1. **Problem:** Factorize the quadratic equation $$z^2 + (4 - 2i)z + (5 - 4i) = 0$$ by completing the square. 2. **Formula and rules:** To complete the square for a quadratic equation $$z^2 + bz + c = 0$$, rewrite it as $$\left(z + \frac{b}{2}\right)^2 = \left(\frac{b}{2}\right)^2 - c$$. 3. **Step-by-step solution:** - Given equation: $$z^2 + (4 - 2i)z + (5 - 4i) = 0$$ - Identify $$b = 4 - 2i$$ and $$c = 5 - 4i$$. - Compute $$\frac{b}{2} = \frac{4 - 2i}{2} = 2 - i$$. - Rewrite equation as: $$\left(z + (2 - i)\right)^2 = (2 - i)^2 - (5 - 4i)$$ - Calculate $$ (2 - i)^2 $$: $$ (2 - i)^2 = 2^2 - 2 \times 2 \times i + i^2 = 4 - 4i + (-1) = 3 - 4i $$ - Substitute back: $$\left(z + 2 - i\right)^2 = (3 - 4i) - (5 - 4i) = 3 - 4i - 5 + 4i = -2$$ - So: $$\left(z + 2 - i\right)^2 = -2$$ - Taking square roots: $$z + 2 - i = \pm \sqrt{-2} = \pm i \sqrt{2}$$ - Therefore: $$z = -2 + i \pm i \sqrt{2}$$ 4. **Final factorization:** $$\left(z + 2 - i - i \sqrt{2}\right)\left(z + 2 - i + i \sqrt{2}\right) = 0$$ or equivalently: $$\left(z + 2 - i(1 + \sqrt{2})\right)\left(z + 2 - i(1 - \sqrt{2})\right) = 0$$ This completes the factorization by completing the square.