JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let α=8−14i, A={z∈C:z2−(zˉ)2−112iαz−αˉzˉ=1} and B={z∈C:∣z+3i∣=4}. Then ∑z∈A∩B(Re z−Im z) is equal to
Choose the correct answer:
- A.
14
(Correct Answer) - B.
13
- C.
12
- D.
11
14
Explanation
Solution
1. Set A ko simplify karna:
Maana z=x+iy.
αz−αˉzˉ=2i⋅Im(αz)=2i⋅Im((8−14i)(x+iy))=2i(8y−14x).
z2−(zˉ)2=4ixy.
Equation: 4ixy−112i2i(8y−14x)=1⟹8y−14x=2xy−56⟹xy+7x−4y−28=0.
Factorizing: (x−4)(y+7)=0. Toh x=4 ya y=−7.
2. Set B (Circle):
x2+(y+3)2=16.
-
Case 1: x=4⟹16+(y+3)2=16⟹y=−3. Point: (4,−3).
-
Case 2: y=−7⟹x2+(−4)2=16⟹x=0. Point: (0,−7).
3. Final Sum:
z1=4−3i (Re z1− Im z1=4−(−3)=7)
z2=0−7i (Re z2− Im z2=0−(−7)=7)
Sum =7+7=14.
Explanation
Solution
1. Set A ko simplify karna:
Maana z=x+iy.
αz−αˉzˉ=2i⋅Im(αz)=2i⋅Im((8−14i)(x+iy))=2i(8y−14x).
z2−(zˉ)2=4ixy.
Equation: 4ixy−112i2i(8y−14x)=1⟹8y−14x=2xy−56⟹xy+7x−4y−28=0.
Factorizing: (x−4)(y+7)=0. Toh x=4 ya y=−7.
2. Set B (Circle):
x2+(y+3)2=16.
-
Case 1: x=4⟹16+(y+3)2=16⟹y=−3. Point: (4,−3).
-
Case 2: y=−7⟹x2+(−4)2=16⟹x=0. Point: (0,−7).
3. Final Sum:
z1=4−3i (Re z1− Im z1=4−(−3)=7)
z2=0−7i (Re z2− Im z2=0−(−7)=7)
Sum =7+7=14.