、复数i/(1+i)在复数平面内的对应点到原点的距离为?
问题描述:
、复数i/(1+i)在复数平面内的对应点到原点的距离为?
答
i/(1+i)
=i(1-i)/(1+i)(1-i)
=(i+1)/(1+1)
=1/2+i/2
所以距离是√[(1/2)²+(1/2)²]=√2/2
答
写成x+yi 距离为:(x^2+y^2)^(1/2)
i/(1+i)
=i(1-i)/(1+i)(1-i)
=i-i^2)/(1-i^2)
=(i+1)/(1+1)
=1/2(1+i)=1/2+i/2
x=1/2 y=/12
S=1/2*(1+1)^(1/2)=根号2/2