Q.If $\omega\ne1$ is a cube root of unity, show that $\dfrac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2}+\dfrac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}=-1$.

Lists the four values of the fractional power by de Moivre's theorem and multiplies them, using that the sum of their arguments is a multiple of $2\pi$.

1. Write $z=\cos\dfrac\pi3+i\sin\dfrac\pi3=e^{i\pi/3}$.
2. By the generalised de Moivre theorem, the four values of $z^{3/4}$ are $w_k=\cos\left(\dfrac{3\left(\frac\pi3+2k\pi\right)}4\right)+i\sin\left(\dfrac{3\left(\frac\pi3+2k\pi\right)}4\right)$ for $k=0,1,2,3$, i.e. arguments $\theta_k=\dfrac{\pi+6k\pi}4$.
3. $\theta_0=\dfrac\pi4$, $\theta_1=\dfrac{7\pi}4$, $\theta_2=\dfrac{13\pi}4$, $\theta_3=\dfrac{19\pi}4$.
4. The product of the four (unit-modulus) values is $w_0w_1w_2w_3=\cos(\theta_0+\theta_1+\theta_2+\theta_3)+i\sin(\theta_0+\theta_1+\theta_2+\theta_3)$, since moduli are all $1$.
5. $\theta_0+\theta_1+\theta_2+\theta_3=\dfrac{1+7+13+19}4\pi=\dfrac{40}4\pi=10\pi$.
6. $\cos10\pi+i\sin10\pi=\cos0+i\sin0=1$ (since $10\pi$ is an integer multiple of $2\pi$).

Writing the square root as $x+iy$ and equating real and imaginary parts of $(x+iy)^2=i$ pins down $x=y=1/\sqrt2$.

1. Let $\sqrt i=x+iy$ for real $x,y$. Then $(x+iy)^2=i$.
2. Expand: $x^2-y^2+2xyi=0+1\cdot i$.
3. Equate real parts: $x^2-y^2=0\Rightarrow x^2=y^2\Rightarrow x=\pm y$.
4. Equate imaginary parts: $2xy=1$, so $xy>0$, meaning $x$ and $y$ have the same sign — hence $x=y$ (not $x=-y$).
5. Substitute $x=y$ into $2xy=1$: $2x^2=1\Rightarrow x^2=\dfrac12\Rightarrow x=\pm\dfrac{1}{\sqrt2}$, and correspondingly $y$ has the same sign.
6. So $\sqrt i=\pm\left(\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt2}i\right)=\pm\dfrac{1}{\sqrt2}(1+i)$.

Rewriting each bracket using $1+\omega+\omega^2=0$ as $-2\omega$ and $-2\omega^2$, then simplifying powers with $\omega^3=1$, gives the value $-16$.

1. Recall the key property of a cube root of unity $\omega\neq1$: $1+\omega+\omega^2=0$ and $\omega^3=1$.
2. Rewrite $1-\omega+\omega^2 = (1+\omega+\omega^2) - 2\omega = 0-2\omega = -2\omega$.
3. Rewrite $1+\omega-\omega^2 = (1+\omega+\omega^2) - 2\omega^2 = 0-2\omega^2 = -2\omega^2$.
4. So the expression becomes $(-2\omega)^4 + (-2\omega^2)^4 = 16\omega^4 + 16\omega^8$.
5. Reduce the powers using $\omega^3=1$: $\omega^4=\omega^3\cdot\omega=\omega$, and $\omega^8=\omega^6\cdot\omega^2=(\omega^3)^2\omega^2=\omega^2$.
6. So the expression is $16\omega+16\omega^2 = 16(\omega+\omega^2)$.
7. Since $1+\omega+\omega^2=0$, we have $\omega+\omega^2=-1$, so the expression $=16(-1)=-16$.

Reduce all powers of $\omega$ mod 3, use $(1-\omega)(1-\omega^2)=3$ (a standard cube-root-of-unity identity), then square it.

1. $\omega$ is a complex cube root of unity, so $\omega^3=1$ and $1+\omega+\omega^2=0$.
2. Reduce the exponents modulo 3: $\omega^4=\omega^{3+1}=\omega$, and $\omega^8=\omega^{6+2}=\omega^2$.
3. So the product becomes $(1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2) = \big[(1-\omega)(1-\omega^2)\big]^2$.
4. Expand $(1-\omega)(1-\omega^2) = 1 - \omega^2 - \omega + \omega^3 = 1 - (\omega+\omega^2) + 1$.
5. Using $\omega+\omega^2=-1$: this equals $1-(-1)+1 = 3$.
6. Therefore the full product is $3^2 = 9$.
7. This matches option (a).

The expression equals $\omega+\omega^2=-1$, where $\omega$ is a complex cube root of unity.

1. Let $\omega=\dfrac{-1+i\sqrt3}{2}$; then $\omega^2=\dfrac{-1-i\sqrt3}{2}$, with $\omega^3=1$ and $1+\omega+\omega^2=0$.
2. $\omega^{100}=\omega^{99}\cdot\omega=(\omega^3)^{33}\cdot\omega=1^{33}\cdot\omega=\omega$.
3. $(\omega^2)^{100}=\omega^{200}=\omega^{198}\cdot\omega^2=(\omega^3)^{66}\cdot\omega^2=\omega^2$.
4. So the expression $=\omega^{100}+(\omega^2)^{100}=\omega+\omega^2$.
5. Since $1+\omega+\omega^2=0$, we get $\omega+\omega^2=-1$.
6. Options (a) $2$, (b) $0$, (d) $1$ do not match this identity.