If z= 12-2i is such that |z+1|= z+ (1+i), i= -1 and , R, then + is equal to
- 2
- -4
- 3
- -1
Answer
(3) 3
Solution
z+1= 32-2i |z+1|= 94+4= 52
RHS = ( 12-2i )+ (1+i)= ( 2+ )+i( -2 )
∵ LHS is real: -2 =0 and 2+ = 52
2+2 = 52 =1, =2
+ =3
Answer: (3)
If z= 12-2i is such that |z+1|= z+ (1+i), i= -1 and , R, then + is equal to
(3) 3
z+1= 32-2i |z+1|= 94+4= 52
RHS = ( 12-2i )+ (1+i)= ( 2+ )+i( -2 )
∵ LHS is real: -2 =0 and 2+ = 52
2+2 = 52 =1, =2
+ =3
Answer: (3)
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