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 $-1$.

Convert $\sin\frac{\pi}{8}$ and $\cos\frac{\pi}{8}$ using co-function identities to write the fraction as $\dfrac{1+\cos\varphi+i\sin\varphi}{1+\cos\varphi-i\sin\varphi}$, simplify to $e^{i\varphi}$, then apply De Moivre's theorem for the fractional power.

Let $\theta=\dfrac{\pi}{8}$ and set $\varphi = \dfrac{\pi}{2}-\theta = \dfrac{3\pi}{8}$. Using co-function identities, $\sin\theta=\cos\varphi$ and $\cos\theta=\sin\varphi$, so the given fraction becomes

$\frac{1+\cos\varphi+i\sin\varphi}{1+\cos\varphi-i\sin\varphi}$

Use the half-angle identities $1+\cos\varphi = 2\cos^2\frac{\varphi}{2}$ and $\sin\varphi = 2\sin\frac{\varphi}{2}\cos\frac{\varphi}{2}$:

Numerator:

$1+\cos\varphi+i\sin\varphi = 2\cos^2\frac{\varphi}{2} + 2i\sin\frac{\varphi}{2}\cos\frac{\varphi}{2} = 2\cos\frac{\varphi}{2}\left(\cos\frac{\varphi}{2}+i\sin\frac{\varphi}{2}\right)$

Denominator:

$1+\cos\varphi-i\sin\varphi = 2\cos\frac{\varphi}{2}\left(\cos\frac{\varphi}{2}-i\sin\frac{\varphi}{2}\right)$

Ratio:

$\frac{2\cos\frac\varphi2\left(\cos\frac\varphi2+i\sin\frac\varphi2\right)}{2\cos\frac\varphi2\left(\cos\frac\varphi2-i\sin\frac\varphi2\right)} = \frac{\cos\frac\varphi2+i\sin\frac\varphi2}{\cos\frac\varphi2-i\sin\frac\varphi2} = \frac{e^{i\varphi/2}}{e^{-i\varphi/2}} = e^{i\varphi}$

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