Answer to Question #168226 in Complex Analysis for Festus

Question

Express (6+i )(2-i )/(4+3i )(1-2i) in the form a + ib

Answer 1

$\frac{{(6 + i)(2 - i)}}{{\left( {4 + 3i} \right)\left( {1 - 2i} \right)}} = \frac{{12 + 2i - 6i - {i^2}}}{{4 + 3i - 8i - 6{i^2}}} = \frac{{12 + 1 - 4i}}{{4 + 6 - 5i}} = \frac{{13 - 4i}}{{10 - 5i}}=$

$= \frac{{(13 - 4i)(10 + 5i)}}{{(10 - 5i)(10 + 5i)}} = \frac{{130 - 40i + 65i - 20{i^2}}}{{100 + 25}} = \frac{{130 + 20}}{{125}} + \frac{{25}}{{125}}i = \frac{6}{5} + \frac{1}{5}i$

Answer: $\frac{6}{5} + \frac{1}{5}i$