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Question
If \[z_1 = 2 - i, z_2 = - 2 + i,\] find
Re \[\left( \frac{z_1 z_2}{z_1} \right)\]
Solution
\[ z_1 = 2 - i, z_2 = - 2 + i, z_1 = 2 + i\]
\[ \therefore \left( \frac{z_1 z_2}{z_1} \right) = \left( \frac{\left[ 2 - i \right]\left[ - 2 + i \right]}{2 + i} \right)\]
\[ = \left( \frac{- 4 + 2i + 2i - i^2}{2 + i} \right)\]
\[ = \left( \frac{- 3 + 4i}{2 + i} \right)\]
\[ = \left[ \frac{- 3 + 4i}{2 + i} \times \left( \frac{2 - i}{2 - i} \right) \right]\]
\[ = \left( \frac{- 6 + 3i + 8i - 4 i^2}{2^2 - i^2} \right)\]
\[ = \left( \frac{- 2 + 11i}{4 - \left( - 1 \right)} \right)\]
\[ = \left( \frac{- 2 + 11i}{5} \right)\]
\[Re\left( \frac{z_1 z_2}{z_1} \right) = \frac{- 2}{5}\]
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