Good evening everyone, i come to you today with a question I don't know whether I've approached correctly and also don't know if i got the correct answers. Thank you for taking a look :D.
By writing $z = x + iy$, where $x, y \in \mathbb{R}$, find all the values of $x$ and $y$ that solve:
$$z(1+z^*) + \frac{5\operatorname{Im}(iz)}{1-2i} - 2\operatorname{Re}(z^*) - 4 = 0$$
To start:
$$z = x + iy$$
$$z^*=x-iy$$
$$5\operatorname{Im}(iz) = 5\operatorname{Im}(ix-y) = 5x$$
$$2\operatorname{Re}(z^*)=-2y$$
When we sub in for these values of $z$ and $z^*$:
$$(x+iy)(1+x-iy) + \frac{5x}{1-2i} + 2y - 4 = 0$$
Times everything by $1-2i$:
$$(1-2i)(x+iy)(1+x-iy) + 5x + 2y(1-2i) - 4(1-2i) = 0$$
Multiply out the brackets and regroup:
$$(1-2i)(x+x^2-ixy+iy+iyx-i^2y^2) + 5x + 2y - 4yi - 4 + 8i = 0$$
$$(1-2i)(x+x^2+iy+y^2) + 5x + 2y - 4yi - 4 + 8i = 0$$
$$(x+x^2+iy+y^2-2ix-2ix^2-2i^2y-2iy^2) + 5x + 2y - 4yi - 4 + 8i = 0$$
$$x^2-2ix^2+y^2-2iy^2+6x-2ix+2y-3iy-4+8i=0$$
Now to separate into two different quadratic equations s.t. $\operatorname{Im}=0$ and $\operatorname{Re}=0$:
Real Quadratic (1): $$x^2+y^2+6x+2y-4=0$$
and
Imaginary Quadratic (2): $$-2x^2-2y^2-2x-3y+8 = 0 = 2x^2+2y^2+2x+3y-8$$
Finding $y$ in terms of $x$:
$$(2)-2\cdot(1) \implies -10x-y = 0$$
So $$y = -\frac{x}{10}$$
Now whack this into equation 1:
(1): $$x^2+\left(-\frac{x}{10}\right)^2+6x+2\left(-\frac{x}{10}\right)-4=0$$
$$\implies x^2+\frac{x^2}{100}+6x-\frac{2x}{10}-4=0$$
$$\implies 101x^2+580x-400=0$$
From which we get $$x = \frac{-290\pm10\sqrt{1245}}{101} \implies x=0.6222\dots, x=-6.3648\dots$$
Which we can sub back into $y = -\frac{x}{10}$ to find:
$$y = \frac{290\pm10\sqrt{1245}}{1010} \implies y = -0.0622\dots, y=0.6364$$
Now for the other quadratic (2). We know $y = -\frac{x}{10}$:
$$2x^2+2\left(-\frac{x}{10}\right)^2+2x+3\left(-\frac{x}{10}\right)-8 = 0$$
$$\implies 101x^2+85x-400=0$$
From which we get $$x=\frac{-85\pm5\sqrt{6753}}{202} \implies x=1.6133\dots, x=-2.4549$$
Now, again, sub back into $y=-\frac{x}{10}$ to find:
$$y=\frac{85\pm5\sqrt{6753}}{2020} \implies y=-0.1613\dots, y=0.2455$$
Now, with these values for $x$ and $y$ in mind, we can answer the question:
$$z=x+iy$$
Where $x=0.6222, x=-6.3648, x=1.6133, x=-2.4549$
And $y=-0.0622, y=0.6365, y=-0.1613, y=0.2455$
Using this i should be able to sub the corresponding $x$ and $y$ values into the original equation and it should equal $0$.
It doesn't, and I have no idea what to do. I've done the question twice and gotten the same result, and a third time to type up here and still got the same answer. I really don't know what to do. You can see I've given it my best shot but to absolutely no avail.
Could someone please either tell me I'm going in the complete wrong direction and point me to where i should be going, or correct the small mistakes i may have made throughout, or just anything to help me get my answer really.
Thank you for your time in reading this essay of a problem if you got this far too, it's hugely appreciated :P
Alternative approach, using $\,\operatorname{Re}(z)=\frac{1}{2}(z+\bar z)\,$ and $\,\operatorname{Im}(z)=\frac{1}{2i}(z-\bar z)\,$ the equation rewrites as:
$$ \require{cancel} z(1+\bar z) + \frac{5(iz+i\bar z)}{2i(1-2i)} \cdot \frac{1+2i}{1+2i} - (z+\bar z) - 4 = 0 $$
$$ \require{cancel} z(1+\bar z) + \frac{\bcancel{5}\cancel{i}(z+\bar z)(1+2i)}{2 \cancel{i} \cdot \bcancel{5}} - (z+\bar z) - 4 = 0 $$
$$ \require{cancel} 2z(1+\bar z) - (1-2i)(z + \bar z) - 8 = 0 $$
$$ \require{cancel} 2z \bar z + (1+2i)z -(1-2i) \bar z - 8 = 0 \tag{1} $$
It follows that $\,2i \cdot \operatorname{Im}\big((1+2i)z\big) = (1+2i)z -(1-2i) \bar z = - 2 z \bar z + 8 \in \mathbb{R}\,$, but the LHS is a purely imaginary number if non-zero, which in turn implies $\,\operatorname{Im}\big((1+2i)z\big)=0\,$.
Therefore $\,z=(1-2i)\cdot \lambda\,$ for some real $\,\lambda \in \mathbb{R}\,$, then substituting back in $(1)$ gives:
$$ 2 \cdot 5 \cdot \lambda^2 + \cancel{5 \cdot \lambda} - \cancel{5 \cdot \lambda} - 8 = 0 \quad \iff \quad \lambda^2 = \frac{4}{5} \quad \iff \quad z = \pm\frac{2}{\sqrt{5}}(1-2i) $$