Verified Solution • 2025–26 Session moderate QJEEM22Ma2403: JEE Main 2022 Mathematics - Class 11 JEEM22Ma2403 Solution: Let the minimum value of is attained at . Then is equal to : Let the minimum value v0v_{0}v0 of v=∣z∣2+∣z−3∣2+∣z−6i∣2,z∈Cv=|z|^{2}+|z-3|^{2}+|z-6 i|^{2}, z \in \mathbb{C}v=∣z∣2+∣z−3∣2+∣z−6i∣2,z∈C is attained at z=z0{ }{z}=z_{0}z=z0 . Then ∣2z02−zˉ03+3∣2+v02\left|2 z_{0}^{2}-\bar{z}_{0}^{3}+3\right|^{2}+v_{0}^{2}2z02−zˉ03+32+v02 is equal to : Multiple Choice Options A 1000 B 1024 C 1105 D 1196 Step-by-Step Verified Solution Let z=x+iyz = x + iyz=x+iy v=x2+y2+(x−3)2+y2+x2+(y−6)2v = {x^2} + {y^2} + {(x - 3)^2} + {y^2} + {x^2} + {(y - 6)^2}v=x2+y2+(x−3)2+y2+x2+(y−6)2 =(3x2−6x+9)+(3y2−12y+36)= (3{x^2} - 6x + 9) + (3{y^2} - 12y + 36)=(3x2−6x+9)+(3y2−12y+36) =3(x2+y2−2x−4y+15)= 3({x^2} + {y^2} - 2x - 4y + 15)=3(x2+y2−2x−4y+15) =3[(x−1)2+(y−2)2+10]= 3[{(x - 1)^2} + {(y - 2)^2} + 10]=3[(x−1)2+(y−2)2+10] vmin{v_{\min }}vmin at z=1+2i=z0z = 1 + 2i = {z_0}z=1+2i=z0 and v0=30{v_0} = 30v0=30 so ∣2(1+2i)2−(1−2i)3+3∣2+900|2{(1 + 2i)^2} - {(1 - 2i)^3} + 3{|^2} + 900∣2(1+2i)2−(1−2i)3+3∣2+900 =∣2(−3+4i)−(1−8i3−6i(1−2i)+3∣2+900= |2( - 3 + 4i) - (1 - 8{i^3} - 6i(1 - 2i) + 3{|^2} + 900=∣2(−3+4i)−(1−8i3−6i(1−2i)+3∣2+900 =∣−6+8i−(1+8i−6i−12)+3∣2+900= | - 6 + 8i - (1 + 8i - 6i - 12) + 3{|^2} + 900=∣−6+8i−(1+8i−6i−12)+3∣2+900 =∣8+6i∣2+900= |8 + 6i{|^2} + 900=∣8+6i∣2+900 =1000= 1000=1000 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring