F.4 Maths question

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?

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