If 1 - i is a root of the equation {x^2} - ax | vedclass

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(D) Since the coefficients of the quadratic equation ${x^2} - ax + b = 0$ are real, the complex roots must occur in conjugate pairs.Given that $1 - i$

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$2$

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(D) Since the coefficients of the quadratic equation ${x^2} - ax + b = 0$ are real, the complex roots must occur in conjugate pairs.Given that $1 - i$ is a root, its conjugate $1 + i$ must also be a root of the equation.For a quadratic equation ${x^2} - (	ext{sum of roots})x + (	ext{product of roots}) = 0$, we have:Sum of roots $= (1 - i) + (1 + i) = 2$.Product of roots $= (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2$.Comparing this with ${x^2} - ax + b = 0$, we get $a = 2$ and $b = 2$.

If $1 - i$ is a root of the equation ${x^2} - ax + b = 0$, then $b = $

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