Geometric Series Summation: The Sum of ( sum_{{k0}}^{{n-1}} (-1 - i)^k )

Understanding the sum of a geometric series can be quite illuminating, especially when dealing with complex numbers. In this article, we will delve into the process of finding the sum of the series ( sum_{{k0}}^{{n-1}} (-1 - i)^k ), providing a detailed step-by-step explanation using both direct application of formulas and elementary algebraic manipulations.

Direct Application of the Geometric Series Formula

For a geometric series of the form ( sum_{{k0}}^{{n-1}} r^k ), the sum can be derived using the formula:

[ S_n frac{1 - r^n}{1 - r} ]

In our case, the series is ( sum_{{k0}}^{{n-1}} (-1 - i)^k ). Here, the common ratio ( r ) is ( -1 - i ). Let's apply the formula:

[ S_n frac{1 - (-1 - i)^n}{1 - (-1 - i)} ]

First, simplify the denominator:

[ 1 - (-1 - i) 1 1 i 2 i ]

Now, let's handle the numerator. To simplify ( (-1 - i)^n ), let's express it in polar form. Recall that for a complex number ( z a bi ), the polar form is given by:

[ z r(cos theta isin theta) quad text{where} quad r sqrt{a^2 b^2} quad text{and} quad theta tan^{-1}left(frac{b}{a}right) ]

For ( z -1 - i ):

[ r sqrt{(-1)^2 (-1)^2} sqrt{2} ]

[ theta tan^{-1}left(frac{-1}{-1}right) tan^{-1}(1) pi frac{pi}{4} pi frac{5pi}{4} ]

Thus, ( -1 - i sqrt{2} left(cos frac{5pi}{4} isin frac{5pi}{4}right) ). Now, raise this to the power of ( n ):

[ (-1 - i)^n (sqrt{2})^n left(cos frac{5npi}{4} isin frac{5npi}{4}right) ]

Substitute back into the formula for the sum:

[ S_n frac{1 - (sqrt{2})^n left(cos frac{5npi}{4} isin frac{5npi}{4}right)}{2 i} ]

Therefore:

[ S_n frac{2 - i}{5} left(1 - (sqrt{2})^n left(cos frac{5npi}{4} isin frac{5npi}{4}right)right) ]

Algebraic Manipulation Approach

Let ( S ) equal the sum. Then we see that:

[ (-1 - i)S sum_{k0}^{n-1} (-1 - i)^k ]

We can reindex this sum to write:

[ (-1 - i)S sum_{k1}^{n-1} (-1 - i)^k ]

Then by subtracting the original sum from both sides of this equation, we find:

[ (-1 - i)S - S sum_{k1}^{n-1} (-1 - i)^k - sum_{k0}^{n-1} (-1 - i)^k ]

Observe that nearly every term on the right-hand side in the first sum cancels a term in the second sum. Only the last term in the first series and the first term in the second series remain:

[ -2 - iS -1 - i^{n-1} - 1 ]

So we see that:

[ S frac{-1 - i^{n-1} - 1}{-2 - i} ]

We can write this as:

[ S frac{1 - i^{n-1}i^{n-1}}{2i} ]

or as:

[ S frac{(1 - i^{n-1}i^{n-1})(2 - i)}{5} ]

Simplified Final Answer

In conclusion, the sum of the series ( sum_{{k0}}^{{n-1}} (-1 - i)^k ) can be expressed as:

[ S_n frac{1 - (-1 - i)^n}{1 - (-1 - i)} ]

or

[ S_n frac{2 - i}{5} left(1 - (-1 - i)^n cdot (sqrt{2})^n left(cos frac{5npi}{4} isin frac{5npi}{4}right)right) ]

This result gives us a comprehensive understanding of how to calculate the sum of a geometric series involving complex numbers, which can be particularly useful in various mathematical and engineering applications.