Finding the Modulus of a Complex Number: Step-by-Step Solution

The question asks for the modulus of the complex number given by the expression \(i^{2n+1} (-i)^{2n-1}\), where \(n\) is a natural number (\(n \in \mathbb{N}\)) and \(i\) is the imaginary unit, with \(i = \sqrt{-1}\).

Understanding the Complex Expression

The given complex number is \(Z = i^{2n+1} (-i)^{2n-1}\). To find its modulus, we first need to simplify this expression. Let's break down the terms:

• The first term is \(i^{2n+1}\).
• The second term is \((-i)^{2n-1}\).

Simplifying the Expression

We can use the properties of exponents and the powers of \(i\):

• Recall that \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). The powers of \(i\) repeat every four powers.
• Also, for any integer \(k\), \(i^{4k} = (i^4)^k = 1^k = 1\).

Let's rewrite the second term \((-i)^{2n-1}\):

Using the exponent rule \((ab)^m = a^m b^m\):

Since \(n \in \mathbb{N}\), \(n\) is a positive integer (1, 2, 3, ...). This means \(2n\) is always an even integer (2, 4, 6, ...), and \(2n-1\) is always an odd integer (1, 3, 5, ...). Therefore, \((-1)^{2n-1}\) will always be \(-1\).

So, \((-i)^{2n-1} = -1 \cdot i^{2n-1} = - i^{2n-1}\).

Now substitute this back into the original expression for \(Z\):

Using the exponent rule \(a^m \cdot a^p = a^{m+p}\):

Now we simplify \(i^{4n}\):

Since \(n \in \mathbb{N}\), \(1^n = 1\).

So, the expression for \(Z\) becomes:

The given complex number simplifies to the real number \(-1\).

Calculating the Modulus

The modulus of a complex number \(z = x + yi\) is given by \(|z| = \sqrt{x^2 + y^2}\). In our case, the simplified complex number is \(Z = -1\). We can write this in the form \(x+yi\) as \(-1 + 0i\). Here, \(x = -1\) and \(y = 0\).

The modulus of \(Z\) is:

Alternatively, the modulus of a real number \(x\) is simply its absolute value, \(|x|\). The complex number is \(-1\), which is a real number. Its modulus is \(|-1| = 1\).

Final Answer

The modulus of the complex number \(i^{2n+1} (-i)^{2n-1}\) is 1.

Revision Table: Complex Numbers and Modulus

Here's a quick table summarizing key concepts related to this problem:

Additional Information: Modulus Properties

The modulus of a complex number has several useful properties that can sometimes simplify calculations involving complex number expressions. For any complex numbers \(z_1\) and \(z_2\):

• Modulus of a product: \(|z_1 z_2| = |z_1| |z_2|\)
• Modulus of a quotient: \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1}}{|z_2|}\) (where \(z_2 \neq 0\))
• Modulus of a power: \(|z^n| = |z|^n\) for integer \(n\).
• Triangle inequality: \(|z_1 + z_2| \le |z_1| + |z_2|\)
• Difference inequality: \(|z_1 - z_2| \ge ||z_1| - |z_2||\)

In this problem, we could have potentially used the product property \(|z_1 z_2| = |z_1| |z_2|\) at an earlier stage. Let \(z_1 = i^{2n+1}\) and \(z_2 = (-i)^{2n-1}\). Then \(|Z| = |i^{2n+1} (-i)^{2n-1}| = |i^{2n+1}| \cdot |(-i)^{2n-1}|\).

• \(|i^{2n+1}| = |(i^4)^{n/2} \cdot i|\) or \(|(i^2)^n \cdot i| = |(-1)^n \cdot i|\). Since \(|ab| = |a||b|\), \(|(-1)^n \cdot i| = |(-1)^n| \cdot |i| = 1 \cdot 1 = 1\).
• \(|(-i)^{2n-1}| = |(-1)^{2n-1} i^{2n-1}| = |-1 \cdot i^{2n-1}| = |-1| \cdot |i^{2n-1}|\). Since \(|-1|=1\), this is \(1 \cdot |i^{2n-1}|\). Similarly, \(|i^{2n-1}| = |(i^4)^{(2n-2)/4} \cdot i^{1}|\) or \(|(i^2)^{n-1} \cdot i| = |(-1)^{n-1} i|\). \(|(-1)^{n-1} i| = |(-1)^{n-1}| \cdot |i| = 1 \cdot 1 = 1\).

So, \(|Z| = |i^{2n+1}| \cdot |(-i)^{2n-1}| = 1 \cdot 1 = 1\). This confirms the result obtained through simplification.

Using simplification first often makes the process clearer, especially when dealing with exponents involving variables like \(n\).