Question

Find the conjugate of

$\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}$

Answer 1

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$\bold{\underline{Question-}}$
Find the conjugate of

$\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}$
$\bold{\underline{Answer-}}$

$\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)}$

$= \frac{3(2 + 3i) -2i (2 + 3i)}{1(2 - i) + 2i(2 - i)} \\ \\ \\ \\ = \frac{6 + 9i - 4i - {6i}^{2} }{2 - i + 4i - {2i}^{2} } \\ \\ \\ \\ = \frac{6 + 5i + 6}{2 + 3i + 2} \\ \\ \\ \\ = \frac{12 + 5i}{4 + 3i} \times \frac{4 - 3i}{4 - 3i} \\ \\ \\ \\ = \frac{12(4 - 3i) + 5i(4 - 3i)}{ {(4)}^{2} - {(3i)}^{2} } \\ \\ \\ \\ = \frac{48 - 36i + 20i - 15i {}^{2} }{16 - {9i}^{2} } \\ \\ \\ \\ = \frac{48 - 16i + 15}{16 + 9} \\ \\ \\ \\ = \frac{63 - 16i}{25} \\ \\ \\ \\ = \frac{63}{25} - \frac{16}{25} i$

Hence the conjugate of
$\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)} \: \: \: \: \: \: \: \: \: = \frac{63}{25} - \frac{16}{25} i$

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Answer 2

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Here is your answer

$\frac{(3 - 2i)(2 + 3i)}{(1 + 2i)(2 - i)} \\ \\ = \frac{6 + 9i - 4i - 6 {i}^{2} }{2 - i + 4i - 2 {i}^{2} } \\ \\ = \frac{6 + 5i - 6 {i}^{2} }{2 + 3i - 2 {i}^{2} }$

Now
$\frac{6 + 5i + 6}{2 + 3i + 2} \\ \\ = \frac{12 + 5i}{4 + 3i} \\ \\ = \frac{12 + 5i}{4 + 3i} \times \frac{4 - 3i}{4 - 3i} \\ \\ = \frac{48 - 36i + 20i - 15 {i}^{2} }{ {4}^{2} - (3i) {}^{2} } \\ \\ = \frac{48 - 16i + 15}{16 - 9 {i}^{2} } \\ \\ = \frac{63 - 16i}{16 + 9} \\ \\ = \frac{63 - 16i}{25} \\ = \frac{63}{25} - \frac{16i}{25}$


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