貌似不对。。。
追答1=cos0+sin(0)i √2/2(1+i)=cos(π/4)+sin(π/4)i
arccos1=0 arccos(√2/2+√2/2i)=π/4
-1=cos(π)+sin(π)i
arccos(-1)=π
呃,大神,答案应该是二分之派加罗格根号2减1 i
追答谢谢提示,修正如下:
复数z
arccosh (z)=ln[z+√(z^2-1)]
z=i
arccosh(i)=ln[i+√2i]=lni+ln(√2+1)
i=cos(π/2)+isin(π/2)=e^(π/2)i
arccos(i)=lne^(π/2)i +ln(√2+1)
=(π/2)i+ln(√2+1)