The conjugate of frac{(2 + i)^2}{3 + i} in th | vedclass

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Question

(C) Let $z = \frac{(2 + i)^2}{3 + i}$.First,expand the numerator: $(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$.So,$z = \frac{3 + 4i}{3 + i}$.To si

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$\frac{13}{10} + i\left(-\frac{9}{10}\right)$

Vedclass · Answer

(C) Let $z = \frac{(2 + i)^2}{3 + i}$.First,expand the numerator: $(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$.So,$z = \frac{3 + 4i}{3 + i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(3 - i)$:$z = \frac{(3 + 4i)(3 - i)}{(3 + i)(3 - i)} = \frac{9 - 3i + 12i - 4i^2}{3^2 + 1^2} = \frac{9 + 9i + 4}{9 + 1} = \frac{13 + 9i}{10} = \frac{13}{10} + i\frac{9}{10}$.The conjugate of $z = a + ib$ is $\bar{z} = a - ib$.Therefore,the conjugate is $\frac{13}{10} - i\frac{9}{10}$.

The conjugate of $\frac{(2 + i)^2}{3 + i}$ in the form of $a + ib$ is

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