MathematicsGeometrical Applications of Complex NumbersJEE Advanced 2008Difficult
Visualized Solution (English)
Defining the Region A
- Set A={z:Im(z)≥1}
- Let z=x+iy, then Im(z)=y
- The condition becomes y≥1
- This represents the half-plane above and including the line y=1.
Visualizing the Circle B
- Set B={z:∣z−(2+i)∣=3}
- This is the standard form of a circle ∣z−z0∣=r
- Center: z0=2+i⟹(2,1)
- Radius: r=3
- Cartesian equation: (x−2)2+(y−1)2=9
Deriving the Line C
- Set C={z:Re((1−i)z)=2}
- (1−i)(x+iy)=x+iy−ix−i2y=(x+y)+i(y−x)
- Taking the real part: x+y=2
- This is a straight line with intercepts (2,0) and (0,2).
Finding the Intersection A∩B∩C
- To find B∩C, substitute y=2−x into (x−2)2+(y−1)2=9
- (x−2)2+(2−x−1)2=9
- Solving this quadratic gives two values of x.
- Constraint from A: y≥1⟹2−x≥1⟹x≤2−1
- Only one root satisfies this condition. Thus, n(A∩B∩C)=1.
Question 2: Geometric Interpretation
- Evaluate ∣z−(−1+i)∣2+∣z−(5+i)∣2 for z∈A∩B∩C
- Let Q=−1+i⟹(−1,1) and R=5+i⟹(5,1)
- The expression is PQ2+PR2
- Midpoint of QR=(2−1+5,21+1)=(2,1), which is the center of circle B.
- Thus, QR is a diameter of length 6.
Question 2: Applying Pythagoras Theorem
- Since QR is a diameter, ∠QPR=90∘ (Angle in a semi-circle).
- By Pythagoras Theorem: PQ2+PR2=QR2
- QR=∣5−(−1)∣=6
- Value =62=36
- 36 lies between 35 and 39.
Question 3: Bounds for ∣z∣−∣w∣+3
- z is a fixed point P on circle B.
- w satisfies ∣w−2i∣<3.
- Using Triangle Inequality: ∣∣z∣−∣w∣∣≤∣z−w∣
- The maximum possible value of ∣z−w∣ is approximately 6 based on the diameters.
- Thus, −6<∣z∣−∣w∣<6.
- Adding 3: −3<∣z∣−∣w∣+3<9.
- The value lies between 3 and 9.
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