Solve the system z_1+bar z_2=1+2i , 2+iz_1+(z_2)/(i)=0…

Solve the system $z_1+\bar z_2=1+2i$, $\ 2+iz_1+\dfrac{z_2}{i}=0$, given that $\text{Im}(z_2)=-\dfrac{6+\sqrt6}{2}$.

Step 1 · Simplify the second equation

Since $\dfrac1i=-i$:

$$ 2+iz_1-iz_2=0 \quad\Longrightarrow\quad i(z_1-z_2)=-2 \quad\Longrightarrow\quad z_1-z_2 = \frac{-2}{i} = 2i $$

Step 2 · The reduced system

$$ \begin{cases} z_1+\bar z_2 = 1+2i \\ z_1-z_2 = 2i \end{cases} $$

Step 3 · Eliminate $z_1$

From the second equation, $z_1=z_2+2i$. Substitute into the first:

$$ (z_2+2i)+\bar z_2 = 1+2i \quad\Longrightarrow\quad z_2+\bar z_2 = 1 $$

Step 4 · Extract $\text{Re}(z_2)$

With $z_2=x+iy$: $\ z_2+\bar z_2=2x=1 \implies x=\dfrac12$.

Step 5 · Apply the given constraint

$$ y = \text{Im}(z_2) = -\frac{6+\sqrt6}{2} $$

$$ \boxed{z_2 = \frac12 - \frac{6+\sqrt6}{2}i = \frac{1-i(6+\sqrt6)}{2}} $$

Step 6 · Find $z_1$

$$ z_1 = z_2+2i = \frac12 + i\left(2-\frac{6+\sqrt6}{2}\right) = \frac12 + i\cdot\frac{4-6-\sqrt6}{2} = \frac12 - \frac{2+\sqrt6}{2}i $$

$$ \boxed{z_1 = \frac{1-i(2+\sqrt6)}{2}} $$

Step 7 · Verify the first equation

$$ \bar z_2 = \frac12+\frac{6+\sqrt6}{2}i \quad\Longrightarrow\quad z_1+\bar z_2 = 1 + i\left(\frac{6+\sqrt6-2-\sqrt6}{2}\right) = 1+i\cdot2 = 1+2i \quad \checkmark $$

Step 8 · Verify the second equation

$$ iz_1 = \frac{i-i^2(2+\sqrt6)}{2} = \frac{(2+\sqrt6)+i}{2}, \qquad \frac{z_2}{i}=-iz_2 = \frac{-i-(-i^2)(6+\sqrt6)}{2}\cdot(-1)\ \Rightarrow\ -iz_2 = -\frac{(6+\sqrt6)+i}{2} $$

$$ 2+iz_1+\frac{z_2}{i} = 2+\frac{(2+\sqrt6)+i}{2}-\frac{(6+\sqrt6)+i}{2} = 2+\frac{2+\sqrt6-6-\sqrt6}{2} = 2+\frac{-4}{2} = 0 \quad \checkmark $$

Step 9 · Numerical values

With $\sqrt6\approx2.449$:

$$ z_1 \approx 0.5-2.225i, \qquad z_2\approx0.5-4.225i $$