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How Many z Satisfy |z - 2i| = 2 and z(1-i) - conj(z)(1+i) = 4i | JEE

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Question
The number of values of zz satisfying ∣z−2i∣=2|z - 2i| = 2 and z(1−i)−zˉ(1+i)=4iz(1-i) - \bar{z}(1+i) = 4i simultaneously is
A00
B22correct
C33
D44
Solution
Step 1: Identify the first condition. ∣z−2i∣=2|z - 2i| = 2 is a circle with centre 2i2i, i.e. the point (0,2)(0,2), and radius 22. Step 2: Simplify the second condition by writing z=x+iyz = x + iy, so zˉ=x−iy\bar z = x - iy.
z(1−i)=(x+iy)(1−i)=x−ix+iy+y=(x+y)+i(y−x),z(1-i) = (x+iy)(1-i) = x - ix + iy + y = (x+y) + i(y-x),
zˉ(1+i)=(x−iy)(1+i)=x+ix−iy+y=(x+y)+i(x−y).\bar z(1+i) = (x-iy)(1+i) = x + ix - iy + y = (x+y) + i(x-y).
 Step 3: Subtract.
z(1−i)−zˉ(1+i)=i[(y−x)−(x−y)]=2i(y−x).z(1-i) - \bar z(1+i) = i\left[(y-x) - (x-y)\right] = 2i(y-x).
 Step 4: Set this equal to 4i4i.
2i(y−x)=4i  ⟹  y−x=2,2i(y-x) = 4i \implies y - x = 2 ,
 which is the straight line y=x+2y = x + 2. Step 5: Test whether the line passes through the centre (0,2)(0,2) of the circle:
2=0+2.2 = 0 + 2 .
 Step 6: A line through the centre of a circle is a diameter, so it cuts the circle in exactly 22 points. (They are (±2, 2±2)(\pm\sqrt2,\ 2\pm\sqrt2), but the count is all that is asked.) Answer: (2).
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