Let a eq b be two non-zero real numbers. Then the number of elements in the set X=\z \in \mathbbC: \operatornameRe(a z^2+b z)=a. and .\operatornameRe(b z^2+a z)=b\ is equal to :

Complex Numbers PYQ — JEE Main 2023 · 6 Apr · Shift 2 · Q27 with Solution

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Correct answer: NO SINGLE OPTION IS UNIVERSALLY CORRECT., CORRECT CLASSIFICATION: $|X|=0$ WHEN $A+B\NE 0$, AND $|X|=\INFTY$ WHEN $A+B=0$., SO THE PROBLEM STATEMENT OR STORED ANSWER IS LIKELY INCOMPLETE/INCORRECT.

\quad x,y\in\mathbb{R}.$$ Then $$z^2=(x+iy)^2=x^2-y^2+2ixy,$$ so $$\operatorname{Re}(z^2)=x^2-y^2,\qquad \operatorname{Re}(z)=x.$$ 2. Now use the given conditions: $$\operatorname{Re}(az^2+bz)=a$$ and $$\operatorname{Re}(bz^2+az)=b.$$ Since $a,b$ are real, $$\operatorname{Re}(az^2+bz)=a\operatorname{Re}(z^2)+b\operatorname{Re}(z)=a(x^2-y^2)+bx,$$ $$\operatorname{Re}(bz^2+az)=b\operatorname{Re}(z^2)+a\operatorname{Re}(z)=b(x^2-y^2)+ax.$$ Hence the system becomes $$a(x^2-y^2)+bx=a \quad \cdots (1)$$ $$b(x^2-y^2)+ax=b \quad \cdots (2)$$ 3. Put $$u=x^2-y^2.$$ Then (1) and (2) are linear equations in $u$ and $x$: $$au+bx=a$$ $$bu+ax=b.$$ Subtract the two equations: $$(a-b)u+(b-a)x=a-b.$$ Since $a\ne b$, $$(a-b)(u-x)=a-b,$$ so $$u-x=1.$$ Therefore, $$u=x+1.$$ 4. Substitute into the first equation: $$a(x+1)+bx=a$$ $$ax+a+bx=a$$ $$(a+b)x=0.$$ Now consider two cases. ### Case 1: $a+b\ne 0$ Then $$x=0.$$ Hence $$u=x^2-y^2=-y^2,$$ and also $u=x+1=1$. So $$-y^2=1,$$ which is impossible for real $y$. So in this case, there is **no solution**. ### Case 2: $a+b=0$ Then $$b=-a,$$ and since $a,b$ are non-zero, $a\ne 0$. From $$u=x+1,$$ we get $$x^2-y^2=x+1.$$ Rearrange: $$x^2-x-y^2=1,$$ $$\left(x-\frac12\right)^2-y^2=\frac54.$$ This is a hyperbola in the $(x,y)$-plane, so it has infinitely many real points $(x,y)$. Thus there are infinitely many complex numbers $$z=x+iy$$ that satisfy both conditions. 5. Therefore: - if $a+b\ne 0$, number of elements is $0$; - if $a+b=0$ (i.e. $b=-a$), number of elements is infinite. Since the question asks for the number of elements under only the condition $a\ne b$ and non-zero real, the answer is **not uniquely determined** unless one additionally assumes $a+b=0$. So option D is not always true. 6. Checking the stored answer: The stored answer is **D (Infinite)**. But this holds only when $a+b=0$. For example, take $a=1$, $b=2$: $$u+2x=1,\qquad 2u+x=2.$$ Subtracting gives $u-x=1$, and solving gives $x=0$, $u=1$, impossible since $u=x^2-y^2=-y^2\le 0$. So there is no solution. Hence the stored answer cannot be correct as stated.

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\quad x,y\in\mathbb{R}.$$ Then $$z^2=(x+iy)^2=x^2-y^2+2ixy,$$ so $$\operatorname{Re}(z^2)=x^2-y^2,\qquad \operatorname{Re}(z)=x.$$ 2. Now use the given conditions: $$\operatorname{Re}(az^2+bz)=a$$ and $$\operatorname{Re}(bz^2+az)=b.$$ Since $a,b$ are real, $$\operatorname{Re}(az^2+bz)=a\operatorname{Re}(z^2)+b\operatorname{Re}(z)=a(x^2-y^2)+bx,$$ $$\operatorname{Re}(bz^2+az)=b\operatorname{Re}(z^2)+a\operatorname{Re}(z)=b(x^2-y^2)+ax.$$ Hence the system becomes $$a(x^2-y^2)+bx=a \quad \cdots (1)$$ $$b(x^2-y^2)+ax=b \quad \cdots (2)$$ 3. Put $$u=x^2-y^2.$$ Then (1) and (2) are linear equations in $u$ and $x$: $$au+bx=a$$ $$bu+ax=b.$$ Subtract the two equations: $$(a-b)u+(b-a)x=a-b.$$ Since $a\ne b$, $$(a-b)(u-x)=a-b,$$ so $$u-x=1.$$ Therefore, $$u=x+1.$$ 4. Substitute into the first equation: $$a(x+1)+bx=a$$ $$ax+a+bx=a$$ $$(a+b)x=0.$$ Now consider two cases. ### Case 1: $a+b\ne 0$ Then $$x=0.$$ Hence $$u=x^2-y^2=-y^2,$$ and also $u=x+1=1$. So $$-y^2=1,$$ which is impossible for real $y$. So in this case, there is **no solution**. ### Case 2: $a+b=0$ Then $$b=-a,$$ and since $a,b$ are non-zero, $a\ne 0$. From $$u=x+1,$$ we get $$x^2-y^2=x+1.$$ Rearrange: $$x^2-x-y^2=1,$$ $$\left(x-\frac12\right)^2-y^2=\frac54.$$ This is a hyperbola in the $(x,y)$-plane, so it has infinitely many real points $(x,y)$. Thus there are infinitely many complex numbers $$z=x+iy$$ that satisfy both conditions. 5. Therefore: - if $a+b\ne 0$, number of elements is $0$; - if $a+b=0$ (i.e. $b=-a$), number of elements is infinite. Since the question asks for the number of elements under only the condition $a\ne b$ and non-zero real, the answer is **not uniquely determined** unless one additionally assumes $a+b=0$. So option D is not always true. 6. Checking the stored answer: The stored answer is **D (Infinite)**. But this holds only when $a+b=0$. For example, take $a=1$, $b=2$: $$u+2x=1,\qquad 2u+x=2.$$ Subtracting gives $u-x=1$, and solving gives $x=0$, $u=1$, impossible since $u=x^2-y^2=-y^2\le 0$. So there is no solution. Hence the stored answer cannot be correct as stated.

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