Solving z−2iz−6i=1⇒y=4 .....(1)
(where z=x+iy)
Now solving z+2iz−8+2i=53
⇒x2+y2−25x+4y+104=0 .....(2)
Solving (1) & (2) ⇒z=17+4i & 8+4i
⇒∑∣z∣2=(17)2+(4)2+(8)2+(4)2=385
JEE Main 2026 — Mathematics Coordinate Geometry
Let S={z∈C:z−2iz−6i=1 and z+2iz−8+2i=53}.
Then z∈ s∑∣z∣2 is equal to
Held on 24 Jan 2026 · Verified 6 Jul 2026.
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