The problem is the following:

Determine conditions for $a,b,c\in\mathbb{C}$ such that $az+b\overline{z}+c=0$ has unique solution in $\mathbb{C}$.

Teacher answer:

If $az+b\overline{z}+c=0$, then $\overline{b}z+\overline{a}\overline{z}+\overline{c}=0$. So $\left(\begin{array}\\ a & b \\ \overline{b}&\overline{a} \end{array}\right)\left(\begin{array}\\ z \\ \overline{z} \end{array}\right)=\left(\begin{array}\\ -c \\ -\overline{c} \end{array}\right)$. Like in linear algebra, we have only one solution iff $|a|^2-|b|^2\neq 0$.

My answer is too long (equaling real and imaginary parts and taking a 2x2 system of linear equations), and teacher answer didn't convince me because teacher is solving a system where one component $z$ depends of another component which is $\overline{z}$ (if we obtain $z$ we automatically obtain $\overline{z}$ and viceversa).

Edit: Being more precise I have two conceptual questions, about validity and why the teacher's procedure works:

1) Why is solved for the pair $(z, \overline{z})$? if we obtain $z$, we also obtain its conjugate automatically. That has not sense.

2) Why do you need the conjugate of the first equation (is the same equation)?

3) Why condition $|a|^2-|b|^2\neq 0$ is necessary for unicity of solutions in the pair $z,\overline{z}$? I only know that there is unique solution for $(z,w)$ but we could have unique solution for $(z,\overline{z})$ and multiple solutions for $(z,w)$.

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    $\begingroup$ What is your question? You want an explanation of why the teacher's answer is valid? Do you have a question about real and imaginary parts? $\endgroup$ Aug 20, 2013 at 0:35
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    $\begingroup$ @GastónBurrull: Thank you. Interesting; it seems clear how your teacher gave the necessary and sufficient condition for $\left(\begin{array}\\ a & b \\ \overline{b}&\overline{a} \end{array}\right)\left(\begin{array}\\ x \\ y \end{array}\right)=\left(\begin{array}\\ -c \\ -\overline{c} \end{array}\right)$ to have a unique solution $\begin{pmatrix}x\\y\end{pmatrix}\in\mathbb C^2$, but for the relevant solutions you need the restriction $y=\overline x$, which was not addressed in what is written here. $\endgroup$ Aug 20, 2013 at 2:29
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    $\begingroup$ I'm pretty sure the best answer to this equation is "complexification": you know there is a linear equation of real vector spaces, you take the tensor product $- \otimes_\mathbb{R} \mathbb{C}$ to convert it into a linear equation of complex vector spaces with the same sort of solution space, then you make a change of basis to make the complexified version look like what the teacher did. $\endgroup$
    – user14972
    Aug 26, 2013 at 1:52
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    $\begingroup$ @Hurkyl, that might be the best answer for this question for someone who does not have to ask this question... $\endgroup$ Aug 26, 2013 at 3:28
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    $\begingroup$ @Mariano: ... but shouldn't be difficult for someone in practice with such things to translate into something suitable for the OP. I'm not in practice, and at the time I had given up on writing an answer I was happy with, and was offering it as an idea if someone else wanted to do it. But I've since worked out a way to write it I was happy with. (see my answer below) $\endgroup$
    – user14972
    Aug 26, 2013 at 3:33

7 Answers 7


1) The teacher's answer explains that there is only one solution for $x$, $y$ in


when $|a|^2-|b|^2\neq0$. It remains to see that that unique solution automatically has $y=\bar{x}$. Because if you actually solve for $x$ and $y$, you have

$$\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{|a|^2-|b|^2}\begin{bmatrix}\bar{a}&-b\\-\bar{b}&a\end{bmatrix}\begin{bmatrix}-c\\-\bar{c}\end{bmatrix}=\frac{1}{|a|^2-|b|^2}\begin{bmatrix}-\bar{a}c+b\bar{c}\\\bar{b}c-a\bar{c}\end{bmatrix} $$

and now it's clear that $x$ and $y$ are automatically conjugates.

2) The equation found from taking the conjugate is not "the same" equation, in a linear algebra sense. It is not a scalar multiple of the original equation, so in a linear algebra sense, it is a linearly independent equation.

3) If the determinant is $0$, then of course one possibility is that there are no solutions for $x$ and $y$. But there might be infinitely many solutions, and as you suggest, maybe exactly one of them also satisfies $y=\bar{x}$. I think you are right to worry about this when your teacher has not.

If there are infinitely many solutions, then the two equations' left-hand sides are linearly dependent, and we need only examine the first equation $ax+by=c$, leading us to $y=(c-ax)/b$ (assuming for now that $b\neq0$, where $b=0$ is a trivial case to consider). How could this work out to equal $\bar{x}$?

$$\begin{align}\bar{x}&=(c-ax)/b\\\implies x&=(\bar{c}-\bar{a}\bar{x})/\bar{b}\\&=(\bar{c}-\bar{a}(c-ax)/b)/\bar{b}\\\implies \bar{b}bx&=b\bar{c}-c\bar{a}+a\bar{a}x\\\implies \bar{b}bx-a\bar{a}x&=b\bar{c}-c\bar{a}\\\implies (|b|^2-|a|^2)x&=b\bar{c}-c\bar{a}\\\implies 0&=b\bar{c}-c\bar{a}\end{align}$$

So $b\bar{c}=c\bar{a}$. But then $$\begin{align}ax+b\bar{x}&=-c\\\implies a\bar{c}x+b\bar{c}\bar{x}&=-c\bar{c}\\\implies a\bar{c}x-\bar{a}c\bar{x}&=-|c|^2\\\implies a\bar{c}x-\overline{a\bar{c}x}&\phantom{=}\text{is a real number}\end{align}$$

This is only possible if in fact $c=0$. Then the equation becomes $ax+by=0$, a homogeneous equation. The solution vectors $(x,y)$ are all scalar multiples of each other, and so if one solution is indeed a pair of the form $(x,\bar{x})$, then all real multiples of that solution will be too.

  • $\begingroup$ We have $a\bar{c}x+b\bar{c}\bar{x}=a\bar{c}x+c\bar{a}\bar{x}\in\mathbb{R}$ always since $a\bar{c}x+c\bar{a}\bar{x}=2\operatorname{Re}(c\bar{a}\bar{x})$, we didn't need that $c=0$. $\endgroup$ Aug 28, 2013 at 4:57

$(1)$ and $(2)$: Certainly, $\bar{z}$ is a function of $z$, however, $z$ and $\bar{z}$ are linearly independent. That is, there are no $a$ and $b$ so that $az+b\bar{z}=0$ for all $z\in\mathbb{C}$. However, we know that $\mathrm{Re}(z)=\frac12(z+\bar{z})$ and $\mathrm{Im}(z)=\frac1{2i}(z-\bar{z})$, and that lets us compute both the real and complex parts of $z$ as a linear combination of $z$ and $\bar{z}$. So, although $az+b\bar{z}+c=0\iff \bar{a}\bar{z}+\bar{b}z+\bar{c}=0$, in order to solve $az+b\bar{z}+c=0$ as a linear equation, we employ $\bar{a}\bar{z}+\bar{b}z+\bar{c}=0$.

$(3)$ Perhaps this might help to explain why these two equations relate to a real matrix equation. Note that $$ \begin{bmatrix} {\small1/2}&{\small1/2}\\ {\small-i/2}&{\small i/2} \end{bmatrix} \begin{bmatrix} z\\ \bar{z} \end{bmatrix} = \begin{bmatrix} z_r\\ z_i \end{bmatrix} $$ Here is the equation from your instructor $$ \begin{bmatrix} a&b\\ \bar{b}&\bar{a} \end{bmatrix} \begin{bmatrix} z\\ \bar{z} \end{bmatrix} = \begin{bmatrix} -c\\ -\bar{c} \end{bmatrix} $$ Multiply $\begin{bmatrix}{\small1/2}&{\small1/2}\\{\small-i/2}&{\small i/2}\end{bmatrix}$ on the left and insert the identity $$ \begin{align} \begin{bmatrix} {\small1/2}&{\small1/2}\\ {\small-i/2}&{\small i/2} \end{bmatrix} \begin{bmatrix} a&b\\ \bar{b}&\bar{a} \end{bmatrix} \overbrace{ \begin{bmatrix} 1&i\\ 1&-i \end{bmatrix} \begin{bmatrix} {\small1/2}&{\small1/2}\\ {\small-i/2}&{\small i/2} \end{bmatrix} }^\text{identity} \begin{bmatrix} z\\ \bar{z} \end{bmatrix} &= \begin{bmatrix} {\small1/2}&{\small1/2}\\ {\small-i/2}&{\small i/2} \end{bmatrix} \begin{bmatrix} -c\\ -\bar{c} \end{bmatrix}\\ \begin{bmatrix} {\small1/2}&{\small1/2}\\ {\small-i/2}&{\small i/2} \end{bmatrix} \begin{bmatrix} a+b&i(a-b)\\ \overline{a+b}&\overline{i(a-b)} \end{bmatrix} \begin{bmatrix} z_r\\ z_i \end{bmatrix} &= \begin{bmatrix} -c_r\\ -c_i \end{bmatrix}\\ \begin{bmatrix} a_r+b_r&b_i-a_i\\ a_i+b_i&a_r-b_r \end{bmatrix} \begin{bmatrix} z_r\\ z_i \end{bmatrix} &= \begin{bmatrix} -c_r\\ -c_i \end{bmatrix} \end{align} $$ The determinant of the last matrix is $(a_r^2+a_i^2)-(b_r^2+b_i^2)=|a|^2-|b|^2$.

  • $\begingroup$ $z$ and $\bar z$ are «certainly» linearly independent over $R$. But $a$, $b$, $c$ and everything else in this problem is in $\mathbb C$... $\endgroup$ Aug 26, 2013 at 3:29
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    $\begingroup$ @MarianoSuárez-Alvarez: If we are talking about a particular $z$, then $2$ and $\bar{2}$ are not linearly independent over $\mathbb{R}$. However, I tried to explain that I was talking about $z$ and $\bar{z}$ in general (there are no $a$ and $b$ so that $az+b\bar{z}=0$ for all $z\in\mathbb{C}$) $\endgroup$
    – robjohn
    Aug 26, 2013 at 4:27
  • $\begingroup$ Is really nice that multiplying by an invertible matrix and an identity you proved that my solution is equivalent to the teacher's one! Thank you very much. $\endgroup$ Aug 28, 2013 at 5:23
  • $\begingroup$ @GastónBurrull: I am not complaining as it is completely up to you, but I am curious about why you changed the accepted answer. $\endgroup$
    – robjohn
    May 27, 2015 at 5:08

Consider the following fundamental property of complex numbers. If $z = x+i y$, where $x,y\in\mathbb{R}$, i.e., $$z = x+i y, \quad \bar z = x-i y,$$ then $$x = \frac{1}{2}(z+\bar z), \quad y = \frac{1}{2i}(z-\bar z).$$ That is, if we know $z,\bar z$ we can find $x,y$ uniquely and vice versa. Thus, it doesn't matter which way we solve the system. We can write it in terms of $x,y$ and solve for $x,y$ or write it in terms of $z,\bar z$ and solve for $z,\bar z$. For this particular problem, even if we're interested in $x,y$ it is easiest to solve for $z,\bar z$ and then find $x,y$.

A reader may claim we need only $z$ to find $x$ and $y$. Indeed, $x=\mathrm{Re}(z)$ and $y=\mathrm{Im}(z)$. But these operations are not linear. (For example, $\mathrm{Re}(i z) \ne i\mathrm{Re}(z)$.) To take advantage of linear algebra we must relate $x,y$ to $z,\overline z$ via linear transformations. These transformations are given above and in many of the other answers. Since the transformation is invertible, it is appropriate to treat $z,\overline z$ as independent, just as $x,y$ are independent. (This is made very explicit in @robjohn's answer.) Thus, starting from a linear equation in $z,\overline z$ we conjugate and treat the resulting two-by-two system as if $z,\overline z$ are independent. Of course, the fact that $z,\overline z$ are related by conjugation will restrict the space of solutions to a subspace of the space of solutions to the more general $z,w$ system.

This method is well known and often used in physics.

To answer your questions:

(1) If indeed we could solve the equation $az+b\bar z=c=0$ for $z$ directly, we would not need to introduce the conjugate equation. Since this is not possible we introduce the conjugate equation, arriving at a two-by-two system that we can solve.

(2) See the answer to (1). When thinking in terms of $z,\bar z$ we should consider the equation and its conjugate as "different equations" not "the same equation."

(3) The condition $$\mathrm{det}\left(\begin{array}{cc}a&b\\ \bar b&\bar a\end{array}\right) = |a|^2-|b^2|\ne 0$$ is the condition that the matrix $\left(\begin{array}{cc}a&b\\ \bar b&\bar a\end{array}\right)$ be invertible. If the matrix is not invertible the equation may have no solution or an infinite number of solutions. For example:

(a) The equation $z+\bar z-2i = 0$ has no solution.

(b) The equation $z+\bar z-2 = 0$ has an infinite number of solutions, $z = 1+i y$ where $y\in\mathbb{R}$.

  • $\begingroup$ (3) I know that $|a|^2-|b|^2\neq 0$ is asufficient condition for unicity of the pair $(z,\overline{z})$ (since there is a unique solution for the more general pair $(z,w)$), but, why is a necessary condition? $\endgroup$ Aug 22, 2013 at 4:51
  • $\begingroup$ This is a fundamental theorem of linear algebra. For an $n\times n$ linear system $A x = b$, $A$ is invertible if and only if the system has a unique solution. Note that the more general system you're considering in terms of $z,w$ is not related to the original problem unless $w=\bar z$. $\endgroup$
    – user26872
    Aug 22, 2013 at 5:06
  • $\begingroup$ Yes, the system has unique solution in $\mathbb{C}\times\mathbb{C}$ iff $|a|^2-|b|^2\neq 0$, but this is more powerful than existence of an unique solution of the form $(z,\overline{z})$. $\endgroup$ Aug 22, 2013 at 5:15
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    $\begingroup$ There can be solutions not of the form $(z,\bar{z})^T$. For example, when $a=b=1$ and $c=0$, the solution space consists of everything of the form $(w, -w)^T$; most of those are very much not of the form $(z, \bar{z})^T$. In fact, as both sets are two-dimensional real subspaces of the four-dimensional real vector space $\mathbb{C}^2$, one would expect them to only have a single point in common with rare exceptions. That that doesn't happen means some nontrivial additional phenomenon is involved. $\endgroup$
    – user14972
    Aug 22, 2013 at 23:15
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    $\begingroup$ @oen: If the intersection of the space of all solutions to the matrix equation and the space of pairs of the form $(z,\bar{z})^T$ is a single point, then the original problem has a unique solution, despite the matrix equation having infinitely many solutions. We need to explain why that can't happen. $\endgroup$
    – user14972
    Aug 23, 2013 at 8:53

You know you can set this up as a real linear equation by splitting the complex variable $z$ into two real variables $x + iy$, and splitting the one complex equation into its real and imaginary parts to get two real equations.

So, we have a real system of equations $T\vec{v} = \vec{w}$. We know that the solution space is either:

  • Empty
  • Has a unique solution
  • Has a one-dimensional family of solutions $\vec{v}_0 + t \vec{u}$ for $t \in \mathbb{R}$
  • Has a two-dimensional family of solutions: i.e. every $\vec{v}$ is a solution.

We can take the same equation $T \vec{v} = \vec{w}$ and look for its complex solutions. Based on how solving linear equations works, it's clear the complex solution space will be of the same form as the real solution space, just with complex variables allowed. (e.g. in the third case above, we allow $t \in \mathbb{C}$ but keep the same $\vec{v}_0$ and $\vec{u}$)

But now that we've moved to complex linear algebra, we have a bit more freedom with our scalars! We can, for example, make the invertible change of basis:

$$ \left( \begin{matrix} 1 & i \\ 1 & -i \end{matrix} \right) \cdot \left( \begin{matrix}x \\ y \end{matrix} \right) = \left( \begin{matrix} z \\ \bar{z} \end{matrix} \right)$$

The use of the names $z$ and $\bar{z}$ are somewhat less than literal, since $x$ and $y$ are now allowed to be complex numbers.

Our complex system of equations have to be equivalent to the original complex equation and its conjugate (e.g. it comes from making substitutions like $z = x+iy$ and combining the equation and its conjugate linearly. Remember $x,y$ are still allowed to be complex!), which is the system of equations the teacher set up. But since this system of equations is just the complexification of a real system of equations in disguise, the complex solution space has to be of the same form as the real solution space, as described above, and so we can use the teacher's argument.

If one didn't believe things had to work out, you could explicitly compute the change of basis on the matrix:

$$ \left( \begin{matrix} 1 & i \\ 1 & -i \end{matrix} \right) \cdot \left( \begin{matrix} \Re a + \Re b & -\Im a + \Im b \\ \Im a + \Im b & \Re a - \Re b \end{matrix} \right) \cdot \left( \begin{matrix} 1 & i \\ 1 & -i \end{matrix} \right)^{-1} \\= \left( \begin{matrix} a + b & ia - ib \\ \bar{a} + \bar{b} & -i\bar{a} + i\bar{b} \end{matrix} \right) \cdot \left( \begin{matrix} 1 & 1 \\ -i & i \end{matrix} \right) \cdot \frac{1}{2} = \left( \begin{matrix} a & b \\ \bar{b} & \bar{a} \end{matrix} \right) $$

  • $\begingroup$ Although I did not full understand everything (I only recently learned tensor product ), at least I could capture the idea behind robjohn solution (or your change of basis). I know it works. The fact that works is not a coincidence. Is the fact that the complexification of reals is possible behind it. $\endgroup$ Aug 28, 2013 at 5:34

Funny problem. There is a number of ways it can be treated, starting from elementary (working with real & imaginary parts) & proceeding to more advanced (including some more geometric approaches).

Your questions 1 & 2 read pretty much the same to me, & the only sensible answer I can provide is 'because working w/ z & $\bar{z}$ concurrently allows us to use Linear Algebra' - without getting messy, at least, which is something that the solution using real & imaginary parts cannot claim. That aside, it's clearly a trick (but then again, what isn't?).

I understand your question about the determinant condition working out great in $\mathbb{C}^2$ but possibly being an overkill here, & it's a valid question. I propose to explain why the condition is in fact both sufficient & necessary using subspaces. I'll warn you that this boils down to exactly working w/ real & imaginary parts, albeit after having recast your problem in the $2\times2$ matrix form you give. From a certain POV, that's keeping the worst of both worlds: passing to that form is unintuitive & working with real/imaginary parts is cumbersome. Nevertheless, I find it far more intuitive than either solution, for reasons that might become clear once I stop running my mouth & do some math.

Let's consider any two complex numbers $z = z_R + {\rm i} z_I$ & $z' = z'_R + {\rm i} z'_I$. I embed (z,z') into $\mathbb{R}^4$ by writing $(z_R,z_I,z'_R,z'_I)$ for $(z,z')$ - this is a homomorphism between $\mathbb{C}^2$ and $\mathbb{R}^4$. Then, the pair $(z,\bar{z})$ becomes $(z_R,z_I,z_R,-z_I) = z_R(1,0,1,0) + z_I(0,1,0,-1) = z_R {\rm e_1} + z_I {\rm e_i}$, with ${\rm e_{1,i}}$ the obvious vectors in $\mathbb{R}^4$. Hence, all possible pairs $(z,\bar{z})$ constitute a $2-$D subspace $\mathbb{E}$ of $\mathbb{R}^4$, spanned by ${\rm e_{1,i}}$.

Write, now, $T : \mathbb{C}^2 \to \mathbb{C}^2$ for the linear operator represented by the matrix $\left(\begin{array}{cc}a&b\\\bar{b}&\bar{a}\end{array}\right)$. Then, the restriction $\left.T\,\right\vert_\mathbb{E} : \mathbb{E} \to \mathbb{E}$ is well defined, that is, $T$'s range falls indeed within $\mathbb{E}$. Indeed, we have (easy calculation)

$$ T {\rm e_1} = (a_R+b_R){\rm e_1} + (a_I+b_I){\rm e_i} , \\ T {\rm e_i} = (b_I-a_I){\rm e_1} + (a_R-b_R){\rm e_i} . $$

Equipping $\mathbb{E}$ with the basis ${\rm e_{1,i}}$ and expressing $T$ in that basis, then, we obtain the $2\times2$ matrix

$$ \left(\begin{array}{cc} a_R+b_R & b_I-a_I \\ a_I+b_I & a_R-b_R \end{array}\right) , $$

which is precisely what you'd obtain had you chosen to recast the original equation in terms of real & imaginary parts. (The reason is obvious, I believe.) Its determinant is, naturally, $|a|^2-|b|^2$.

Note, finally, that the equation for $(z,\bar{z})$ you report has a right-hand side that also falls in $\mathbb{E}$ (also for obvious intuitive reasons). Hence, the equation is solvable as long as $\left.T\,\right\vert_\mathbb{E} : \mathbb{E} \to \mathbb{E}$ is a bijection, i.e., as long as the aforementioned determinant is nonzero. & that's why the determinant condition is also necessary, not just sufficient.

  • $\begingroup$ Thanks, your explanation for (3) was the best! $\endgroup$ Aug 28, 2013 at 5:16
  • $\begingroup$ Thank you too, that problem really threw me for a while. Good luck w/ the rest of your studies! $\endgroup$ Aug 28, 2013 at 6:55

Here is an answer to the original question that sweeps linear algebra under the rug. The only assumption is that $a$ and $b$ are not both equal to $0$ (when it is clear that the cardinality of the solution set depends on $c$ being $0$.)

If there are two solutions $z_1$ and $z_2$, then using these in the equation and subtracting equations leads to

$$a(z_1-z_2) = -b(\overline{z_1-z_2})$$

Comparing absolute values, $|a|=|b|$.

Conversely if $|a|=|b|$, then after rescaling the equation by the right complex number, we can have


which we can view as the equation


where $X=e^{it}x$. Apparently $c'$ is real, and at that point there are many apparent solutions: $X=\frac{c'}{2}+is$ for any $s\in\mathbb{R}$.

So aside from when $a=b=0$, there are multiple solutions if and only if $|a|=|b|$. This is equivalent to "there is one solution or zero solutions if and only if $|a|\neq|b|$". But zero solutions is not an option, especially not when $|a|\neq|b|$. We can demonstrate one: $\frac{-\bar{a}c+b\bar{c}}{|a|^2-|b|^2}$

  • $\begingroup$ Why cardinality of solution depends of $c$ being $0$ or not? I think dimension of solution space depends of dimension of the kernel associated to the real linear system (which is independent of the value of $c$. The only difference is that if $c$ is $0$ the solution space is exactly the kernel) $\endgroup$ Aug 28, 2013 at 5:19
  • $\begingroup$ @Gaston I think you are mis-reading my assertion that you see up there in parentheses. I'm only sweeping aside the situation when $a=b=0$, and pointing out how trivial that situation would be. In the case where $a$ and $b$ both equal $0$, then the cardinality of the solution set is $0$ if and only if $c\neq0$, and the cardinality is infinite if and only if $c=0$. $\endgroup$
    – 2'5 9'2
    Aug 29, 2013 at 0:51

With the answers of all, I can see the answer for (2) clearly:

For solving $p(z,\overline{z})=0$ one needs to solve $\operatorname{Re}p(z,\overline{z})=0$ and $\operatorname{Im}p(z,\overline{z})=0$

In order to do this in linear algebra with $(z,\overline{z})$ instead of solving for $(\operatorname{Re}(z),\operatorname{Im}(z))$, just add $\overline{p(z,\overline{z})}=0$. Since $\operatorname{Re}p(z,\overline{z})=\frac{1}{2}(p(z,\overline{z})+\overline{p(z,\overline{z}))}$ and $\operatorname{Im}p(z,\overline{z})=\frac{1}{2i}(p(z,\overline{z})-\overline{p(z,\overline{z})})$ and $w,\overline{w}$ is a basis equivalent to real and imaginary part basis.


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