Solving Equations Involving the Conjugate by Splitting z = x + iy

An equation in a complex unknown $z$ is a genuine polynomial in $z$ only when it is built from $z$ alone (so the Fundamental Theorem of Algebra applies and a degree-$n$ equation has exactly $n$ roots). The moment the conjugate $\bar z$, the parts $\mathrm{Re}(z)$ / $\mathrm{Im}(z)$, or the modulus $|z|$ appear, that count is off the table — those operations are not analytic, so you cannot reason by degree. Instead, treat $z$ as a point $(x, y)$ in $\mathbb{R}^2$ and convert the one complex equation into two real equations.

Why one complex equation = two real equations

Two complex numbers are equal exactly when their real parts match and their imaginary parts match:

$A + Bi = C + Di \iff A = C \ \text{and}\ B = D \quad (A,B,C,D \in \mathbb{R}).$

So a single equation $f(z) = w$ becomes a $2\times 2$ real system in the unknowns $x, y$. Each real solution pair $(x, y)$ is one root $z = x + iy$.

The method (4 steps)

1. Substitute $z = x + iy$ (with $x, y \in \mathbb{R}$) everywhere, using the building blocks below.
2. Separate the equation into its real part and its imaginary part, and set each side equal — giving two real equations.
3. Solve the system. The imaginary equation often factors ($PQ = 0$), splitting the work into a few branches; solve each branch in the real equation.
4. Collect & count all distinct $(x, y)$ pairs — that is the number of complex roots.

Substitution toolkit ($z = x + iy$)

Note $\mathrm{Im}(z) = y$ is the real number $y$, not $iy$. Watch where the stray $i$ lands.

Worked mini-example

Find all $z \in \mathbb{C}$ with $\;z^2 = \bar z$.

Put $z = x + iy$. Then $z^2 = (x^2 - y^2) + 2xy\,i$ and $\bar z = x - iy$. Match parts:

• Real: $\;x^2 - y^2 = x.$
• Imag: $\;2xy = -y \Rightarrow y(2x + 1) = 0 \Rightarrow y = 0 \ \text{or}\ x = -\tfrac12.$

Branch $y = 0$: real equation gives $x^2 = x \Rightarrow x = 0, 1$ — roots $z = 0,\ 1$.

Branch $x = -\tfrac12$: real equation gives $\tfrac14 - y^2 = -\tfrac12 \Rightarrow y^2 = \tfrac34 \Rightarrow y = \pm\tfrac{\sqrt 3}{2}$ — roots $z = -\tfrac12 \pm \tfrac{\sqrt 3}{2} i$.

That is 4 roots total (the four cube roots of unity, in fact, since $z^2 = \bar z \Rightarrow z^3 = |z|^2 = 1$ for $z \ne 0$). A naive "degree 2, so 2 roots" guess would be wrong — exactly the trap these problems set.

Exam trick: never count roots by degree once $\bar z$, $\mathrm{Re}$, $\mathrm{Im}$, or $|z|$ is present. Split into $x, y$, factor the imaginary equation to branch, and count every real $(x,y)$ pair. Don't forget to recombine branches — and drop any spurious pair that violates the other equation.