Find the value of the real number a such that there exists at least one complex number Z satisfying the conditions |Z + 
| = a 2 –3a + 2 and |Z + i 
| < a 2 .
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. |Z + 
| = a 2 –3a + 2 and |Z + i 
| = a 2 are two circles in the Argand diagram with centres A (– 
, 0), B(0, – 
) and radii a 2 –3a + 2 and a 2 respectively.
∴ a 2 –3a + 2 > 0, i.e., a < 1 or a > 2 … (i)
and Z lies on the first circle and lies inside the second circle. Hence, |Z + 
| = a 2 –3a + 2 and |Z + i 
| < a 2 ⇒ |Z + 
| = |Z + 
I + 
–i 
| ≤ |Z + 
i| + | 
–i 
| i.e., a
2 –3a + 2 < a 2 + 2
⇒ –3a < 0 ⇒ a > 0 … (ii)
Also the given conditions imply that the distance between centers is less than the sum or difference between the radii.
∴ 2 < a 2 + a 2 –3a + 2 or 2 < a 2 – (a 2 –3a + 2)
i.e., 2a 2 – 3a > 0 or 3a – 2 > 
a > 
or a > 
… (iii)
Combining (i), (ii) and (iii), a > 2
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