早教吧作业答案频道 -->数学-->
复数z=6+8i(4+3i)(1+i),则|z|=.
题目详情
复数z=
,则|z|=___.
6+8i |
(4+3i)(1+i) |
▼优质解答
答案和解析
z=
=
=
-
i.
∴|z|=
=
.
故答案为:
.
(6+8i)(4-3i)(1-i) |
(4+3i)(4-3i)(1+i)(1-i) |
62-34i |
25×2 |
31 |
25 |
17 |
25 |
∴|z|=
(
|
2 |
故答案为:
2 |
看了复数z=6+8i(4+3i)(...的网友还看了以下:
(2010•松江区二模)已知集合A={z|(a+bi).z+(a−bi)z+2=0,a,b∈R,z 2020-05-14 …
设z=z(x,y)是由方程x2+y2-z=φ(x+y+z)所确定的函数,其中φ具有2阶导数且φ′≠ 2020-06-12 …
(1)已知复数(m2-5m+6)+(m2-3m)i是纯虚数,求实数m的值;(2)把复数z的共轭复数 2020-08-01 …
计算下列各题(Ⅰ)已知函数f(x)=ln(2x+1)x,求f′(2);(Ⅱ)求∫π2−π2(xco 2020-08-02 …
复数z=1+i,.z为z的共轭复数,则z.z−z−1=. 2020-08-02 …
设复数Z满足(1+i)Z=2,其中i为虚数单位,则Z=[]A.1+iB.1﹣iC.2+2iD.2﹣ 2020-08-02 …
设复数Z满足(1-Z)/(1+Z)=i,则|1+Z|=?(1-z)/(1+z)=i.===>1-z= 2020-10-31 …
x,y,z>0x^2/(1+x^2)+y^2/(1+y^2)+z^2/(1+z^2)=2求证x/(1 2020-10-31 …
正数x,y,zxyz=1证明:1/[x*x*(y+1)+1]+1/[y*y*(z+1)+1]+1/[ 2020-10-31 …
1、已知x+y+z=0求x(1/y+1/z)+y(1/x+1/z)+z(1/x+1/y)+62、x- 2020-10-31 …