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Question

# If at least one value of complex number z=x+iy satisfies the condition ∣∣z+√2∣∣=√a2−3a+2 and the inequality ∣∣z+i√2∣∣<a2, then

A
a>2
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B
a=2
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C
a<2
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D
None of these
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Solution

## The correct option is A a>2Let z=x+iy be a complex number satisfying the given condition.then a2−3a+2=∣∣z+√2∣∣=∣∣z+i√2−i√2+√2∣∣⇒a2−3a+2≤∣∣z+i√2∣∣+√2|1−i|<a2+2⇒a>0Since a2−3a+2=∣∣z+√2∣∣ represents a circle with center at A (−√2,0) and radius √a2−3a+2and ∣∣z+i√2∣∣<a2 represents the interior of the circle with center at B (0,√2) and radius a. Thus there will be a complex number satisfying the given condition and the given inequality if the distance AB is less than the sum or difference of the two radii of 2 circles. i.e.⇒√(−√2−0)2+(0+√2)2<√a2−3a+2±a⇒2±a<√a2−3a+2Square both sides to get, −a<−2or7a<−2⇒a>2,a<−27But a>0 thus a>2

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