Chapter 2: Problem 53
Find the image of the region \(|z| \leq 8, \pi / 2 \leq \arg (z) \leq 3 \pi / 4\), under each of the following principal \(n\) th root functions. (a) \(f(z)=z^{1 / 2}\) (b) \(f(z)=z^{1 / 3}\) (c) \(f(z)=z^{1 / 4}\)
Short Answer
Expert verified
(a) \(|w| \leq 4, \pi/4 \leq \arg(w) \leq 3\pi/8\).
(b) \(|w| \leq 2, \pi/6 \leq \arg(w) \leq \pi/4\).
(c) \(|w| \leq \sqrt{2}, \pi/8 \leq \arg(w) \leq 3\pi/16\).
Step by step solution
01
Understanding the Region
The region is defined as \(|z| \leq 8\) with the arg of \(z\) ranging from \(\pi/2\) to \(3\pi/4\). This is a sector of a circle centered at the origin with radius 8, starting from the positive imaginary axis (\(\pi/2\)) to the line that forms a \(45^\circ\) angle past the imaginary axis (\(3\pi/4\)).
02
Transformation for f(z)=z^{1/2}
The function \(f(z) = z^{1/2}\) takes a complex number and finds its principal square root. The modulus \(|z|\) will become \(|z|^{1/2}\), and the argument \(\arg(z)\) will become \(\arg(z)/2\). Thus, the new modulus is \(8^{1/2} = 4\), and the new argument range is from \(\pi/4 = (\pi/2)/2\) to \(3\pi/8 = (3\pi/4)/2\). Hence, the image under this function is \(|w| \leq 4, \pi/4 \leq \arg(w) \leq 3\pi/8\).
03
Transformation for f(z)=z^{1/3}
For the function \(f(z) = z^{1/3}\), the modulus becomes \(|z|^{1/3}\), and the argument becomes \(\arg(z)/3\). Thus, the new modulus of any point in the image is \(8^{1/3} = 2\). The new argument range transforms to \(\pi/6 = (\pi/2)/3\) to \(\pi/4 = (3\pi/4)/3\). Thus, the image is \(|w| \leq 2, \pi/6 \leq \arg(w) \leq \pi/4\).
04
Transformation for f(z)=z^{1/4}
For the function \(f(z) = z^{1/4}\), the modulus is transformed to \(|z|^{1/4}\) and the argument to \(\arg(z)/4\). The new modulus becomes \(8^{1/4} = \sqrt[4]{8}\), which simplifies to \(\sqrt{2}\). The new argument range transforms from \(\pi/8 = (\pi/2)/4\) to \(3\pi/16 = (3\pi/4)/4\). Hence, the image is \(|w| \leq \sqrt{2}, \pi/8 \leq \arg(w) \leq 3\pi/16\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Principal nth Root
The concept of the principal nth root is a fundamental element in complex analysis. When you take the nth root of a complex number, you are essentially finding a complex number w such that when raised to the power n, it yields the original number z, i.e., \( w^n = z \). Complex numbers have multiple nth roots, but the principal nth root is the one with the smallest non-negative argument, making it standardized and predictable.
For a complex number \( z = re^{i\theta} \), where \( r \) is the modulus and \( \theta \) is the argument, the principal nth root\( w = z^{1/n} \) is given by
For a complex number \( z = re^{i\theta} \), where \( r \) is the modulus and \( \theta \) is the argument, the principal nth root\( w = z^{1/n} \) is given by
- The modulus of \( w \) is \( r^{1/n} \).
- The argument of \( w \) is \( \theta / n \).
Argument of a Complex Number
The argument of a complex number is an essential part of expressing complex numbers in polar forms. It is actually the angle that a line representing the complex number makes with the positive real axis in the complex plane.
If you have a complex number \( z = a + bi \), its argument \( \arg(z) \) is given by \( \theta = \tan^{-1}(b/a) \). The standard range for the argument is \( 0 \) to \( 2\pi \), although any equivalent angle can also be considered.
When transforming a complex number through functions like principal nth roots, the argument is divided by n. This effectively scales down the argument, reducing the angle by n times, and therefore compressing angular space.
If you have a complex number \( z = a + bi \), its argument \( \arg(z) \) is given by \( \theta = \tan^{-1}(b/a) \). The standard range for the argument is \( 0 \) to \( 2\pi \), although any equivalent angle can also be considered.
When transforming a complex number through functions like principal nth roots, the argument is divided by n. This effectively scales down the argument, reducing the angle by n times, and therefore compressing angular space.
- For example, if \( \arg(z) \) was \( \pi/2 \) to \( 3\pi/4 \), under a square root transformation \( \arg(w) = \arg(z)/2 \), making it \( \pi/4 \) to \( 3\pi/8 \).
- Similarly, dividing by 3 (as in a cube root function) or 4 (a fourth root function), will further compress the range.
Image of a Region
The image of a region in complex analysis refers to the transformed area when a region of the complex plane undergoes a mapping, typically through a complex function. When considering transformations such as taking the principal nth root, you transform the modulus and argument of each point within a region, which results in a new image region.
For example, consider a region of the complex plane defined by
Under the square root transformation \( f(z) = z^{1/2} \), the circle's image with radius 8 becomes a circle with radius 4, and the angle range scales down by half, turning into \( \pi/4 \leq \arg(w) \leq 3\pi/8 \).
Similarly, cube root and fourth root transformations further modify these regions:
For example, consider a region of the complex plane defined by
- \( |z| \leq 8 \)
- \( \pi/2 \leq \arg(z) \leq 3\pi/4 \)
Under the square root transformation \( f(z) = z^{1/2} \), the circle's image with radius 8 becomes a circle with radius 4, and the angle range scales down by half, turning into \( \pi/4 \leq \arg(w) \leq 3\pi/8 \).
Similarly, cube root and fourth root transformations further modify these regions:
- Cube root: Results in a smaller circle and tighter angle range.
- Fourth root: Compresses the region even further.