Let [a b] be two non-zero real numbers.

Question

Let $$a \neq b$$ be two non-zero real numbers. Then the number of elements in the set $$X=\left\{z \in \mathbb{C}: \operatorname{Re}\left(a z^{2}+b z\right)=a\right.$$ and $$\left.\operatorname{Re}\left(b z^{2}+a z\right)=b\right\}$$ is equal to :

Accepted Answer

Let $z=x+i y$

$$
\[\begin{gathered} \Rightarrow  z^2=x^2-y^2+2 i x y \ \therefore  a z^2+b z \ =a\left(x^2-y^2+2 i x y\right)+b(x+i y) \ =a\left(x^2-y^2\right)+b x+2 a i x y+b i y \end{gathered}$$
\]

$\begin{gathered} \operatorname{Re}\left(a z^2+b z\right)=b \ \Rightarrow a\left(x^2-y^2\right)+b x=a \ \Rightarrow x^2-y^2+\frac{b}{a} x=1 .........(i) \end{gathered}$

$\begin{aligned} & \text { and } b z^2+a z \\ & =b\left(x^2-y^2+2 i x y\right)+a(x+i y) \\ & =b\left(x^2-y^2\right)+a x+2 b i x y+a i y \\ & \operatorname{Re}\left(b z^2+a z\right)=b \\ & \Rightarrow  b\left(x^2-y^2\right)+a x=b \\ & \Rightarrow x^2-y^2+\frac{a}{b} x=1 ..........(ii)\end{aligned}$

On subtracting Equation (ii) from Equation (i), we get

$$
\begin{aligned}
& \frac{b}{a} x-\frac{a}{b} x=0 \\
& \Rightarrow \left(\frac{b}{a}-\frac{a}{b}\right) x=0 \\
& \Rightarrow x=0 \text { or } \frac{b}{a}-\frac{a}{b}=0 \\
& \Rightarrow b^2-a^2=0 \\
& \Rightarrow a= \pm b \\
& \Rightarrow a=-b(\text { since } a \neq b)
\end{aligned}
$$

From (i), when $x=0$, then

$$
\begin{aligned}
& 0-y^2=1 \Rightarrow y^2=-1 \\
& \Rightarrow y \in \phi \Rightarrow z \in \phi \text { has no solution. }
\end{aligned}
$$

When, $a=-b$, then $x^2-y^2-x=1$ has infinitely many solutions.

Solutions · 1