Problem 139
Question
If \(n>1\), then the roots of \(z^{n}=(z+1)^{n}\) lie on a (A) circle (B) straight line (C) parabola (D) None of these
Step-by-Step Solution
Verified Answer
The roots lie on a circle.
1Step 1: Understanding the Equation
We begin by analyzing the equation \(z^n = (z+1)^n\). This can be rewritten by using properties of powers: \(\frac{z^n}{(z+1)^n} = 1\). This is equivalent to saying that the quotients \((\frac{z}{z+1})^n\) equals 1.
2Step 2: Expanding the Quotient
From \(\left(\frac{z}{z+1}\right)^n = 1\), we recognize that \(\frac{z}{z+1}\) must be an \(n\)th root of unity. Therefore, \(\frac{z}{z+1} = \omega^k\) where \(\omega\) is a primitive \(n\)th root of unity and \(k = 0, 1, 2, \ldots, n-1\).
3Step 3: Expressing in Terms of z
Solving \(\frac{z}{z+1} = \omega^k\) for \(z\) gives us \(z = \omega^k (z+1)\). This can be rewritten as \(z - \omega^k z = \omega^k\) which simplifies to \(z(1 - \omega^k) = \omega^k\).
4Step 4: Finding the Root Locations
From \(z(1 - \omega^k) = \omega^k\), solve for \(z\) to get \(z = \frac{\omega^k}{1-\omega^k}\). This defines a set of roots which can be seen to lie in a specific geometric location depending on properties of \(\omega\).
5Step 5: Recognizing the Geometric Shape
Examining \(z = \frac{\omega^k}{1-\omega^k}\), knowing that \(\omega^k\) is a rotation on the unit circle in the complex plane, shows that all roots generated by \(\omega^k\) lead to points distributed evenly around a circle.
Key Concepts
nth roots of unitygeometric representationcomplex plane
nth roots of unity
When we talk about the "nth roots of unity," we're referring to a special set of numbers in the complex plane. These are solutions to the equation \(x^n = 1\). They lie evenly spaced along the unit circle, which is the circle centered at the origin with a radius of 1.
- For each positive integer \(n\), there are \(n\) complex numbers that solve \(x^n = 1\).
- The main property of these roots is that they are equally spaced around the circle.
- The first one, usually denoted \(\omega\), is called the primitive nth root of unity.
geometric representation
Geometric representation refers to how complex numbers can be visually mapped on the plane, offering insights into their behavior and relationships with each other.
- Each complex number \(a + bi\) can be represented as a point \((a, b)\) in a 2D coordinate system.
- The real part \(a\) is plotted on the horizontal axis, while the imaginary part \(b\) is plotted on the vertical axis.
- This method makes it simple to visualize operations like addition and multiplication.
- Rotation and symmetry play significant roles here.
- Visualizing these roots helps in understanding the nature of the solutions, in this case, evenly distributed along a circle.
complex plane
The complex plane is an essential tool for mathematicians in visualizing and manipulating complex numbers. Think of it as an upgraded version of the regular XY plane, tailored to handle numbers that have both real and imaginary parts.
- The complex plane has a horizontal axis (real axis) and a vertical axis (imaginary axis).
- Every point \((a, b)\) on this plane represents the complex number \(a + bi\).
- This setup enables operations like plotting, rotation, and scaling of complex numbers through straightforward geometric means.
- The unity roots form a perfect circle centered at the origin with a radius of one.
- Operations that involve complex numbers become visually intuitive, often turning complex algebraic manipulation into simple diagrammatic movements.
Other exercises in this chapter
Problem 137
If \(|z-4+3 i| \leq 2\), then the least and the greatest values of \(|z|\) are (A) 3,7 (B) 4,7 (C) 3,9 (D) None of these
View solution Problem 138
If \(|z-4+3 i| \leq 1\) and \(m\) and \(n\) are the least and greatest values of \(|z|\) and \(k\) is the least value of \(\frac{x^{4}+x^{2}+4}{x}\) on the inte
View solution Problem 140
Let \(z\) be a complex number satisfying \(z^{2}+z+1=0\). If \(n\) is not a multiple of 3 , then the value of \(z^{n}+z^{2 n}=\) (A) 2 (B) \(-2\) (C) 0 (D) \(-1
View solution Problem 141
If \(1, \alpha_{1}, \alpha_{2}, \alpha_{3}\) and \(\alpha_{4}\) be the roots of \(x^{5}-1=0\), then \(\frac{\omega-\alpha_{1}}{\omega^{2}-\alpha_{1}} \cdot \fra
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