问题
解答题
(Ⅰ)已知z∈C,且|z|-i=
(Ⅱ)已知z1=a+2i,z2=3-4i(i为虚数单位),且
|
答案
(Ⅰ)设z=x+yi,代入方程|z|-i=
+2+3i,得出. z
-i=x-yi+2+3i=(x+2)+(3-y)i,x2+y2
故有
,解得
=x+2x2+y2 3-y=-1
,x=3 y=4
∴z=3+4i,复数
=z 2+i
=2+i,虚部为13+4i 2+i
(Ⅱ)
=z1 z2
=a+2i 3-4i
,且3a-8+(4a+6)i 25
为纯虚数则3a-8=0,且4a+6≠0,解得a=z1 z2 8 3