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Q.

If Arg(3Z + 2i) = π2  then the locus of Z is

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a

x2+y2+3x−2=0,y=0

b

X = 0 such that y > −23

c

x2+y2+2x−4y=0  such that y < 0, x2 +y2 > 1

d

X2+ y2 + 2x – 4y = 0 such that 2x – y + 4 > 0

answer is B.

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Detailed Solution

Arg[3x + 3iy + 2i] = π2                  ⇒tan−13y+23x=π2 ⇒3y+2>0,x=0
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