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Question

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

A
a > 2
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B
a = 12
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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 > 2If z = x + iy is a complex number satisfying the given conditions, then a2−3a+2=|z+√2|=|z+i√2+√2−i√2|≤|z+i√2|+√2|1−i|<a2+2 ⇒−3a<0⇒a>0 .....(i) Since |z+√2=a2−3a+2| represents a circle with centre at A(−√2,0) and radius √a2−3a+2,and|z+√2i|<a2 represents the interior of the circle with centre at B(0,−√2) and radius a, therefore 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 radii of the two circles, i.e., if √(−√2−0)2+(0+√2)2<√a2−3a+2±a ⇒ 2±a<√a2−3a+2 ⇒4+a2±4a<a2−3a+2 ⇒ -a < -2 or 7a < -2 ⇒ a > 2 or a < 72 But a > 0 from (i), therefore a > 2. Area of the triangle = 12 |z|2

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