\frac{\left(3-6i\right)\cdot \left(-\frac{1}{3}i+1\right)\:}{-2+i\:}\cdot \frac{2-i}{2-i}

asked by guest
on Nov 16, 2024 at 8:26 am



You asked:

Evaluate the expression: \(\frac{\left(3 - 6 i\right) \cdot \left(-\left( \frac{1}{3} \cdot i \right) + 1\right)}{-2 + i} \cdot \frac{2 - i}{2 - i}\)

MathBot Answer:

Evaluated



\(\displaystyle \frac{\left(3 - 6 i\right) \cdot \left(-\left( \frac{1}{3} \cdot i \right) + 1\right)}{-2 + i} \cdot \frac{2 - i}{2 - i} = \frac{\left(-2 - i\right) \left(1 - \frac{i}{3}\right) \left(2 - i\right) \left(2 + i\right) \left(3 - 6 i\right)}{25} \)

Expanded

\[\frac{\left(3 - 6 i\right) \cdot \left(-\left( \frac{1}{3} \cdot i \right) + 1\right)}{-2 + i} \cdot \frac{2 - i}{2 - i} = - \frac{9}{5} + \frac{13 i}{5}\]

Factored

\[\frac{\left(3 - 6 i\right) \cdot \left(-\left( \frac{1}{3} \cdot i \right) + 1\right)}{-2 + i} \cdot \frac{2 - i}{2 - i} = \left(-2 - i\right) \left(\frac{1}{25} - \frac{2 i}{25}\right) \left(2 - i\right) \left(2 + i\right) \left(3 - i\right)\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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