfy the Cauchy-Riemann equations at some points but still not be analytic?
  4. How do the Cauchy-Riemann equations relate to the concept of differentiability in complex analysis?
  5. Are there other methods besides the Cauchy-Riemann equations to check if a function is analytic?

Tip:

Remember that satisfying the Cauchy-Riemann equations is necessary but not always sufficient for analyticity; the function must also be continuously differentiable.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Complex Analysis
Cauchy-Riemann Equations

Formulas

-

Theorems

Cauchy-Riemann Equations

Suitable Grade Level

College