JEE Main 2023 — Mathematics Algebra

Given,

$$X={z\in C:\mathrm{Re}(a{z}^{2}+bz)=a\mathrm{and}\mathrm{Re}(b{z}^{2}+az)=b}$$X=z∈C:Re(az2+bz)=aandRe(bz2+az)=b

Now, let $z=x+iy$z=x+iy

So, $\mathrm{Re}(a{z}^{2}+bz)=a$Re(az2+bz)=a

$$\Rightarrow \mathrm{Re}(a{(x+iy)}^{2}+b(x+iy))=a$$⇒Re(a(x+iy)2+b(x+iy))=a

$$\Rightarrow \mathrm{Re}(a({x}^{2}-{y}^{2}+2ixy)+b(x+iy))=a$$⇒Re(a(x2−y2+2ixy)+b(x+iy))=a

$$\Rightarrow a({x}^{2}-{y}^{2})+bx=a...(i)$$⇒a(x2−y2)+bx=a...(i)

And, $\mathrm{Re}(b{z}^{2}+az)=b$Re(bz2+az)=b

$$\Rightarrow \mathrm{Re}(b{(x+iy)}^{2}+a(x+iy))=b$$⇒Re(b(x+iy)2+a(x+iy))=b

$$\Rightarrow \mathrm{Re}(b({x}^{2}-{y}^{2}+2ixy)+a(x+iy))=b$$⇒Re(b(x2−y2+2ixy)+a(x+iy))=b

$$\Rightarrow b({x}^{2}-{y}^{2})+ax=b...(\mathrm{ii})$$⇒b(x2−y2)+ax=b...(ii)

From $(i)$(i) and $(\mathrm{ii}),(i)-(\mathrm{ii})$(ii),(i)−(ii) we get,

$$({x}^{2}-{y}^{2})(a-b)-x(a-b)=a-b$$(x2−y2)(a−b)−x(a−b)=a−b

$$\Rightarrow {x}^{2}-{y}^{2}-x=1...(\mathrm{iii})$$⇒x2−y2−x=1...(iii)

From $(i)$(i) and $(\mathrm{ii}),(i)+(\mathrm{ii})$(ii),(i)+(ii)

$(({x}^{2}-{y}^{2})+x-1)(a+b)=0$((x2−y2)+x−1)(a+b)=0 (Note: here $a+b\neq 0$a+b=0 is considered but it is not clear from the question)

$$\Rightarrow {x}^{2}-{y}^{2}+x-1=0...(\mathrm{iv})$$⇒x2−y2+x−1=0...(iv)

Now from $(\mathrm{iii})$(iii) and $(\mathrm{iv})$(iv) we get,

$x=0,{y}^{2}=-1$x=0,y2=−1 (No solution)