Question

Find the conjugate of $\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}.$

Answer 1

Step 1 : Given that
$\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}.$

Step 2: Multiplication of two complex numbers
$\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}=\frac{6+9i-4i-6(i{)}^2}{2-i+4i-2(i{)}^2}.$

$=\frac{6+9i-4i+6}{2-i+4i+2}$

$=\frac{12+5i}{4+3i}$

Step 3: Do rationalisation
$\frac{12+5i}{4+3i}=\frac{12+5i}{4+3i}\times \frac{4-3i}{4-3i}$

$=\frac{48-36i+20i+15}{(4{)}^2+(3{)}^2}$

$=\frac{63-16i}{25}$

$=\frac{63}{25}$-$\frac{16}{25}$i

Step 4: Write conjugate.
Conjugate of $\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}=\frac{63}{25}+\frac{16}{25}i$