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Value of |z+z̄| When |z|=min{|z−1|,|z+1|} | JEE

JEE Maths question with a full step-by-step solution.

Question
If z=min{z1,z+1}|z| = \min\{|z - 1|,\, |z + 1|\}, then the value of z+zˉ|z + \bar{z}| is
A11correct
B22
C33
D44
Solution
Step 1: Simplify the target quantity. With z=x+iyz = x + iy, we have z+zˉ=2x=2Re(z)z + \bar{z} = 2x = 2\,\mathrm{Re}(z), so:
z+zˉ=2x=2x|z + \bar{z}| = |2x| = 2|x|
Step 2: Interpret the condition z=min{z1,z+1}|z| = \min\{|z - 1|, |z + 1|\}. The minimum equals one of the two distances, so either z=z1|z| = |z - 1| or z=z+1|z| = |z + 1| (whichever is smaller); both extreme cases occur on the boundary where the chosen distance is achieved. Step 3: Solve z=z1|z| = |z - 1| (distance from origin equals distance from 11). This is the perpendicular bisector of 00 and 11:
x2+y2=(x1)2+y2    0=2x+1    x=12x^2 + y^2 = (x - 1)^2 + y^2 \implies 0 = -2x + 1 \implies x = \frac{1}{2}
Step 4: Solve z=z+1|z| = |z + 1| (distance from origin equals distance from 1-1):
x2+y2=(x+1)2+y2    0=2x+1    x=12x^2 + y^2 = (x + 1)^2 + y^2 \implies 0 = 2x + 1 \implies x = -\frac{1}{2}
Step 5: In either case x=12|x| = \dfrac{1}{2}, so:
z+zˉ=2x=212=1|z + \bar{z}| = 2|x| = 2\cdot\frac{1}{2} = 1
Correct answer: (1) Note: This question can solve be by triangle inequality.
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