If (2x - y + 1) + i(x - 2y - 1) = 2 - 3i,then | vedclass

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(D) Given $(2x - y + 1) + i(x - 2y - 1) = 2 - 3i$. Comparing the real and imaginary parts,we get: $2x - y + 1 = 2 \implies 2x - y = 1$ (Equation $1$)

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$\frac{12}{41} + \frac{15}{41}i$

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(D) Given $(2x - y + 1) + i(x - 2y - 1) = 2 - 3i$. Comparing the real and imaginary parts,we get: $2x - y + 1 = 2 \implies 2x - y = 1$ (Equation $1$) $x - 2y - 1 = -3 \implies x - 2y = -2$ (Equation $2$) Multiplying Equation $2$ by $2$,we get $2x - 4y = -4$ (Equation $3$). Subtracting Equation $3$ from Equation $1$: $(2x - y) - (2x - 4y) = 1 - (-4) 3y = 5 \implies y = \frac{5}{3}$. Substituting $y = \frac{5}{3}$ in Equation $1$: $2x - \frac{5}{3} = 1 2x = 1 + \frac{5}{3} = \frac{8}{3} x = \frac{4}{3}$. We need the multiplicative inverse of $(x - iy) = (\frac{4}{3} - i\frac{5}{3})$. The inverse is $\frac{1}{\frac{4}{3} - i\frac{5}{3}} = \frac{3}{4 - 5i}$. Rationalizing the denominator: $\frac{3(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{12 + 15i}{16 + 25} = \frac{12 + 15i}{41} = \frac{12}{41} + \frac{15}{41}i$.

If $(2x - y + 1) + i(x - 2y - 1) = 2 - 3i$,then the multiplicative inverse of $(x - iy)$ is

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