I am having a hard time solving this equation. A know the solution from WolframAlpha, but the setp-by-step solution is not available.
$$z^2+(1+i)\,\overline z+4i=0$$
http://www.wolframalpha.com/input/?i=z%5E2%2B(1%2Bi)conj(z)%2B4i%3D0
How can I solve this?
Hint:
$$z^2+(1+i)\,\overline z+4i=0$$ $$z:=x+y\cdot i$$
$$(x+y i)^2+(1+i)\,(x-y i)+4i=0$$
$$\dots$$
Write $z = z_r + i z_i$, and $z^* = z_r - i z_i$ and set up two equations (for the real and for the imaginary) in two variables ($z_r$ and $z_i$).
Where $\Re$ is the real part and $\Im$ is the imaginary part. (So $\Re+\mathrm{i}\Im=z$)
$$(\Re+i\Im)^2+(1+\mathrm{i})(\Re-i\Im)+4\mathrm{i}=0$$
$$(\Re^2-\Im^2+2\mathrm{i}\Re\Im)+(1+\mathrm{i})(\Re-\mathrm{i}\Im)+4\mathrm{i}=0$$
$$(\Re^2-\Im^2+2\mathrm{i}\Re\Im)+(\Re+\Im+\mathrm{i}\Re-\mathrm{i}\Im)+4\mathrm{i}=0$$
$$\Re^2-\Im^2+2\mathrm{i}\Re\Im+\Re+\Im+\mathrm{i}\Re-\mathrm{i}\Im+4\mathrm{i}=0$$
Complex parts are equal:
$$2\Im\Re+\Re-\Im=-4$$
Real parts are equal:
$$\Re^2-\Im^2+\Re+\Im=0$$
Then I couldn't move further. But WolframAlpha could: http://www.wolframalpha.com/input/?i=2xy+%2Bx-y%3D-4;x%5E2-y%5E2%2Bx%2By%3D0
Too long for a comment.
$$z^2+(1+i)\,\bar z+4i=0$$
Let $z=x+iy,\;x,y\in\mathbb{R}$
$$(x+iy)^2+(1+i)(x-iy)+4i=0$$
$$x^2+x-y^2+y+i (2 x y+x-y+4)=0$$ which translates in the system $$\left\{ \begin{array}{l} x^2+x-y^2+y=0 \\ 2 xy+x-y+4=0 \\ \end{array} \right. $$ Solve the second equation for $y$ we get $y=-\dfrac{x+4}{2 x-1}$
and plug in the first
$$x^2+x-\frac{(x+4)^2}{(2 x-1)^2}-\frac{x+4}{2 x-1}=0$$ from which
$4 x^4-6 x^2-14 x-12=0$
factor $2 (x-2) (x+1) \left(2 x^2+2 x+3\right)=0$
$x_1=2;\;x_2=-1$ and substituting in $y=-\dfrac{x+4}{2 x-1}$
$y_1=-2;\;y_2=1$
So the solutions to the given equation are
$z_1=2-2i;\;z_2=-1+i$
Hope this helps
$z = x+iy\\ z^2 + (1+i)\bar z + 4i = 0\\ x^2 -y^2 + (2xy)i + (x-y) + (x+y)i + 4i = 0$
Separate the real and "imaginary" components
$x^2 -y^2 + (x-y)= 0\\ 2xy + x+y + 4= 0$
rotate the plane
$u = x+y\\ v = x-y$
making that substitution into $x^2 -y^2 + (x-y)= 0$ equation.
$uv + v = 0\\ (u+1)v = 0\\ u = -1 \text { or } v=0$
This will be useful
The other equation transforms to
$\frac 12 u^2 - \frac 12 v^2 + u + 4 = 0$
case: $u = -1$
$-\frac 12 v^2 + \frac 72= 0\\ v = \pm \sqrt {7}\\ x = \frac 12 (u+v)\\ y = \frac 12 (u-v)\\ z = -\frac 12 + i\frac {7}{2}, -\frac 12 - i \frac {7}{2}$
case: $v = 0$
$\frac 12 u^2 + u + 4 = 0$
has no real roots and we are requiring $x,y$ to be real.
The following is based on the often useful observation that $(1+i)^2=2i\,$, so the equation can be written as:
$$ z^2+(1+i)\,\overline z+2(1+i)^2=0 $$
Dividing by $(1+i)^2\,$:
$$\require{cancel} \left(\frac{z}{1+i}\right)^2 + \frac{\bar z}{1+i} \color{red}{\cdot \frac{(1-i)^2}{(1-i)^2}} + 2 = 0 \;\;\iff\;\; \left(\frac{z}{1+i}\right)^2 + \frac{\bar z}{1-i} \cdot \frac{-\bcancel{2}i}{\bcancel{2}} + 2 = 0 $$
With the substitution $\displaystyle\;w = \frac{z}{1+i} \iff \bar w = \frac{\bar z}{1 - i}\,$ the latter can be written as:
$$ w^2 - i\,\bar w + 2 = 0 \tag{1} $$
Taking conjugates on both sides:
$$ \bar w^2 + i\,w + 2 = 0 \tag{2} $$
Subtracting $(1)-(2)\,$:
$$ w^2 - \bar w^2 - i\,(w + \bar w) = 0 \;\;\iff\;\;(w+\bar w)(w - \bar w - i)=0 \tag{3} $$
The factorization $(3)$ leaves two cases to check:
$\;w-\bar w -i=0 \iff \operatorname{Im}(w) = \frac{1}{2}\,$, so $w = x + \frac{1}{2}i$ for some $x \in \mathbb{R}\,$, but substituting back into $(1)$ gives no solutions.
$\;w+\bar w=0 \iff \operatorname{Re}(w) = 0\,$, so $w=i y$ for some $y \in \mathbb{R}\,$, then $(1)$ gives $-y^2-y+2=0$ with solutions $y = 1, -2$ corresponding to $w = i, -2i$ and $z=(1+i)w=-1+i,2-2i\,$.
