Question

Verify the statements indicated in Examples 1 to 5 above. For each of the following sets, either verify (as in Example 1) that it is a vector space, or show which requirements are not satisfied. If it is a vector space, find a basis and the dimension of the space.

Short answer

$\begin{array}{l}u\left(x,y\right)=x^3-3x{y}^2\\ v(x,y)=3x^{2}y-y^3\end{array}$

Step-by-step solution

Given Solution

The real and the imaginary parts of this complex function:

$f\left(z\right)=z^3$

Real and Imaginary Part

The square root of a negative number is an imaginary number, which has no measurable value. While an imaginary number is not a real number in the sense that it cannot be quantified on a number line, it is "real" in the sense that it exists and is used in mathematics.

Real and Imaginary Part

Any complex function could be written as:

$$\begin{gathered} f\left(z\right)=f(x+iy) \\ =u\left(x,y\right)+iv\left(x,y\right) \end{gathered}$$

Where, u(x,y) is called the real part of the function,

and v(x,y) is called the imaginary part of the function.

Binomial Theorem

Since, z = x + iy

$$\begin{gathered} f\left(z\right)=z^3 \\ ={\left(x+iy\right)}^3 \end{gathered}$$

Using Binomial theorem

$$\begin{gathered} {\left(a+b\right)}^n{=}_{k=0}^n\frac{n!}{(n-k)!k!}{a}^{n-k}{b}^k \\ f\left(z\right)=x^3+3i{x}^{2}y-3x{y}^2-i{y}^3 \\ =\left(x^3-3x{y}^2\right)+\left(3x^{2}y-y^3\right) \\ =u\left(x,y\right)+iv(x,y) \end{gathered}$$

Therefore,

$$\begin{gathered} u\left(x,y\right)=x^3-3y^2 \\ v\left(x,y\right)=3x^{2}y-y^3 \end{gathered}$$