Proving the Mathematical Identity: 1 i^n 1 - i^n 2^{(n/2 - 1)} cos (nπ/4)

In this article, we will explore the proof of the mathematical identity involving complex numbers. Specifically, we will demonstrate that the expression 1 i^n 1 - i^n is equivalent to 2^{(n/2 - 1)} cos (nπ/4).

Introduction

This identity involves complex numbers and Euler's formula, which relates complex exponentials to trigonometric functions. The proof will be conducted in several steps, making use of polar form, de Moivre's theorem, and Euler's formula. By the end of this article, you will have a clear understanding of the proof and its underlying principles.

Step 1: Expressing 1 i and 1 - i in Polar Form

To begin, we need to express the complex numbers 1 i and 1 - i in polar form. The polar form of a complex number z a bi is given by z r e^{iθ}, where r is the modulus and θ is the argument.

$1 i sqrt{2} left( cos frac{π}{4} i sin frac{π}{4} right)$

$1 - i sqrt{2} left( cos left(-frac{π}{4}right) i sin left(-frac{π}{4}right) right)$

Step 2: Raising Both Expressions to the Power n

Now, we will raise both expressions to the power of n. Using the properties of exponents and the polar form, we obtain:

$1 i^n left(sqrt{2} e^{i frac{π}{4}}right)^n 2^{frac{n}{2}} e^{i frac{nπ}{4}}$

$1 - i^n left(sqrt{2} e^{-i frac{π}{4}}right)^n 2^{frac{n}{2}} e^{-i frac{nπ}{4}}$

Step 3: Adding the Two Results

We now add the two results obtained in the previous step:

$1 i^n 1 - i^n 2^{frac{n}{2}} e^{i frac{nπ}{4}} 2^{frac{n}{2}} e^{-i frac{nπ}{4}}$

Factoring out the common term 2^{frac{n}{2}}, we get:

$ 2^{frac{n}{2}} left(e^{i frac{nπ}{4}} e^{-i frac{nπ}{4}}right)$

Application of Euler's Formula

Euler's formula states that (e^{i θ} e^{-i θ} 2 cos θ). Applying this to our expression, we find:

$e^{i frac{nπ}{4}} e^{-i frac{nπ}{4}} 2 cos left(frac{nπ}{4}right)$

Therefore, the expression simplifies to:

$1 i^n 1 - i^n 2^{frac{n}{2}} cdot 2 cos left(frac{nπ}{4}right) 2^{frac{n}{2} - 1} cdot 2 cos left(frac{nπ}{4}right)$

Thus, we have:

$1 i^n 1 - i^n 2^{frac{n}{2} - 1} cos left(frac{nπ}{4}right)$

Conclusion

We have successfully demonstrated the identity 1 i^n 1 - i^n 2^{frac{n}{2} - 1} cos left(frac{nπ}{4}right), thus completing the proof. This proof involves the use of polar form, de Moivre's theorem, and Euler's formula, which are fundamental concepts in complex number theory.