Solve the following system of linear equations with complex coefficients: (1−i)z + (1+i)w = 3, 2z − (2+5i)w = 2+3i.

Step 1: Write the given system of equations.

$$(1-i)z + (1+i)\omega = 3 \quad \text{(1)}$$ $$2z - (2+5i)\omega = 2+3i \quad \text{(2)}$$

Step 2: Eliminate $z$. Multiply Equation (1) by 2 and Equation (2) by $(1-i)$.
Multiply Equation (1) by 2:

$$2(1-i)z + 2(1+i)\omega = 2(3)$$ $$(2-2i)z + (2+2i)\omega = 6 \quad \text{(3)}$$

Multiply Equation (2) by $(1-i)$:

$$(1-i)(2z) - (1-i)(2+5i)\omega = (1-i)(2+3i)$$ $$(2-2i)z - (2+5i-2i-5i^2)\omega = (2+3i-2i-3i^2)$$ $$(2-2i)z - (2+3i+5)\omega = (2+i+3)$$ $$(2-2i)z - (7+3i)\omega = 5+i \quad \text{(4)}$$

Step 3: Subtract Equation (4) from Equation (3).

$$[(2-2i)z + (2+2i)\omega] - [(2-2i)z - (7+3i)\omega] = 6 - (5+i)$$ $$(2+2i)\omega + (7+3i)\omega = 6 - 5 - i$$ $$(2+2i+7+3i)\omega = 1 - i$$ $$(9+5i)\omega = 1 - i$$

Step 4: Solve for $\omega$.

$$\omega = \frac{1-i}{9+5i}$$

Multiply the numerator and denominator by the conjugate of the denominator, $9-5i$:

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