Proving the Equation: 1i^n / (1 - i^(n-2)) 2i^(n-1)

In this article, we will go through the detailed steps of proving the equation 1in / (1 - in-2) 2in-1. This proof involves the use of complex numbers, their exponential form, and properties of modulus and argument. Understanding the steps will provide valuable insights into working with complex numbers and manipulating equations involving exponents.

Step-by-Step Proof

Let's break down the proof into several stages, starting from the basic properties and moving through the complex number manipulations required.

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

First, we express 1i and 1 - i in their polar forms. This helps us to manipulate the numbers using trigonometric functions and exponents.

1i: The modulus of 1i is 1i √(12 12) √2, and its argument is arg(1i) π/4. In polar form, 1i √2(cos(π/4) i sin(π/4)) √2eiπ/4. 1 - i: The modulus of 1 - i is 1 - i √(12 (-1)2) √2, and its argument is arg(1 - i) -π/4. In polar form, 1 - i √2(cos(-π/4) i sin(-π/4)) √2e-iπ/4.

Using these expressions, we can rewrite the original equation.

Step 2: Substitute into the Equation

Now, we substitute the expressions obtained in polar form into our original equation:

1i^n √2ei(nπ/4) 2n/2ei(nπ/4) 2n/2cos(nπ/4) i sin(nπ/4) 1 - in-2 √2e-i(n-2)π/4 2(n-2)/2e-i(n-2)π/4 2(n-2)/2cos((n-2)π/4) i sin((n-2)π/4)

Now, we proceed to the next step.

Step 3: Compute the Ratio

The next step is to compute the ratio:

1i^n / (1 - in-2) [2n/2ei(nπ/4)] / [2(n-2)/2e-i(n-2)π/4]

This can be simplified as follows:

The modulus part: 2n/2 / 2(n-2)/2 2(n/2 - (n-2)/2) 2(n/2 - n/2 1) 2 The exponential part: ei(nπ/4) / e-i(n-2)π/4 ei(nπ/4 (n-2)π/4) ei((n n - 2)π/4) ei(2n - 2)π/4 ei(n-1)π/2

Combining both parts, we get:

1i^n / (1 - in-2) 2ei(n-1)π/2

Step 4: Express ei(n-1)π/2

Using the property of exponentials and powers of i, we can further simplify:

ei(n-1)π/2 in-1

This means:

2ei(n-1)π/2 2in-1

Conclusion

We have shown that:

1i^n / (1 - in-2) 2in-1

This completes the proof. This method can be applied to similar problems involving complex numbers and their polar forms.

Note

It is also worth noting that:

(1 ± i) √2(1/√2 ± i/√2) √2e±πi/4

Therefore:

(1i^n / (1 - in-2)) (sqrt{2}^2(cot(nπ/4) i sin(nπ/4)) / (e^{-n-2πi/4})) 2(e^π/2)^{n-1} 2in-1

This concludes the proof and provides a comprehensive understanding of the manipulation required.