Q.Show that one value of $\left[\dfrac{1+\sin\frac{\pi}{8} + i\cos\frac{\pi}{8}}{1+\sin\frac{\pi}{8} - i\cos\frac{\pi}{8}}\right]^{8/3}$ is $-1$.

Rewrite $1+\sin\theta\pm i\cos\theta$ using $\sin\theta=\cos(\tfrac\pi2-\theta)$ and $\cos\theta=\sin(\tfrac\pi2-\theta)$, so the fraction collapses to $e^{i\phi}$ for $\phi=\tfrac\pi2-\theta$; then raising to the power $8/3$ gives one value directly.

Let $\theta=\dfrac\pi8$ and set $\phi=\dfrac\pi2-\theta=\dfrac{3\pi}{8}$, so $\sin\theta=\cos\phi$ and $\cos\theta=\sin\phi$.

Numerator: $1+\sin\theta+i\cos\theta = 1+\cos\phi+i\sin\phi$. Using $1+\cos\phi=2\cos^2\tfrac{\phi}2$ and $\sin\phi=2\sin\tfrac\phi2\cos\tfrac\phi2$:

$1+\cos\phi+i\sin\phi = 2\cos^2\tfrac\phi2+2i\sin\tfrac\phi2\cos\tfrac\phi2 = 2\cos\tfrac\phi2\left(\cos\tfrac\phi2+i\sin\tfrac\phi2\right).$

Denominator: by the same steps, $1+\sin\theta-i\cos\theta = 2\cos\tfrac\phi2\left(\cos\tfrac\phi2-i\sin\tfrac\phi2\right)$.

So the ratio is

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