Question

For each of the following functions \(w=f(z)=u+i v,\) find \(u\) and \(v\) as functions of \(x\) and \(y .\) Sketch the graphs in the \((x, y)\) plane of the images of \(u=\) const. and \(v=\) const. for several values of \(u\) and several values of \(v\) as was done for \(w=z^{2}\) in Figure \(9.3 .\) The curves \(u=\) const. should be orthogonal to the curves \(v=\) const. $$w=\frac{z+1}{2 i}$$

Short answer

\( u = \frac{y}{2} \) and \( v = -\frac{(x+1)}{2} \). Horizontal and vertical lines for several values of \(u\) and \(v\).

Step-by-step solution

Substitute the Complex Variable

Let’s start by expressing the complex number in terms of the given variables: Multiply and divide by the conjugate. Given function is:\[ w = \frac{z+1}{2i} \] We can rewrite it as:\[ w = \frac{(x+iy + 1)}{2i} \] Then, multiply numerator and denominator by the conjugate of the denominator to simplify:\[ w = \frac{(x + 1) + iy}{2i} = \frac{(x+1)}{2i} + \frac{y}{2} \]

Simplify the Expression

Simplify the terms divided by the imaginary unit, noting that \( \frac{1}{i} = -i \):\[ w = \left( \frac{(x+1)}{2i} \right) + \frac{y}{2} \] Thus:\[ w = -\frac{(x+1)}{2}i + \frac{y}{2} \] So, identifying the real and imaginary parts:\[ u = \frac{y}{2} \] \[ v = -\frac{(x+1)}{2} \]

Express u and v as Functions of x and y

From the earlier step, it is clear that: \[ u = \frac{y}{2} \] and \[ v = -\frac{(x+1)}{2} \]

Determine the Contours

To determine the images of \( u = \text{const.} \) and \( v = \text{const.} \) in the \((x, y)\) plane, consider: For \( u = c_1 \): \[ \frac{y}{2} = c_1 \implies y = 2c_1 \] These are horizontal lines.For \( v = c_2 \): \[ -\frac{(x+1)}{2} = c_2 \implies x + 1 = -2c_2 \implies x = -2c_2 - 1 \] These are vertical lines.

Sketch the Graphs

Sketch the horizontal lines for several values of \( u \) (e.g., \( u = -2, -1, 0, 1, 2 \), hence \( y = -4, -2, 0, 2, 4 \)) and the vertical lines for several values of \( v \) (e.g., \( v = -2, -1, 0, 1, 2 \), hence \( x = 3, 1, -1, -3, -5 \)). The horizontal lines and vertical lines should be orthogonal to each other in the \((x, y)\) plane.

Key concepts

Complex Functions

A complex function maps complex numbers from the domain to the codomain. Here, we have the complex function: The equation given is: \begin{align*} w = \frac{z+1}{2 i} \, \text{where} \, z = x + iy, \end{align*} where \(w\) is the output of the function, \(z\) is the input, and \(i\) is the imaginary unit. By substituting \(z = x + iy\), we can break down the function into its real and imaginary components. In our exercise, the real part of the function is denoted by \(u\), and the imaginary part is denoted by \(v\). This breakdown gives us: \begin{align*} w = u + iv = \frac{(x+1)}{2i} + \frac{y}{2}, \, \text{where} \, u = \frac{y}{2} \end{align*} and \begin{align*} v = -\frac{(x+1)}{2}. \end{align*} Let's move on to these real and imaginary parts, where understanding them is crucial to getting a clear visual of how complex functions behave.

Contour Mapping

Contour mapping in complex analysis helps us visualize how the real and imaginary parts of complex functions behave. In this exercise, we look at the contours or level curves of \(u\) and \(v\). The contours represent constant values of \(u\) and \(v\). For \(u = c_1\), the equation in our exercise simplifies to: \(c_1 = y/2\) or \(y = 2c_1\). These are horizontal lines in the (x, y) plane.Likewise, for \(v = c_2\), the equation becomes: \(c_2 = -(x + 1)/2\) or \(x = -2c_2 - 1\), which are vertical lines. When you sketch these lines for different values of \(u\) and \(v\), you'll notice that they intersect orthogonally. This orthogonal intersection is a significant attribute in complex analysis, as it showcases a harmonic relationship between real and imaginary components of the function.

Imaginary Unit

The imaginary unit \(i\) is fundamental in complex analysis. It is defined as the square root of \(-1\), meaning: \(i^2 = -1\). This special property allows us to extend the real number system to complex numbers, enabling the representation and manipulation of quantities that go beyond the real line. In our exercise, the imaginary unit \(i\) plays a crucial role in converting and understanding the function \(w = (z+1)/2i\). We used the property that \(1/i = -i\) to simplify the expression. Subsequently, we separated the real and imaginary parts and expressed the function in terms of \(u\) and \(v\), which were needed to sketch the graphs. By grasping how \(i\) operates within these functions, we can better analyze and reflect underlying physical phenomena represented by these complex quantities.