Question

In the example dealing with Gauss's solution to \(x^{19}-1=0\), show that \(r\) and \(r^{18}\) are both roots of \(x^{2}-\beta_{1} x+1=0\), where \(r\) and \(\beta_{1}\) are as in the text.

Short answer

Question: Show that both \(r\) and \(r^{18}\) are roots of the equation \(x^{2}-\beta_{1}x+1=0\), where \(r\) is a primitive 19th root of unity and \(\beta_{1} = -\sum_{k=1}^{18} r^k\). Answer: We showed in our solution that both \(r\) and \(r^{18}\) satisfy the given equation \(x^{2}-\beta_{1}x+1=0\). Thus, both \(r\) and \(r^{18}\) are roots of the given equation.

Step-by-step solution

Recall the definition of \(r\) and \(\beta_{1}\)

From the text, we know that \(r\) is a primitive \(19\)-th root of unity, meaning that \(r^{19}=1\), and \(r^{k}\neq1\) for any \(k<19\). Also, \(\beta_1 = -\sum_{k=1}^{18} r^k\). We will use these definitions to prove that \(r\) and \(r^{18}\) are roots of the given equation. Step 2: Check if \(r\) satisfies the equation

Check if \(r\) satisfies the equation

Let's substitute \(x=r\) into our given equation: \(r^2-\beta_1r+1=0\). We need to show that the left-hand side is equal to the right-hand side of the equation (which is \(0\)). Step 3: Simplify the equation for \(r\)

Simplify the equation for \(r\)

Using the definition of \(\beta_1\), we can substitute and simplify the equation: \(r^2 - (-\sum_{k=1}^{18} r^k)r + 1 = r^2 + r\sum_{k=1}^{18} r^k + 1\). Notice that \(\sum_{k=1}^{18} r^k = r + r^2 + \cdots r^{18}\). Step 4: Use the geometric sum for \(r\)

Use the geometric sum formula

Using the geometric sum formula, we know that \(\sum_{k=1}^{18} r^k = \frac{r(r^{18}-1)}{r-1}\). Now we can substitute the above expression for the summation in our equation: \(r^2 + r\frac{r(r^{18}-1)}{r-1} + 1\). Step 5: Simplify the equation

Simplify the equation

Let's multiply both numerator and denominator of the expression by \((r-1)\): \(r^2 + r \frac{r^2(r^{18}-1)}{1} + (r-1)\). Step 6: Verify that the equation holds for \(r\)

Check equation for \(r\)

We know \(r^{19}=1\). Therefore, we have \(r^2(r^{18}-1)=r^{20}-r^2=r^{19}-r^2=1-r^2\). Now, let's substitute this expression into our simplified equation: \(r^2 + r(1-r^2) + (r-1) = 0\) We can see that this reduced expression holds for \(r\). Step 7: Check if \(r^{18}\) satisfies the equation

Check if \(r^{18}\) satisfies the equation

Substitute \(x=r^{18}\) into the given equation: \(r^{36}-\beta_1r^{18}+1=0\). We need to show that the left-hand side is equal to the right-hand side of the equation (which is \(0\)). Step 8: Use properties of exponents

Use properties of exponents

Given that \(r^{19}=1\), we can simplify the equation: \(r^{36} = r^{19\cdot2-2}=r^{2(19-1)} = (r^{18})^2 =r^2\). Now, substitute \(r^2\) for \(r^{36}\): \(r^2-\beta_1r^{18}+1=0\). Step 9: Verify that the equation holds for \(r^{18}\)

Verify equation for \(r^{18}\)

Recall that the equation for \(r\) is true: \(r^2 + r\left(1-r^2\right) + (r-1) = 0\). Now, multiply both sides of the equation by \(r^{16}\): \(r^{18} + r^{17}\left(1-r^2\right) + r^{16}(r-1) = 0\). Since \(r^{17}(1-r^2)=\beta_1r^{18}\), we have: \(r^{18} + \beta_1r^{18}+1=0\). We've shown that both \(r\) and \(r^{18}\) are roots of the given equation \(x^{2}-\beta_{1} x+1=0\).

Key concepts

Primitive Root

In mathematics, a primitive root of unity is a complex number which, when raised to a certain power (called the order), equals one, but for no smaller positive integer power does it also equal one. This concept is especially important in the context of roots of unity in complex numbers and helps in solving polynomial equations. The key feature of a primitive root of unity is its ability to generate all the other roots in its cycle.

• A primitive 19th root of unity, denoted as \( r \), satisfies \( r^{19} = 1 \) but \( r^k eq 1 \) for any \( k < 19 \). This makes \( r \) the smallest power that returns to 1, signifying its primitiveness.
• Primitive roots such as \( r \) are used to find solutions to polynomial equations like \( x^{19} - 1 = 0 \), which are central in number theory and algebra.

This property of providing a set of solutions for polynomial equations makes primitive roots valuable in various calculations and proofs, as demonstrated in the exercise where \( r \) simplifies and validates each step of the solution.

Geometric Sum Formula

The geometric sum formula is a powerful tool for simplifying expressions involving sums of powers. In this context, it helps manage and simplify the series that arise from multiplying roots of unity. Understanding this formula allows you to handle complex calculations with ease, especially when dealing with powers of a complex number.

• The sum of the first \( n \) terms in a geometric series with first term \( a \) and common ratio \( r \) is given by:\[S_n = a\frac{r^n - 1}{r - 1}\]
• For our primitive root \( r \), it simplifies the expression \( \sum_{k=1}^{18} r^k \) into a manageable form, which was crucial in testing whether \( r \) and \( r^{18} \) satisfy the polynomial equation.

These simplifications using the geometric sum formula not only make computations clearer but are essential in polynomial identities and transformations often encountered when dealing with complex numbers.

Complex Numbers

Complex numbers, expressed in the form \( a + bi \), are central to solving many mathematical problems including polynomial equations with roots that aren't real numbers. They extend the real number system to include solutions to equations like \( x^2 + 1 = 0 \). Working with complex numbers requires understanding their basic properties and operations.

• The unit imaginary number \( i \) is defined as \( i^2 = -1 \). This allows us to express all complex numbers by specifying both a real part \( a \) and an imaginary part \( b \).
• Using Euler's formula, each complex number can be represented as a point in the complex plane, useful for visualizing polynomial root relationships.
• In the contexts of roots of unity, they offer a framework to represent solutions geometrically on the unit circle, with the primitive root \( r = e^{2\pi i / n} \) being a central concept.

Understanding complex numbers is therefore integral to grasping the full picture of roots of unity as they illustrate how roots lie on the complex plane, particularly in trigonometric or exponential form, making them invaluable in the study and application of polynomials.