Let the set of all values of k R such that the equation z ( z+2+i )+k(2+3i)=0, z C, has at least one solution, be the interval [ , ]. Then 9( + ) is equal to :
- -10
- -8
- 10 13
- 8 13
Answer
(1) -10
Solution
Put z=x+iy : z z=x2+y2, z(2+i)=(2x-y)+i(x+2y)
(x2+y2+2x-y+2k )+i(x+2y+3k)=0
Real part : x2+y2+2x-y+2k=0 … (i)
circle, centre (-1, 12 ), r2=1+ 14-2k= 54-2k
Imaginary part : x+2y+3k=0 … (ii), a straight line
A solution exists line (ii) meets circle (i)
distance of centre from the line r
|-1+1+3k| 5 54-2k
9k25 54-2k 36k2+40k-25 0
k [ , ], where , are the roots of 36k2+40k-25=0
+ =- 4036=- 109
9( + )=-10
Answer: (1)