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

If L has exactly one element, then |s| e |t|, The number of elements in L \cap \ z:|z - 1 + i| = 5\ is at most 2, If L has more than one element, then L has infinitely many elements

If L has exactly one element, then |s| e |t|

If L has more than one element, then L has infinitely many elements

Complex Numbers PYQ — JEE Advanced 2018 · Shift 2 · Q23 with Solution

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Correct answer: A, C, D

Write the equation in terms of $x,y$

Given $z=x+iy,\qquad \overline z=x-iy.$ So $sz+t\overline z+r=0$ becomes $s(x+iy)+t(x-iy)+r=0.$ Thus $(s+t)x+i(s-t)y+r=0.$

This is one complex equation, hence equivalent to two real linear equations in the two real variables $x,y$ .

Interpret geometrically

Let $\alpha=s+t,\qquad \beta=i(s-t).$ Then the equation is $\alpha x+\beta y=-r.$ Taking real and imaginary parts gives a linear system in $x,y$ : $\Re(\alpha)x+\Re(\beta)y=-\Re(r),$ $\Im(\alpha)x+\Im(\beta)y=-\Im(r).$ So the coefficient matrix is

\Re(\alpha) & \Re(\beta)\\ \Im(\alpha) & \Im(\beta) \end{pmatrix}.$$ Hence: - if $\det M\neq 0$, there is exactly one solution; - if $\det M=0$, then either no solution or infinitely many solutions. So a finite number greater than $1$ is impossible. --- 3. **Compute when unique solution occurs** We compute the determinant using the two complex numbers $\alpha=s+t$ and $\beta=i(s-t)$. For two complex numbers $u,v$, the determinant formed by their real and imaginary parts is $$\Re(u)\Im(v)-\Im(u)\Re(v)=\Im(\overline u v).$$ Thus $$\det M=\Im\big(\overline{(s+t)}\, i(s-t)\big).$$ Now simplify: $$\overline{(s+t)}\, i(s-t)=i(\overline s+\overline t)(s-t).$$ Expand: $$(\overline s+\overline t)(s-t)=|s|^2-|t|^2+\overline t s-\overline s t.$$ Note that $\overline t s-\overline s t$ is purely imaginary, so after multiplication by $i$ it becomes purely real and does not affect the imaginary part. Therefore $$\det M=|s|^2-|t|^2.$$ So: - unique solution $\iff |s|^2-|t|^2\neq 0 \iff |s|\neq |t|$; - non-unique case occurs when $|s|=|t|$. Therefore **Option A is true**: > If $L$ has exactly one element, then $|s|\neq |t|$. --- 4. **Check Option B** Option B says: If $|s|=|t|$, then $L$ has infinitely many elements. But when $|s|=|t|$, the determinant is $0$, so the system may have: - infinitely many solutions, or - no solution. So infinitely many solutions are **not guaranteed**. A counterexample: Take $s=t=1$, $r=1$. Then $$z+\overline z+1=0 \implies 2x+1=0.$$ This actually gives infinitely many solutions, so let us choose a better example where inconsistency occurs. Take $s=1$, $t=-1$, $r=1$. Then $|s|=|t|=1$, and $$z-\overline z+1=0.$$ Since $$z-\overline z=2iy,$$ we get $$2iy+1=0,$$ whose real part is $1$ and imaginary part is $2y$; impossible for real $y$. Hence no solution exists. So **Option B is false**. --- 5. **Check Option D** If $L$ has more than one element, then there are at least two solutions of a linear system of two real equations in two variables. This can only happen in the dependent case, where the two equations represent the same line; hence there are infinitely many solutions. Therefore **Option D is true**. --- 6. **Check Option C** The set $L$ is the solution set of a real linear system in $(x,y)$. Hence $L$ is one of the following: - empty set, - a single point, - a straight line. Now consider the circle $$\{z:|z-1+i|=5\}.$$ Since $$|z-1+i|=|z-(1-i)|,$$ it is the circle centered at $1-i$ with radius $5$. Intersection of: - empty set with a circle: $0$ points, - a point with a circle: at most $1$ point, - a line with a circle: at most $2$ points. Hence the number of elements in $$L\cap\{z:|z-1+i|=5\}$$ is at most $2$. So **Option C is true**. --- 7. **Final conclusion** The true statements are: $$\boxed{A,\ C,\ D}$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: A, C, D

Write the equation in terms of $x,y$

Given $z=x+iy,\qquad \overline z=x-iy.$ So $sz+t\overline z+r=0$ becomes $s(x+iy)+t(x-iy)+r=0.$ Thus $(s+t)x+i(s-t)y+r=0.$

This is one complex equation, hence equivalent to two real linear equations in the two real variables $x,y$ .

Interpret geometrically

\Re(\alpha) & \Re(\beta)\\ \Im(\alpha) & \Im(\beta) \end{pmatrix}.$$ Hence: - if $\det M\neq 0$, there is exactly one solution; - if $\det M=0$, then either no solution or infinitely many solutions. So a finite number greater than $1$ is impossible. --- 3. **Compute when unique solution occurs** We compute the determinant using the two complex numbers $\alpha=s+t$ and $\beta=i(s-t)$. For two complex numbers $u,v$, the determinant formed by their real and imaginary parts is $$\Re(u)\Im(v)-\Im(u)\Re(v)=\Im(\overline u v).$$ Thus $$\det M=\Im\big(\overline{(s+t)}\, i(s-t)\big).$$ Now simplify: $$\overline{(s+t)}\, i(s-t)=i(\overline s+\overline t)(s-t).$$ Expand: $$(\overline s+\overline t)(s-t)=|s|^2-|t|^2+\overline t s-\overline s t.$$ Note that $\overline t s-\overline s t$ is purely imaginary, so after multiplication by $i$ it becomes purely real and does not affect the imaginary part. Therefore $$\det M=|s|^2-|t|^2.$$ So: - unique solution $\iff |s|^2-|t|^2\neq 0 \iff |s|\neq |t|$; - non-unique case occurs when $|s|=|t|$. Therefore **Option A is true**: > If $L$ has exactly one element, then $|s|\neq |t|$. --- 4. **Check Option B** Option B says: If $|s|=|t|$, then $L$ has infinitely many elements. But when $|s|=|t|$, the determinant is $0$, so the system may have: - infinitely many solutions, or - no solution. So infinitely many solutions are **not guaranteed**. A counterexample: Take $s=t=1$, $r=1$. Then $$z+\overline z+1=0 \implies 2x+1=0.$$ This actually gives infinitely many solutions, so let us choose a better example where inconsistency occurs. Take $s=1$, $t=-1$, $r=1$. Then $|s|=|t|=1$, and $$z-\overline z+1=0.$$ Since $$z-\overline z=2iy,$$ we get $$2iy+1=0,$$ whose real part is $1$ and imaginary part is $2y$; impossible for real $y$. Hence no solution exists. So **Option B is false**. --- 5. **Check Option D** If $L$ has more than one element, then there are at least two solutions of a linear system of two real equations in two variables. This can only happen in the dependent case, where the two equations represent the same line; hence there are infinitely many solutions. Therefore **Option D is true**. --- 6. **Check Option C** The set $L$ is the solution set of a real linear system in $(x,y)$. Hence $L$ is one of the following: - empty set, - a single point, - a straight line. Now consider the circle $$\{z:|z-1+i|=5\}.$$ Since $$|z-1+i|=|z-(1-i)|,$$ it is the circle centered at $1-i$ with radius $5$. Intersection of: - empty set with a circle: $0$ points, - a point with a circle: at most $1$ point, - a line with a circle: at most $2$ points. Hence the number of elements in $$L\cap\{z:|z-1+i|=5\}$$ is at most $2$. So **Option C is true**. --- 7. **Final conclusion** The true statements are: $$\boxed{A,\ C,\ D}$$ This matches the stored correct answer.

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