Let $s, t, r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+i y$ $(x, y \in \mathbb{R}, i=\sqrt{-1})$ of the equation $s z+t \bar{z}+r=0$, where $\bar{z}=x-i y$. Then, which of the following statement(s) is (are) TRUE?

(A) If $L$ has exactly one element, then $|s| \neq|t|$; (C) The number of elements in $L \cap\{z:|z-1+i|=5\}$ is at most 2; (D) If $L$ has more than one element, then $L$ has infinitely many elements

(B) If $|s|=|t|$, then $L$ has infinitely many elements

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Complex Numbers

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Let

s, t, r

be non-zero complex numbers and

L

be the set of solutions

z = x + i y

(x, y \in R, i = - 1)

of the equation

sz + t \overset{z}{ˉ} + r = 0

, where

\overset{z}{ˉ} = x - i y

. Then, which of the following statement(s) is (are) TRUE?

(A)If $L$ has exactly one element, then $∣ s ∣ \neq = ∣ t ∣$ ✓ Correct
(B)If $∣ s ∣ = ∣ t ∣$ , then $L$ has infinitely many elements
(C)The number of elements in $L \cap {z : ∣ z - 1 + i ∣ = 5}$ is at most 2✓ Correct
(D)If $L$ has more than one element, then $L$ has infinitely many elements✓ Correct

Answer: (A), (C), (D)

Solution

From

sz + t \overset{z}{ˉ} + r = 0

and its conjugate, eliminating

\overset{z}{ˉ}

z = \frac{r ˉ t - r s ˉ}{∣ s ∣ ^{2} - ∣ t ∣ ^{2}}

(A): If

L

has exactly one element, then

∣ s ∣^{2} - ∣ t ∣^{2} \neq = 0

, so

∣ s ∣ \neq = ∣ t ∣

. ✓

(B): Counterexample:

s = t = i

r = i

gives

L = \emptyset

. ✗

(C): Adding the equation and its conjugate:

(s + \overset{ˉ}{t}) z + (\overset{s}{ˉ} + t) \overset{z}{ˉ} + (r + \overset{r}{ˉ}) = 0

, which represents a line (or no points). A line intersects a circle in at most 2 points. ✓

(D): If

α

and

β

both satisfy the equation, then

λ α + (1 - λ) β

also satisfies it for any

λ \in R

, giving infinitely many solutions. ✓

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · One or more correct options

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JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Complex Numbers

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