Q.Show that $\left(\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2}\right)^{10}+\left(\dfrac{1}{\sqrt2}-\dfrac{i}{\sqrt2}\right)^{10} = 0$

Multiply the three $\text{cis}$ forms; the angles add, and $A+B+C=\pi$ for a triangle.

Since $x=\text{cis}\,A$, $y=\text{cis}\,B$, $z=\text{cis}\,C$ (writing $\text{cis}\,\theta=\cos\theta+i\sin\theta$), and the product of $\text{cis}$ terms adds their angles,

$xyz = \text{cis}(A+B+C).$

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Use $1+\omega+\omega^2=0$ to rewrite each bracket as $-2\omega$ and $-2\omega^2$, then reduce the powers using $\omega^3=1$.

Since $1,\omega,\omega^2$ are the cube roots of unity, $1+\omega+\omega^2=0$ and $\omega^3=1$.

From $1+\omega+\omega^2=0$: $1+\omega^2=-\omega$, so

$1-\omega+\omega^2 = (1+\omega^2)-\omega = -\omega-\omega = -2\omega.$

Also $1+\omega=-\omega^2$, so

$1+\omega-\omega^2 = (1+\omega)-\omega^2 = -\omega^2-\omega^2 = -2\omega^2.$

Hence

$(1-\omega+\omega^2)^5+(1+\omega-\omega^2)^5 = (-2\omega)^5+(-2\omega^2)^5 = -32\omega^5-32\omega^{10}.$

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Use $1+\omega+\omega^2=0$ to rewrite each bracket as $-2\omega$ and $-2\omega^2$, then use $\omega^3=1$ to reduce the powers.

Since $1,\omega,\omega^2$ are the cube roots of unity, $1+\omega+\omega^2=0$ and $\omega^3=1$.

From $1+\omega+\omega^2=0$ we get $1+\omega^2=-\omega$, so

$1-\omega+\omega^2 = (1+\omega^2)-\omega = -\omega-\omega = -2\omega.$

Similarly $1+\omega=-\omega^2$, so

$1+\omega-\omega^2 = (1+\omega)-\omega^2 = -\omega^2-\omega^2 = -2\omega^2.$

Hence

$(1-\omega+\omega^2)^5+(1+\omega-\omega^2)^5 = (-2\omega)^5+(-2\omega^2)^5 = -32\omega^5-32\omega^{10}.$

…

Converting to polar form and using De Moivre gives $a=-1,\ b=\sqrt3$, so $a^2+b^2=4$.

$\sqrt3+i$ has modulus $r=\sqrt{3+1}=2$ and amplitude $\theta$ with $\cos\theta=\tfrac{\sqrt3}{2},\ \sin\theta=\tfrac12$, i.e. $\theta=30^\circ$.

So $\sqrt3+i = 2(\cos30^\circ+i\sin30^\circ)$.

By De Moivre's theorem, $(\sqrt3+i)^{100}=2^{100}(\cos3000^\circ+i\sin3000^\circ)$.

Since $3000^\circ = 8(360^\circ)+120^\circ$, the angle reduces to $120^\circ$: …

The three cube roots of $8$ are $2,\ -1+i\sqrt3,\ -1-i\sqrt3$.

Write $8 = 8(\cos0+i\sin0)$. The cube roots are

$8^{1/3}\!\left(\cos\tfrac{2k\pi}{3}+i\sin\tfrac{2k\pi}{3}\right),\ k=0,1,2$, and $8^{1/3}=2$.

$k=0:\ 2(\cos0+i\sin0)=2.$

$k=1:\ 2(\cos120^\circ+i\sin120^\circ)=2\!\left(-\tfrac12+i\tfrac{\sqrt3}{2}\right)=-1+i\sqrt3.$ …

Get $\alpha+\beta$ and $\alpha\beta$ from the coefficients, build up $\alpha^2+\beta^2$ then $\alpha^4+\beta^4$, and evaluate $\alpha^{-1}\beta^{-1}=1/(\alpha\beta)$.

Since $\alpha,\beta$ are roots of $x^2+x+1=0$:

$\alpha+\beta = -1, \qquad \alpha\beta = 1$

Step 1 — find $\alpha^2+\beta^2$:

$\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (-1)^2 - 2(1) = 1-2 = -1$

Step 2 — find $\alpha^4+\beta^4$:

$\alpha^4+\beta^4 = (\alpha^2+\beta^2)^2 - 2(\alpha\beta)^2 = (-1)^2 - 2(1)^2 = 1-2 = -1$

Step 3 — find $\alpha^{-1}\beta^{-1}$: …

Writing $\sqrt{3}+i$ in polar form and applying De Moivre's theorem reduces $(\sqrt3+i)^{100}$ to $2^{100}(\cos\theta+i\sin\theta)$ for some angle $\theta$; comparing with $2^{99}(a+ib)$ immediately gives $a^2+b^2=4$.

Step 1 — Polar form. $|\sqrt3+i| = \sqrt{(\sqrt3)^2+1^2} = \sqrt{4}=2$, and $\arg(\sqrt3+i) = \tan^{-1}\left(\dfrac{1}{\sqrt3}\right) = \dfrac{\pi}{6}$.

So $\sqrt3+i = 2\left(\cos\dfrac{\pi}{6}+i\sin\dfrac{\pi}{6}\right)$.

Step 2 — Apply De Moivre's theorem.

$(\sqrt3+i)^{100} = 2^{100}\left(\cos\dfrac{100\pi}{6}+i\sin\dfrac{100\pi}{6}\right) = 2^{100}(\cos\theta+i\sin\theta)$, where $\theta=\dfrac{50\pi}{3}$.

Step 3 — Compare with the given form.

We are told $(\sqrt3+i)^{100} = 2^{99}(a+ib)$. So …

Using $1+\omega+\omega^2=0$, both brackets simplify to $-2\omega$ and $-2\omega^2$; raising to the 5th power and using $\omega^3=1$ gives the value $32$.

Step 1 — Simplify each bracket using $1+\omega+\omega^2=0$.

Since $1+\omega^2=-\omega$: $1-\omega+\omega^2 = (1+\omega^2)-\omega = -\omega-\omega=-2\omega$.

Since $1+\omega=-\omega^2$: $1+\omega-\omega^2 = (1+\omega)-\omega^2=-\omega^2-\omega^2=-2\omega^2$.

Step 2 — Raise to the 5th power.

$(1-\omega+\omega^2)^5 = (-2\omega)^5 = -32\omega^5$.

$(1+\omega-\omega^2)^5 = (-2\omega^2)^5 = -32\omega^{10}$.

Step 3 — Reduce powers of $\omega$ using $\omega^3=1$. …