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