1+i1−i=(1+i)(1+i)1+1=1+2i−12=2i2=i=cosπ2+isinπ2\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{1+1} = \frac{1+2i-1}{2} = \frac{2i}{2} = i = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}1−i1+i=1+1(1+i)(1+i)=21+2i−1=22i=i=cos2π+isin2π これから r=1r = 1r=1 θ=π2\theta = \frac{\pi}{2}θ=2π