When does $az+b \bar z +c =0$ have one solution? 
When does $az+b \bar z +c =0$ have one solution? $a,b,c \in \mathbb C$ 

What I did:
I rewrote:
$a=a_1+ia_2$,
$b=b_1+ib_2$,
$c=c_1+ic_2$, and
$z=z_1+iz_2$
Putting everything together we are solving:
$(a_1+ia_2)\times (z_1+iz_2) + (b_1+ib_2) \times (z_1-iz_2) +(c_1+ic_2)=0+i0$ 
To have one solution I assumed that the $i$ terms will cancel out.
$i(a_2z_1+z_2a_1-b_1iz_2+iz_1b_2)+i(c_2)=0$
$iz_1(a_2+b_2)+iz_2(a_1-b_1)+i(c_2)=0$
So $a_1=b_1$ which means that the real part should be the same?
I'm not sure if I am working correctly.
Thanks.
 A: Your approach works. Setting $a=a_1+a_2i$, $b=b_1+b_2i$, $c=c_1+c_2i$ and $z=z_1+z_2i$, the equation becomes $(a_1+a_2i)(z_1+z_2i)+(b_1+b_2i)(z_1-z_2i)+(c_1+c_2i)=0$. This we can also write as $(a_1z_1-a_2z_2+b_1z_1+b_2z_2+c_1)+(a_1z_2+a_2z_1-b_1z_2+b_2z_1+c_2)i=0$. We now have an equation of the form $v+wi=0$, where $v,w\in\mathbb{R}$. For a complex number to be $0$, both the real and imaginary part must be zero; we may conclude $$a_1z_1-a_2z_2+b_1z_1+b_2z_2+c_1=0$$ $$a_1z_2+a_2z_1-b_1z_2+b_2z_1+c_2=0$$ Now we have two linear equations in $z_1$ and $z_2$, and we can solve this! We'll first write the equations to $$(a_1+b_1)z_1+(-a_2+b_2)z_2=-c_1$$ $$(a_2+b_2)z_1+(a_1-b_1)z_2=-c_2$$ Making the coefficient for $z_1$ in both equations equal and subtracting, we get $$(b_2^2-a_2^2-a_1^2+b_1^2)z_2=-c_1(a_2+b_2)+c_2(a_1+b_1)$$ Now if $b_2^2-a_2^2-a_1^2+b_1^2=0$, we have either no or infinitely many solutions for $z_2$, neither of which we want. Thus, a property our system must have is $b_1^2+b_2^2\neq a_1^2+a_2^2$, in other words $|a|\neq |b|$ (using the norm/length of the complex numbers). We can now divide by $|b|^2-|a|^2=b_1^2+b_2^2-a_1^2-a_2^2\neq 0$, and we get $$z_1=\frac{c_1(a_1-b_1)-c_2(-a_2+b_2)}{|b|^2-|a|^2}$$ $$z_2=\frac{(a_1+b_1)c_2-(a_2+b_2)c_1}{|b|^2-|a|^2}$$
The conclusion is, the system has exactly one solution only if $|a|\neq |b|$.
Hope this helps you!
A: An approach that does not involve using the "expanded" form of a complex number.
Let $w$ be a different solution than $z$, such that:
$aw+b \bar w +c=0$ and
$az + b \bar z + c= 0$. 
By subtracting  we get: $a(z-w)+b(\bar z- \bar w)=0$ $⇒ a(z-w)=-b(\bar z- \bar w)$ (1)The conjucate of this expression must also equal zero, thus:
$\bar a(\bar z- \bar w)=-\bar b(z-w)$ (2)
By multyplying (1) and (2) we get:
$a \bar a(z-w)(\bar z- \bar w)=b\bar b (z-w)(\bar z- \bar w) ⇒ a \bar a=b\bar b  $  and finally we get: $|a|=|b|$
A: It's easier than that. What you wrote is a linear system of two equations (collect the real and imaginary terms into separate equations):
$$\begin{align}
a_1 z_1-a_2 z_2 + b_1 z_1+b_2 z_2&=-c_1\\
a_2 z_1+a_1 z_2 + b_2 z_1-b_1 z_2&=-c_2
\end{align}$$
This has a single solution, if the rank of the matrix on the left is full (its determinant nonzero) - that's basics of linear algebra. If the determinant is zero, then you'll have infinite amount of solutions or no solutions at all. Collecting the terms into a matrix and computing the determinant gives you the condition
$$(a_1+b_1)(a_1-b_1)+(a_2-b_2)(a_2+b_2)\neq 0$$
This can be rewritten as
$$a_1^2-b_1^2+a_2^2-b_2^2\neq 0$$
or
$$|a|\neq |b|$$
