(6+i)(2-i)/(4+3i)(1-2i)

asked by guest
on Sep 20, 2024 at 12:07 pm



You asked:

Evaluate the expression: \(\frac{\left(6 + i\right) \cdot \left(2 - i\right)}{\left(4 + 3 i\right) \cdot \left(1 - 2 i\right)}\)

MathBot Answer:

Evaluated



\(\displaystyle \frac{\left(6 + i\right) \cdot \left(2 - i\right)}{\left(4 + 3 i\right) \cdot \left(1 - 2 i\right)} = \frac{\left(1 + 2 i\right) \left(2 - i\right) \left(4 - 3 i\right) \left(6 + i\right)}{125} \)

Expanded

\[\frac{\left(6 + i\right) \cdot \left(2 - i\right)}{\left(4 + 3 i\right) \cdot \left(1 - 2 i\right)} = \frac{6}{5} + \frac{i}{5}\]

Factored

\[\frac{\left(6 + i\right) \cdot \left(2 - i\right)}{\left(4 + 3 i\right) \cdot \left(1 - 2 i\right)} = \left(\frac{1}{125} + \frac{2 i}{125}\right) \left(2 - i\right) \left(4 - 3 i\right) \left(6 + i\right)\]

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

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