虚数 一道非常简单的题~~~

(1+i)/i=1/i+i/i=i/i^2+1=1-i
=√2*(√2/2-√2/2*i)
=√2*[cos(-π/4)+isin(-π/4)]

所以
[(1+i)/i]^2008
=(√2)^2008*[cos(-π/4*2008)+isin(-π/4*2008)]
=2^1004*[cos(-502π)+isin(-502π)]
=2^1004*(1+i*0)
=2^1004

i+1/i=i-(-1)/i=i-i=0
所以原式=0

(i+1/i)^2=i^2+(1/i)^2+2*i*(1/i)=-1-1+2=0
所以(i+1/i)^2008=((i+1/i)^2)^1004=0

(i+1/i)的平方等于0
故(i+1/i)的2008次方等于0

因为1/i=-i，所以（i+1/i)的2008次方等于0