將正切函數表示為正弦與餘弦： $$\dfrac{1+i\tan\dfrac{\pi}{8}}{1-i\tan\dfrac{\pi}{8}} = \dfrac{1+i\dfrac{\sin\dfrac{\pi}{8}}{\cos\dfrac{\pi}{8}}}{1-i\dfrac{\sin\dfrac{\pi}{8}}{\cos\dfrac{\pi}{8}}} = \dfrac{\cos\dfrac{\pi}{8}+i\sin\dfrac{\pi}{8}}{\cos\dfrac{\pi}{8}-i\sin\dfrac{\pi}{8}}$$ 利用負角公式：$\cos\left(-\dfrac{\pi}{8}\right) = \cos\dfrac{\pi}{8}$ 且 $\sin\left(-\dfrac{\pi}{8}\right) = -\sin\dfrac{\pi}{8}$，將分母化為複數極式： $$\dfrac{\cos\dfrac{\pi}{8}+i\sin\dfrac{\pi}{8}}{\cos\left(-\dfrac{\pi}{8}\right)+i\sin\left(-\dfrac{\pi}{8}\right)} = \cos\left(\dfrac{\pi}{8} - \left(-\dfrac{\pi}{8}\right)\right) + i\sin\left(\dfrac{\pi}{8} - \left(-\dfrac{\pi}{8}\right)\right)$$ $$= \cos\left(\dfrac{2\pi}{8}\right)+i\sin\left(\dfrac{2\pi}{8}\right) = \cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}$$ $$= \dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i.$$ 答案為選項 $(5)$。