Let a b be two non-zero real numbers.

Question

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

Accepted Answer

Let $z=x+iy$

$\begin{gathered} \Rightarrow {z}^{2}={x}^{2}-{y}^{2}+2ixy \ 	herefore a{z}^{2}+bz \ =a({x}^{2}-{y}^{2}+2ixy)+b(x+iy) \ =a({x}^{2}-{y}^{2})+bx+2aixy+biy \end{gathered}$

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

$\begin{gathered} b{z}^{2}+az \ =b({x}^{2}-{y}^{2}+2ixy)+a(x+iy) \ =b({x}^{2}-{y}^{2})+ax+2bixy+aiy \ Re(b{z}^{2}+az)=b \ \Rightarrow b({x}^{2}-{y}^{2})+ax=b \ \Rightarrow {x}^{2}-{y}^{2}+\frac{a}{b}x=1..........(ii) \end{gathered}$

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

$\begin{gathered} \frac{b}{a}x-\frac{a}{b}x=0 \ \Rightarrow (\frac{b}{a}-\frac{a}{b})x=0 \ \Rightarrow x=0\frac{b}{a}-\frac{a}{b}=0 \ \Rightarrow {b}^{2}-{a}^{2}=0 \ \Rightarrow a=\pm b \ \Rightarrow a=-b(a
e b) \end{gathered}$

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

$\begin{gathered} 0-{y}^{2}=1\Rightarrow {y}^{2}=-1 \ \Rightarrow y\in \phi \Rightarrow z\in \phi \end{gathered}$

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

Solutions · 1