6. Proof:
$$\begin{array}{l} \frac{1+\sin \theta+i \cos \theta}{1+\sin \theta-i \cos \theta} \\ =\frac{(1+\sin \theta+i \cos \theta)^{2}}{(1+\sin \theta)^{2}+\cos ^{2} \theta} \\ =\frac{(1+\sin \theta)^{2}-\cos ^{2} \theta+2 i(1+\sin \theta) \cos \theta}{1+2 \sin \theta+\sin ^{2} \theta+\cos ^{2} \theta} \\ =\frac{1+2 \sin \theta+\sin ^{2} \theta-\left(1-\sin ^{2} \theta\right)+2 i(1+\sin \theta) \cos \theta}{2(1+\sin \theta)} \\ =\frac{2 \sin \theta(1+\sin \theta)+i 2(1+\sin \theta) \cos \theta}{2(1+\sin \theta)} \\ =\sin \theta+i \cos \theta \end{array}$$

Taking $\theta=\frac{\pi}{5}$ in the above equation, we get
$$\frac{1+\sin \frac{\pi}{5}+i \cos \frac{\pi}{5}}{1+\sin \frac{\pi}{5}-i \cos \frac{\pi}{5}}=\sin \frac{\pi}{5}+i \cos \frac{\pi}{5}$$

Taking the fifth power on both sides and using Example 8 of this chapter, we get
$$\begin{array}{l} \frac{\left(1+\sin \frac{\pi}{5}+i \cos \frac{\pi}{5}\right)^{5}}{\left(1+\sin \frac{\pi}{5}-i \cos \frac{\pi}{5}\right)^{5}}=\left(\sin \frac{\pi}{5}+i \cos \frac{\pi}{5}\right)^{5} \\ \quad=e^{i 5\left(\frac{\pi}{2}-\frac{\pi}{5}\right)}=e^{i \frac{3}{2} \pi} \\ \quad=\cos \frac{3}{2} \pi+i \sin \frac{3}{2} \pi=-i \end{array}$$

Therefore,
$$\begin{aligned} (1 & \left.+\sin \frac{\pi}{5}+i \cos \frac{\pi}{5}\right)^{5}+i\left(1+\sin \frac{\pi}{5}-i \cos \frac{\pi}{5}\right)^{5} \\ & =0 \end{aligned}$$