Consider the function \(u(x, y)=x^{3}-3 x y^{2}\). a. Show that \(u(x, y)\) is harmonic; that is, \(\nabla^{2} u=0\) b. Find its harmonic conjugate, \(v(x, y)\). c. Find a differentiable function, \(f(z)\), for which \(u(x, y)\) is the real part. d. Determine \(f^{\prime}(z)\) for the function in part c. [Use \(f^{\prime}(z)=\frac{\partial_{2}}{\partial x}+i \frac{\partial v}{\partial x}\) and rewrite your answer as a function of \(z .]\)

Short Answer

Expert verified
a. The function \(u(x, y) = x^3 - 3xy^2\) is a harmonic function because \(∇^2 u = 0\). b. Its harmonic conjugate is \(v(x, y) = y^3 - 3x^2y + c\). c. The differentiable function is then \(f(z) = z^3 + c\). d. The derivative of this function is \(f^{\prime}(z) = 3z^2\).

Step by step solution

01

Prove \(u(x, y)\) is harmonic

To check if \(u(x, y)\) is harmonic, we calculate the Laplacian, \(∇^2 u\). This is the sum of the second partial derivatives of u with respect to x and y: \[∇^2 u = \frac{∂^2u}{∂x^2} + \frac{∂^2u}{∂y^2}\]
02

Calculate the Laplacian

Upon calculating, we get: \[∇^2 u = \frac{∂^2u}{∂x^2} + \frac{∂^2u}{∂y^2} = 6x-6x = 0\] as both terms cancel out, proving that \(u(x, y)\) is indeed harmonic.
03

Find the harmonic conjugate

The harmonic conjugate \(v(x, y)\) needs to satisfy the Cauchy-Riemann equations: \[\frac{∂u}{∂x} = \frac{∂v}{∂y} \quad and \quad -\frac{∂u}{∂y} = \frac{∂v}{∂x}\] Now let's solve these equations.
04

Calculate the harmonic conjugate

After solving the Cauchy-Riemann equations, the harmonic conjugate is found to be \(v(x, y) = y^3 - 3x^2y + c\), where c is the constant of integration.
05

Find the differentiable function, \(f(z)\)

To find the complex function \(f(z)\), we combine the original function \(u(x, y)\) and its conjugate \(v(x, y)\) to get: \(f(z) = u(x, y) + iv(x, y) = x^3 - 3xy^2 + i(y^3 - 3x^2y + c)\). This is the intended complex function. Rewriting z as \(x + iy\), we get: \(f(z) = z^3 + c\).
06

Determine \(f^{\prime}(z)\)

The derivative of \(f(z)\) is found by using the definition: \(f^{\prime}(z)=\frac{∂}{∂x} + i \frac{∂v}{∂x}\). Upon doing the calculus, we get: \(f^{\prime}(z) = 3z^2\).

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Most popular questions from this chapter

Show that $$ \int_{C} \frac{d z}{(z-1-i)^{n+1}}=\left\\{\begin{array}{cl} 0, & n \neq 0 \\ 2 \pi i, & n=0 \end{array}\right. $$ for \(C\) the boundary of the square \(0 \leq x \leq 2,0 \leq y \leq 2\) taken counterclockwise. [Hint: Use the fact that contours can be deformed into simpler shapes (like a circle) as long as the integrand is analytic in the region between them. After picking a simpler contour, integrate using parametrization.].

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Find series representations for all indicated regions. a. \(f(z)=\frac{z}{z-1},|z|<1,|z|>1\). b. \(f(z)=\frac{1}{(z-1)(2+2)},|z|<1,1<|z|<2,|z|>2\). [Hint Use partial fractions to write this as a sum of two functions first.]

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