求解关于复数 $z$ 和 $w$ 的下列联立方程组：
$\begin{aligned} 2\mathrm{i}z^* - w &= 1 \\ z + w^* &= 4 \end{aligned}$

1. Make $w$ the subject of the first equation and find its complex conjugate.

$w = 2\mathrm{i}z^* - 1$
$w^* = (2\mathrm{i}z^* - 1)^* = -2\mathrm{i}z - 1$

2. Substitute $w^*$ into the second equation and isolate $z$.

$z + (-2\mathrm{i}z - 1) = 4$
$z(1 - 2\mathrm{i}) = 5$

3. Solve for $z$ using complex division.

$z = \frac{5}{1 - 2\mathrm{i}}$
$z = \frac{5(1 + 2\mathrm{i})}{(1 - 2\mathrm{i})(1 + 2\mathrm{i})}$
$z = \frac{5(1 + 2\mathrm{i})}{1^2 + 2^2}$
$z = \frac{5(1 + 2\mathrm{i})}{5}$
$z = 1 + 2\mathrm{i}$

4. Substitute the value of $z$ back to find $w$.

$w = 2\mathrm{i}(1 + 2\mathrm{i})^* - 1$
$w = 2\mathrm{i}(1 - 2\mathrm{i}) - 1$
$w = 2\mathrm{i} - 4\mathrm{i}^2 - 1$
$w = 2\mathrm{i} + 4 - 1$
$w = 3 + 2\mathrm{i}$