let z=(7-5i/1+i)(k+2I), where k is a real number. If the imaginary part of z is equal to k-5, find the real part of z.
(p.s. 7-5i/1+i=1-6i)
please help me to solve, Thank you!!
F.4 Maths question
2013-09-28 7:00 pm
回答 (2)
2013-09-28 11:15 pm
✔ 最佳答案
z=(7-5i/1+i)(k+2i)z=(1-6i)(k+2i)
z=k-12(i)^2-6ik+2i
z=k+12-(6k-2)i
As the imaginary part of z is equal to k-5,
-(6k-2)i=(k-5)i
6k-2=5-k
7k=7
k=1
sub k=1 into the real part,k+12
k+12=1+12 =13
Is that correct?
2013-09-28 7:36 pm
7-5i/1+i=1-6i is not true, it should be 6 - 6i
7-5i/1+i = (7-5i)(1-i)/(1+i)(1-i)
= (7 - 7i - 5i +5)/(2)
=(12-12i) / 2
=6 - 6i
z=(7-5i/1+i)(k+2I) should be z=(7-5i/1+i)(k+2i) ?
2013-09-28 12:20:07 補充：
(6-6i)(k+2i) = 6k+12i-6ki+12 = (12+6k) + (12-6k)i
if (12-6k)i = (k-5)i ==> 12-6k = k-5
17 = 7k
k = 17/7
12-6k = 12-6(17/7) = real part = -18/7
7-5i/1+i = (7-5i)(1-i)/(1+i)(1-i)
= (7 - 7i - 5i +5)/(2)
=(12-12i) / 2
=6 - 6i
z=(7-5i/1+i)(k+2I) should be z=(7-5i/1+i)(k+2i) ?
2013-09-28 12:20:07 補充：
(6-6i)(k+2i) = 6k+12i-6ki+12 = (12+6k) + (12-6k)i
if (12-6k)i = (k-5)i ==> 12-6k = k-5
17 = 7k
k = 17/7
12-6k = 12-6(17/7) = real part = -18/7
收錄日期: 2021-04-13 19:45:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130928000051KK00049