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Question

# Find the real values of the parameter a for which at least one complex number z=x+iy satisfies both the equality |z+√2|=a2−3a+2 and the inequality |z+i√2|<a2.

A
a(2,)
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B
a(,2]
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C
a[2,)
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D
a(,2)
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Solution

## The correct option is A a∈(2,∞)Let z=x+iy⇒|z+√2|=√x2+2√2x+2+y2=a2−3a+2 .................(1)Also, |z+i√2|=√y2+2√2y+2+x2<a2..........................(2)Since, modulus of a complex number is greater than zero,Hence, a2−3a+2>0 [from (1)]⇒(a−2)(a−1)>0Solving the inequality, we get,⇒a∈(−∞,1)∪(2,∞) ... (3)and because of (2), a cannot be zeroSo, a≠0 .... (4)The distance between −√2 and −i√2 is 2. So, for the first circle to overlap with the interior of the second circle the following two inequalities need to hold:r1+r2>2⇒2a2−3a+2>2⇒(a−34)2>916⇒a∈(−∞,0)∪(32,∞) ... (5)and r1<r2+2a2−3a+2<a2+2a>0 ... (6)From (3), (4), (5), (6), We can say thata>2

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