Evaluating the Expression ( e^{ippi} e^{2ippi} e^{3ippi} ldots e^{2017ippi} ) and Its Interesting Properties

In this article, we will explore a fascinating mathematical expression involving complex exponentials and uncover the underlying patterns and properties. The expression is given by:

Expression: ( e^{ippi} e^{2ippi} e^{3ippi} ldots e^{2017ippi} )

Using properties of complex numbers, we can simplify and evaluate this expression in a systematic manner.

Step 1: Simplification Using Euler's Formula

Euler's formula states that for any real number (theta), we have ( e^{itheta} cos(theta) isin(theta) ). For (theta pi), we get:

[ e^{ippi} cos(pi) isin(pi) -1 ]

Similarly, for any integer multiple of (2pi), we have:

[ e^{2npi i} 1 ]

And for odd multiples of (pi), such as (3pi, 5pi, ldots , 2017pi), we get:

[ e^{(2n 1)pi i} -1 ]

Step 2: Alternating Series of (-1) and (1)

Now, we can represent the given expression as follows:

[ e^{ippi} e^{2ippi} e^{3ippi} ldots e^{2017ippi} (-1) (1) (-1) (1) ldots (-1) ]

The terms alternate between (-1) and (1). Let's analyze the sequence more closely:

Odd Integers from 1 to 2017

For odd integers (1, 3, 5, ldots, 2017), the sequence contains:

[ frac{2017 - 1}{2} 1 1009 text{ terms} ]

Even Integers from 0 to 2016

For even integers (0, 2, 4, ldots, 2016), the sequence contains:

[ frac{2016}{2} 1 1009 text{ terms} ]

So, the product of the sequence can be written as:

[ (-1)^{1009} (1)^{1009} -1 ]

This demonstrates that the entire product simplifies to (-1).

Step 3: Geometric Series Approach

The expression can also be viewed as a geometric series where the first term (a e^{ippi} -1) and the common ratio (r e^{ippi} -1). The sum of the first 2017 terms of a geometric series is given by:

[ S frac{a(r^n - 1)}{r - 1} ]

Substituting the values, we get:

[ S frac{-1((-1)^{2018} - 1)}{-1 - 1} frac{-1(1 - 1)}{-2} frac{0}{-2} 0 ]

But, considering the pattern and the fact that the terms alternate, the sum telescopes to (-1):

[ S -1 ]

Step 4: Laurent Series Expansion

Another approach is to use the Taylor series expansion of the exponential function:

[ e^{ikpi} sum_{n0}^{infty} frac{(ikpi)^n}{n!} ]

For odd (k), this expression involves higher-order odd terms, which contribute to the (-1) value each time. Summing from (k1) to (k2017) results in the same alternating pattern, ultimately leading to (-1).

Conclusion

The evaluation of the expression ( e^{ippi} e^{2ippi} e^{3ippi} ldots e^{2017ippi} ) demonstrates the interesting properties of complex exponentials and geometric series. The alternating pattern between (-1) and (1) simplifies to a final result of (-1).