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This question consists of multiple questions but I am stuck on the very last one but without showing the first two the last one will be hard to understand so I'll show all my work:

13a) $w$ is one of the complex cube roots of $1$

$Show$ $that$ $$ w^2 + w + 1 = 0 $$

I did this question by

$$ w^3 = 1 $$

$$ w = 1 cis \left(\frac{\ 2πk}{3}\right) $$

Setting k = 1

$$ w = 1 cis \left(\frac{\ 2π}{3}\right) $$

$$ w = {-1\over 2} + {\sqrt{3}\over 2}i$$

and by plugging this into the quadratic equation above

$$ w^2 + w + 1 = 0 $$ thus LHS = RHS

b)

$The$ $second$ $part$ $was$ $prove$ $that$

$$ {1\over(w^2 + w^4)} = -1 $$

and by plugging in

$$ w = {-1\over 2} + {\sqrt{3}\over 2}i$$

I get

$$ {1\over(({-1\over 2} - {\sqrt{3}\over 2}) + ({-1\over 2} + {\sqrt{3}\over 2} )} = -1 $$

$$ {1\over(-1)} = -1 $$

$$ -1 = -1 $$

$$ LHS = RHS $$

Therefore the statement is proven.

Now this is the question that I am stuck on:

$c)$ $Given$ $that$ $the$ $conjugate$ $of$ $w$ $is$ $equal$ $to$ $w^2$ $,$ $find$ $the$ $conjugate$ $of$ $1+w$ $in$ $terms$ $of$ $w.$

So from my previous answers above

$$ w = {-1\over 2} + {\sqrt{3}\over 2}i$$

$$\overline{w} = {-1\over 2} - {\sqrt{3}\over 2}i = w^2 $$

$$ 1+w = {1\over 2} + {\sqrt{3}\over 2}i $$

How would I write this in terms of w?

bigfocalchord
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  • You can shorten (b) by observing that $w^2+w^4=w^2+w \cdot w^3=w^2+w$ and then cross multiply the fraction and use (a). And for (c), a hint: the conjugate of $1+ z$ is $1+ \bar{z}$. – Nicky Hekster Dec 17 '15 at 22:39
  • You are making this too complicated. Frome $w^2 + w + 1= 0$. you know $1 + w^2 = -w$. Then $\overline{1+ w} = \overline{1} + \overline{w} = 1 + \overline{w} = 1 + w^2 = -w$, – tharris Dec 17 '15 at 22:41
  • Similarly, for part (b), $w^2 + w^4 = w^2 + w$ (since $w^3 = 1$). And $w^2 + w = -1$ by (a), so $\frac{1}{w^2 + w^4} = \frac{1}{-1} = -1$. – tharris Dec 17 '15 at 22:43
  • Um, w is one of the complex roots of 1. You should this a complex root of 1. You must show it for the other strictly complex roots of 1. – fleablood Dec 17 '15 at 22:55
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    You should really observe that $w$ is a non-real complex cube root of $1$ (just to make it iron-clad that $w$ is not identically $1$). Isn't it easier then to just observe that $(w^3-1)/(w-1) = w^2+w+1$? If $w \not= 1$, then surely $w^2+w+1 = 0$. Then tharris's clue works for (b), and Nicky Hekster's clue works for (c). – Brian Tung Dec 17 '15 at 23:05
  • Also noting that tharris also provided a clue for (c). – Brian Tung Dec 17 '15 at 23:14

2 Answers2

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To enlarge upon Nicky Hekster's and tharris's clues: $w^2+w+1 = 0$, so $w^2+1 = -w$. But $\overline{1+w} = 1+\overline{w}$ (this is true for any complex number $w$). By the hypothesis, that equals $1+w^2 = -w$.

Note that (a) is more easily solved by observing that

$$ w^3-1 = (w-1)(w^2+w+1) $$

Since $w$ is a cube root of $1$, the LHS is $0$. Since $w \not= 1$ (I assume that's what you mean by a "complex cube root of $1$"), the first factor is not zero. Thus the second factor must be zero: $w^2+w+1 = 0$.

Finally, (b) can be solved by first seeing that $0 = w^2+w+1 = w^2+w(w^3)+1 = w^4+w^2+1$, so

$$ -1 = \frac{1}{-1} = \frac{1}{w^4+w^2} $$

Brian Tung
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For part (a), note that you have only proved it works for a specific one of the two non-real cube roots of unity. You have to show that it works for both of them. To do so, it is simplest to note that $(w-1)(w^2+w+1)=w^3-1=0,$ and $w\ne 1,$ so we're done.

For (b), since $w^3=1,$ then $$\frac1{w^4+w^2}=\frac1{w+w^2},$$ so what can we conclude from (a)?

For part (c), since $w^2+w+1=0,$ then $1+w^2=-w.$ Thus, $$\overline{1+w}=1+\overline{w}=1+w^2 =-w.$$

Cameron Buie
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