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

$(a)\;\large\frac{63}{25}- \large\frac{16}{25}i\qquad(b)\;\large\frac{63}{25}- \large\frac{17}{25}i\qquad(c)\;\large\frac{16}{25}- \large\frac{63}{25}i\qquad(d)\;\large\frac{63}{25}+ \large\frac{16}{25}i$

Answer : $\;\large\frac{63}{25}+ \large\frac{16}{25}i$
Explanation :
We have , $\;\large\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$
$=\large\frac{6+9i-4i-6i^{2}}{2-i+4i-2i^{2}}$
$= \large\frac{12+5i}{4+3i}$
$\large\frac{12+5i}{4+3i}= \large\frac{12+5i}{4+3i} \times \large\frac{4-3i}{4-3i}$
$= \large\frac{48-36i+20i-15i^{2}}{4^{2}- (3i)^{2}}$
$= \large\frac{63-16i}{16+9}$
$=\large\frac{63}{25}- \large\frac{16}{25}i$
Therefore , conjugate of $\;\large\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$ is $\;\large\frac{63}{25}+ \large\frac{16}{25}i$