Question

一道弱智的数学题，我不会做

i是虚数单位，若z（i+1）=i，则/z/=?

Answer 1

设z=a+bi
z(i+1)=(a+bi)(i+1)
=ai+a+bi²+bi=ai+a-b+bi
=a-b+(a+b)i
所以实部与虚部对应相等，得a-b=0,a+b=1
解得a=b=1/2
所以|z|²=(1/2)²+(1/2)²=1/2
|z|=√2/2

Answer 2

z=i/(i+1)=i(1-i)/[(1+i)(1-i)]=(i-i^2)/(1-i^2)----------注：i^2=-1
=(i+1)/2=1/2+1/2*i
∴|x|=根号【（1/2)^2+(1/2)^2]=根号（2/4）=根号2/2

Answer 3

设z=a+bi
z(1+i)=(a+bi)(1+i)=a+ai+bi+bi^2=a-b+(a+b)i=i
a-b=0,a+b=1 -> a=b=1/2
|z|=√2/2

Answer 4

z(i + 1) = i
z(i + 1)(i - 1) = i(i - 1)
z(i^2 - 1) = i^2 - i
-2z = -1 - i
z = 1/2 + i/2
/z/ = 根号下【(1/4) + (1/4)】= 二分之根号二

Answer 5

z=i/(i+1)
|z|=|i|/|i+1|=1/√2

Answer 6

设z=a+bi
则z(i+1)=(a-b)+(a+b)i=i
所以a-b=0 且 a+b=1
a=0.5 b=0.5 z=0.5+0.5i
/z/=(2*0.5*0.5)^1/2(开根)=2*根号0.5