题目内容
已知复数z满足:丨z丨=1+3i-z
(1)求z的值;
(2)求
的值.
(1)求z的值;
(2)求
| (1+i)2(3+4i)2 | 2z |
分析:(1)设z=x+yi(x,y∈R),代入丨z丨=1+3i-z可得方程,解出即可;
(2)代入z值,利用复数代数形式的运算可求;
(2)代入z值,利用复数代数形式的运算可求;
解答:解:设z=x+yi(x,y∈R),
(1)由丨z丨=1+3i-z,得|x+yi|=1+3i-(x+yi)=1-x+(3-y)i,
则
=1-x,且3-y=0,解得x=-4,y=3,
所以z=-4+3i;
(2)
=
=
=
=3+4i.
(1)由丨z丨=1+3i-z,得|x+yi|=1+3i-(x+yi)=1-x+(3-y)i,
则
| x2+y2 |
所以z=-4+3i;
(2)
| (1+i)2(3+4i)2 |
| 2z |
| 2i(-7+24i) |
| 2(-4+3i) |
| -24-7i |
| -4+3i |
| (-24-7i)(-4-3i) |
| (-4+3i)(-4-3i) |
点评:本题考查复数代数形式的混合运算、复数求模,考查学生的运算能力,属基础题.
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