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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.

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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!

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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: $|a|=|b|$

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    $\begingroup$ Thank you very much, I must say I much prefer this way. $\endgroup$
    – GRS
    Jan 17 '16 at 19:02
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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 $$|a|\neq |b|$$

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  • $\begingroup$ Why would that yield a single solution? Wouldn't there be a multiple of such $a$ and $b$ to make it non-zero? $\endgroup$
    – GRS
    Jan 12 '16 at 11:03
  • $\begingroup$ Also, is the last line the solution of my problem? $\endgroup$
    – GRS
    Jan 12 '16 at 11:03

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