Sketch the complex function: $z\overline{z}+(1+2i)z+(1-2i)+1=0$

Question

Tried sketching the complex function: $z\overline{z}+(1+2i)z+(1-2i)+1=0$

I first simplified it by converting $z=x+yi$

I got: $(x+yi)(x-yi)+(1+2i)(x+yi)+(1-2i)+1=0$

Which gave me this implicit function: $$(x^2+y^2+x-2y+2)+(2x+y-2)i=0$$

This is where I got confused as I don't know how to go further and sketch this function.

Could it be that there's a typo in the question or is there something I'm missing?

Answer 1

You have $(x^2 +y^2 +x−2y+2)+(2x+y−2)i=0 $ so this complex number is identically zero, meaning that both real and imaginary parts must be zero.

However, looking at the real part, you have$$x^2 +y^2 +x−2y+2=0$$ $$\Rightarrow(x+\frac 12)^2+(y-1)^2=-2+1+\frac 14<0,$$ so there are no real $x$ and $y$ satisfying this equation and hence no locus.

Answer 2

Already you have obtained that $$Z=(x^2+y^2+x-2y+2)+(2x+y-2)i=0$$ Note that $$x^2+y^2+x-2y+2=(x+\frac12)^2+(y-1)^2+\frac34\gt0.$$ Hence $\Re( Z)\gt 0$ and $Z\not=0.$
Therefore, there are no such complex numbers.