Solution to equations of the form $bz + d\bar{z} = t$ where $b, d, t \in \mathbb{C}$

Question

Q: Prove that for all complex numbers $b$, $d$ and $t$ in $\mathbb{C}$, the complex equation $bz + d\bar{z} = t$ represents either the whole plane, or a straight line, or a single point, or the empty set.

My attempt:

Case 1: $ b\bar{b} \neq d\bar{d}$

The matrix$\begin{bmatrix} b & d \\ \bar{d} & \bar{b} \\ \end{bmatrix}$ is invertible and hence the system $\begin{bmatrix} b & d \\ \bar{d} & \bar{b} \\ \end{bmatrix}\begin{bmatrix} z \\ \bar{z}\end{bmatrix} = \begin{bmatrix} t \\ \bar{t}\end{bmatrix}$ has a unique solution for each $t$ in $\mathbb{C}$ and hence we have $$ z = \frac{\bar{b}t - d\bar{t}}{b\bar{b} - d\bar{d}} $$ as the only solution.

Case 2: $b\bar{b} = d\bar{d}$

Case 2.a: $t = 0$

If $t = 0$ and $b = 0$ then the equation is satisfied by all $z \in \mathbb{C}$.

If $ b \neq 0$ then we have the following sub-sub cases

Case 2.a.i: $b = \bar{d}$

We have that $bz+ \bar{bz} = 0$ and so $z = (0,k), k \in \mathbb{R}$

Case 2.a.ii: $b \neq \bar{d}$

We have that $z = t(\bar{b}-d)$

Case 2.b: $t \neq 0$

Note that if $z$ is a solution for $bz+ d\bar{z} = t$ and $w$ is a solution for $bw+ d\bar{w} = 0$, $z + w$ is a solution for $bz+ d\bar{z} = t$.

Case 2.b.1: $b = \bar{d}$ If t is not purely real, no $z \in \mathbb{C}$ satisfies the equation, and if t is purely real, $$ z = \biggl( \frac{t}{b\bar{b}}, k \biggr) \quad\text{for } k \in \mathbb{R} $$

I'm having trouble figuring out the general equation in $t$ and seeing the bigger picture of all of this. The question can be found in pg. 261, exercise 11, Complex Analysis in One Variable by Raghavan Narasimhan.

Answer 1

If we write $z = z_0 + iz_1$, etc. for each of the complex numbers $b$, $d$, and $t$, then the equation $bz + d\bar{z} = t$ is equivalent to the linear system over the reals: $$ \left\{ \begin{aligned} (b_0 + d_0)\, z_0 - (b_1 - d_1)\, z_1 &= t_0 \\ (b_1 + d_1)\, z_0 + (b_0 - d_0)\, z_1 &= t_1 \end{aligned} \right. $$

As such, its solution set is either empty or a subspace of dimension $0$, $1$, or $2$, i.e. a point, line, or plane. The various cases can be distinguished in the usual way by analyzing the augmented matrix that represents the system: $$ \left[ \begin{array}{cc|c} (b_0 + d_0) & -(b_1 - d_1) \; & \; t_0 \\ (b_1 + d_1) & \phantom{-}(b_0 - d_0) \; & \; t_1 \end{array} \right] $$