Find the conjugate of frac{(3-2 i)(2+3 i)}{(1 | vedclass

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Question

(A) Let $z = \frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.First,simplify the numerator: $(3-2 i)(2+3 i) = 6 + 9i - 4i - 6i^2 = 6 + 5i + 6 = 12 + 5i$.Next,simp

Vedclass · Accepted Answer

$\frac{63}{25}+\frac{16}{25} i$

Vedclass · Answer

(A) Let $z = \frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.First,simplify the numerator: $(3-2 i)(2+3 i) = 6 + 9i - 4i - 6i^2 = 6 + 5i + 6 = 12 + 5i$.Next,simplify the denominator: $(1+2 i)(2-i) = 2 - i + 4i - 2i^2 = 2 + 3i + 2 = 4 + 3i$.So,$z = \frac{12+5i}{4+3i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(4-3i)$:$z = \frac{(12+5i)(4-3i)}{(4+3i)(4-3i)} = \frac{48 - 36i + 20i - 15i^2}{16 - 9i^2} = \frac{48 - 16i + 15}{16 + 9} = \frac{63 - 16i}{25} = \frac{63}{25} - \frac{16}{25}i$.The conjugate of $z = a + bi$ is $\bar{z} = a - bi$.Therefore,the conjugate of $\frac{63}{25} - \frac{16}{25}i$ is $\frac{63}{25} + \frac{16}{25}i$.

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

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