Question

Let $w$ be a complex number such that $w^5 = 1$ but $w \neq -1$. Then $(1 + w)^{-1}$ is equal to

• A. $\frac{1}{2}(1 - w + w^2 - w^3 + w^4)$
• B. $1 - w + w^2 - w^3 + w^4$
• C. $1 + w + w^2 + w^3 + w^4$
• D. $\frac{1}{2}(1 + w - w^2 + w^3 - w^4)$

Answer 1

Answer: (A)

A

Answer 2

The problem requires finding the value of $(1 + w)^{-1}$ given that $w$ is a complex number satisfying $w^5 = 1$ and $w \neq -1$.

The condition $w^5 = 1$ indicates that $w$ is a fifth root of unity. The roots of the equation $z^5 - 1 = 0$ are given by $w_k = e^{i \frac{2\pi k}{5}}$ for $k = 0, 1, 2, 3, 4$. These roots are $1, e^{i 2\pi/5}, e^{i 4\pi/5}, e^{i 6\pi/5}, e^{i 8\pi/5}$. The condition $w \neq -1$ is satisfied by all fifth roots of unity, as $(-1)^5 = -1 \neq 1$.

We can factor the polynomial $w^5 - 1$ as $(w-1)(w^4 + w^3 + w^2 + w + 1) = 0$.

If $w = 1$, then $(1 + w)^{-1} = (1 + 1)^{-1} = \frac{1}{2}$. Let's check the options for $w=1$: A. $\frac{1}{2}(1 - 1 + 1^2 - 1^3 + 1^4) = \frac{1}{2}(1) = \frac{1}{2}$. B. $1 - 1 + 1^2 - 1^3 + 1^4 = 1$. C. $1 + 1 + 1^2 + 1^3 + 1^4 = 5$. D. $\frac{1}{2}(1 + 1 - 1^2 + 1^3 - 1^4) = \frac{1}{2}(1) = \frac{1}{2}$. Both A and D yield $\frac{1}{2}$ when $w=1$.

If $w \neq 1$, then $w$ is one of the other four fifth roots of unity, and it satisfies $w^4 + w^3 + w^2 + w + 1 = 0$.

Consider option A: $\frac{1}{2}(1 - w + w^2 - w^3 + w^4)$. This expression is $\frac{1}{2}$ times the sum of a geometric series with first term $1$, ratio $-w$, and $5$ terms: Sum $= \frac{1 \cdot (1 - (-w)^5)}{1 - (-w)} = \frac{1 - (-w^5)}{1+w}$. Since $w^5 = 1$, the sum is $\frac{1 - (-1)}{1+w} = \frac{2}{1+w}$. Therefore, $\frac{1}{2}(1 - w + w^2 - w^3 + w^4) = \frac{1}{2} \left( \frac{2}{1+w} \right) = (1+w)^{-1}$. This confirms that option A is correct for all $w$ such that $w^5 = 1$.

Consider option D: $\frac{1}{2}(1 + w - w^2 + w^3 - w^4)$. Let's multiply $(1+w)$ by this expression: $(1+w) \cdot \frac{1}{2}(1 + w - w^2 + w^3 - w^4)$ $= \frac{1}{2} [(1+w)(1 + w - w^2 + w^3 - w^4)]$ $= \frac{1}{2} [1(1 + w - w^2 + w^3 - w^4) + w(1 + w - w^2 + w^3 - w^4)]$ $= \frac{1}{2} [(1 + w - w^2 + w^3 - w^4) + (w + w^2 - w^3 + w^4 - w^5)]$ $= \frac{1}{2} [1 + (w+w) + (-w^2+w^2) + (w^3-w^3) + (-w^4+w^4) - w^5]$ $= \frac{1}{2} [1 + 2w - w^5]$ Since $w^5 = 1$, this simplifies to $\frac{1}{2} [1 + 2w - 1] = \frac{1}{2} [2w] = w$. For this to be equal to $(1+w)^{-1}$, we would need $w = (1+w)^{-1}$, which implies $w(1+w) = 1$, so $w + w^2 = 1$. This is not generally true for all fifth roots of unity. However, if we want $\frac{1}{2}(1 + w - w^2 + w^3 - w^4)$ to be equal to $(1+w)^{-1}$, then we need $w = (1+w)^{-1}$, which means $w(1+w)=1$. This is only true if $w=1$ (since $1+1^2=2 \neq 1$, this is actually not true even for $w=1$).

Let's re-evaluate the product for option D: $(1+w) \times \frac{1}{2}(1 + w - w^2 + w^3 - w^4) = \frac{1}{2} (1 + 2w - w^5) = \frac{1}{2}(1+2w-1) = w$. So, $(1+w)^{-1} = \frac{1}{w} \times \frac{1}{2}(1 + w - w^2 + w^3 - w^4)$. For option D to be correct, we need $\frac{1}{2}(1 + w - w^2 + w^3 - w^4) = (1+w)^{-1}$. This means $\frac{1}{w} \times \frac{1}{2}(1 + w - w^2 + w^3 - w^4) = \frac{1}{2}(1 + w - w^2 + w^3 - w^4)$, which implies $\frac{1}{w}=1$, or $w=1$. Thus, option D is only correct when $w=1$.

Since option A is correct for all valid values of $w$, it is the correct answer for this single-choice question.