If x = frac{5}{1-2i},where i = sqrt{-1},then | vedclass

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(A) Given $x = \frac{5}{1-2i}$. Multiplying numerator and denominator by the conjugate $(1+2i)$:$x = \frac{5(1+2i)}{(1-2i)(1+2i)} = \frac{5(1+2i)}{1+4

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$7$

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(A) Given $x = \frac{5}{1-2i}$. Multiplying numerator and denominator by the conjugate $(1+2i)$:$x = \frac{5(1+2i)}{(1-2i)(1+2i)} = \frac{5(1+2i)}{1+4} = \frac{5(1+2i)}{5} = 1+2i$.Now,calculate $x^2$:$x^2 = (1+2i)^2 = 1^2 + (2i)^2 + 2(1)(2i) = 1 - 4 + 4i = -3 + 4i$.Now,calculate $x^3$:$x^3 = x^2 \cdot x = (-3+4i)(1+2i) = -3 - 6i + 4i + 8i^2 = -3 - 2i - 8 = -11 - 2i$.Substitute these values into the expression $x^3 + x^2 - x + 22$:$(-11 - 2i) + (-3 + 4i) - (1 + 2i) + 22$$= (-11 - 3 - 1 + 22) + (-2i + 4i - 2i)$$= 7 + 0i = 7$.

If $x = \frac{5}{1-2i}$,where $i = \sqrt{-1}$,then the value of $x^3 + x^2 - x + 22$ is

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