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JEE Advanced 2018 · Complex Numbers and Quadratic Equations

Let s,t,rs, t, r be non-zero complex numbers and LL be the set of solutions z=x+iyz = x + iy (x,y∈Rx, y \in \mathbb{R}, i=−1i = \sqrt{-1}) of the equation sz+tzˉ+r=0sz + t\bar{z} + r = 0, where zˉ=x−iy\bar{z} = x - iy. Then, which of the following statement(s) is (are) TRUE?

Answers: A. If LL has exactly one element, then ∣s∣≠∣t∣|s| \neq |t| · C. The number of elements in L∩{z:∣z−1+i∣=5}L \cap \{z : |z - 1 + i| = 5\} is at most 2 · D. If LL has more than one element, then LL has infinitely many elements

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Given: s,t,r non-zero complex; sz+tzˉ+r=0\text{Given: } s, t, r \text{ non-zero complex; } sz + t\bar{z} + r = 0 z=x+iy,x,y∈R,zˉ=x−iyz = x + iy,\quad x, y \in \mathbb{R},\quad \bar{z} = x - iy To find: which of the four statements are TRUE\text{To find: which of the four statements are TRUE} sz+tzˉ+r=0sz + t\bar{z} + r = 0 sˉ zˉ+tˉ z+rˉ=0\bar{s}\,\bar{z} + \bar{t}\,z + \bar{r} = 0 sz+tzˉ=−rsz + t\bar{z} = -r tˉz+sˉzˉ=−rˉ\bar{t}z + \bar{s}\bar{z} = -\bar{r} Δ=ssˉ−ttˉ=∣s∣2−∣t∣2\Delta = s\bar{s} - t\bar{t} = |s|^2 - |t|^2 If ∣s∣≠∣t∣: Δ≠0⇒unique solution\text{If } |s| \neq |t|:\ \Delta \neq 0 \Rightarrow \text{unique solution} ⇒L has exactly one element — statement (1) TRUE\Rightarrow L \text{ has exactly one element — statement (1) TRUE} If ∣s∣=∣t∣: Δ=0⇒no unique solution\text{If } |s| = |t|:\ \Delta = 0 \Rightarrow \text{no unique solution} consistent⇒line;inconsistent⇒empty\text{consistent} \Rightarrow \text{line};\quad \text{inconsistent} \Rightarrow \text{empty} s=t=1, r=i: 2x=−i (impossible)⇒L=∅s = t = 1,\ r = i:\ 2x = -i\ (\text{impossible}) \Rightarrow L = \varnothing ∣s∣=∣t∣⇏infinitely many elements|s| = |t| \not\Rightarrow \text{infinitely many elements} ⇒statement (2) is FALSE\Rightarrow \text{statement (2) is FALSE} L is ∅, a single point, or a line in CL \text{ is } \varnothing,\ \text{a single point, or a line in } \mathbb{C} more than one element⇒line⇒infinitely many — (4) TRUE\text{more than one element} \Rightarrow \text{line} \Rightarrow \text{infinitely many — (4) TRUE} {z:∣z−1+i∣=5}=circle, radius 5, centre 1−i\{z : |z - 1 + i| = 5\} = \text{circle, radius } 5,\ \text{centre } 1 - i L∩circle has at most 2 points — (3) TRUEL \cap \text{circle has at most } 2 \text{ points — (3) TRUE} TRUE: (1), (3), (4)\text{TRUE: } (1),\ (3),\ (4)

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