Problem 11 - Complex numbers from A to...Z

Question

Compute: a) \((1-\cos a+i \sin a)^{n}\) for \(a \in[0,2 \pi)\) and \(n \in \mathbb{N}\); b) \(z^{n}+\frac{1}{z^{n}}\), if \(z+\frac{1}{z}=\sqrt{3}\).

Step-by-Step Solution

Answer

Question: Find the power of the complex number $(1-\cos a+i \sin a)^{n}$, where $a \in[0,2 \pi)$ and $n \in \mathbb{N}$. Answer: The power of the complex number $(1-\cos a+i \sin a)^{n}$ is equal to $\text{cos}(n(\pi - a)) + i\cdot \text{sin}(n(\pi - a))$. Question: Compute the expression $z^{n}+\frac{1}{z^{n}}$, given that $z+\frac{1}{z}=\sqrt{3}$. Answer: The expression $z^{n}+\frac{1}{z^{n}}$ can be simplified to $(\sqrt{3})^n - {n \choose 1} (\sqrt{3})^{n-2}(1) + {n \choose 2}(\sqrt{3})^{n-4}(1)^2 - \cdots$.

Step 1: Problem 1: Find the power of a complex number

To compute \((1-\cos a+i \sin a)^{n}\), where \(a \in[0,2 \pi)\) and \(n \in \mathbb{N}\), we need to apply the formula \((\text{cos}(p) + i \cdot \text{sin}(p))^n = \text{cos}(np) + i\cdot \text{sin}(np)\). We will first rewrite \((1-\cos a+i \sin a)^{n}\) in the form of \((\text{cos}(p) + i \cdot \text{sin}(p))^n\). Let \(z = 1-\cos a+i \sin a\). Notice that \(|z|^2 = (1-\cos a)^2 + (\sin a)^2\), by using the Pythagorean identity we get: \(|z|^2 = 1- 2\cos a + \cos^2 a + \sin^2 a = 2 - 2\cos a = 2(1-\cos a)\) Now, we need to find the angle \(p\), such that \(\cos(p) = 1-\cos a\) and \(\sin(p) = \sin a\). Since \(z\) is in the second quadrant or possibly on the real axis, \(p = \pi - a\). Therefore: \(z = (\cos(p) + i \cdot \sin(p)) = \cos(\pi - a) + i \cdot \sin(\pi - a)\) Now we can apply the formula to find the power of \(z^n\): \((1-\cos a+i \sin a)^{n} = (\text{cos}(\pi - a) + i\cdot \text{sin}(\pi - a))^n = \text{cos}(n(\pi - a)) + i\cdot \text{sin}(n(\pi - a))\)

Step 2: Problem 2: Simplify the given expression

We need to compute the expression \(z^{n}+\frac{1}{z^{n}}\), given that \(z+\frac{1}{z}=\sqrt{3}\). Let's first rewrite the given constraint as a quadratic equation: \(z^2 - \sqrt{3}z + 1 = 0\) Note that the roots of this quadratic equation are \(z\) and its conjugate \(\bar{z}\), because the coefficients are real. The sum of these roots is equal to the coefficient of the linear term (with opposite sign), and the product of these roots is equal to the constant term: \(z + \bar{z} = \sqrt{3} \qquad z \cdot \bar{z} = 1\) Now, we can apply the binomial theorem to the expression, to get: \((z^{n}+\frac{1}{z^{n}}) = (z+\bar{z})^{n} - {n \choose 1} (z+\bar{z})^{n-2}(z\bar{z}) + {n \choose 2}(z+\bar{z})^{n-4}(z \bar{z})^2 - \cdots\) Plug in the values we have found: \((z^{n}+\frac{1}{z^{n}}) = (\sqrt{3})^n - {n \choose 1} (\sqrt{3})^{n-2}(1) + {n \choose 2}(\sqrt{3})^{n-4}(1)^2 - \cdots\) This is the simplified expression for \(z^{n}+\frac{1}{z^{n}}\).

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a formula used to compute powers of complex numbers in a very elegant way. According to this theorem, for any real number \( a \) and any integer \( n \), we have:

\[ (\cos a + i \cdot \sin a)^n = \cos(na) + i \cdot \sin(na) \]
It essentially tells us that raising a complex number to a power is the same as raising the magnitude of that number to the power and multiplying the angle by the power, when the complex number is represented in polar form. This theorem is incredibly useful when dealing with complex numbers raised to large powers, as it avoids the need for lengthy multiplication. Let's take an example from the exercise:

\((1 - \cos a + i \sin a)^n\)

This expression can be daunting at first glance. However, with De Moivre's Theorem, we can rewrite the complex number \(1 - \cos a + i \sin a\) in polar form and then apply the theorem to find its power easily. The power \(n\) scales the angle, and the magnitude is raised to the same power.

Complex Number Algebra

Complex number algebra involves operations such as addition, subtraction, multiplication, division, and the computation of powers and roots with complex numbers. Complex numbers are written in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part. When doing algebra with complex numbers, it's crucial to keep in mind that \(i^2 = -1\). This simple rule underlies all of complex arithmetic.

For instance, when solving the quadratic equation from the exercise:

\(z^2 - \sqrt{3}z + 1 = 0\)

The roots of this equation are complex numbers because the coefficients are real. Algebra with complex numbers allows us to deal with equations that have no real solutions, by introducing the imaginary unit \(i\). This is indispensable when we dive into more advanced mathematics or physics problems that involve complex solutions.

Binomial Theorem for Complex Numbers

Expanding Powers of Binomials

The binomial theorem is a useful tool for expanding expressions that are raised to a power. It works for complex numbers much the same way as for real numbers, with the difference that we work with the real and imaginary parts separately.

The general form of the binomial theorem is:

\[(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k\]
When applied to complex numbers where \(a\) and \(b\) could both have real and imaginary components, we must be cautious to respect the properties of the imaginary unit \(i\). In our exercise, we leverage this theorem to simplify the expression \(z^n + \frac{1}{z^n}\), knowing that \(z + \frac{1}{z} = \sqrt{3}\). To correctly use the binomial theorem in complex scenarios, proper substitution and simplification are key steps.

Quadratic Equations in Complex Numbers

Finding Complex Roots

Quadratic equations in the complex number system sometimes have solutions that cannot be found among the real numbers. Complex numbers provide us with a complete set of solutions to these equations.

In our exercise, the given equation is:

\(z^2 - \sqrt{3}z + 1 = 0\)

To solve this, we note that the roots must be complex since the discriminant is negative. The sum and product of the roots, labeled \(z\) and \(\bar{z}\), can be obtained from the coefficients of the quadratic equation using Vieta's formulas:

\[z + \bar{z} = \sqrt{3}\] and \[z \cdot \bar{z} = 1\]
These lead to the two roots being complex conjugates of each other. Understanding how to manipulate these forms allows us to solve for the roots of the equation and then apply them for further computational needs such as those in the binomial theorem application.