Find the real numbers x and y if (x-iy)(3+5i) | vedclass

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(A) Let $z = (x-iy)(3+5i)$.$z = 3x + 5xi - 3yi - 5yi^2 = 3x + 5xi - 3yi + 5y = (3x+5y) + i(5x-3y)$.Therefore,the conjugate $\bar{z} = (3x+5y) - i(5x-3

Vedclass · Accepted Answer

$x=3, y=-3$

Vedclass · Answer

(A) Let $z = (x-iy)(3+5i)$.$z = 3x + 5xi - 3yi - 5yi^2 = 3x + 5xi - 3yi + 5y = (3x+5y) + i(5x-3y)$.Therefore,the conjugate $\bar{z} = (3x+5y) - i(5x-3y)$.It is given that $\bar{z} = -6 - 24i$.Equating real and imaginary parts,we get:$3x + 5y = -6$ $(i)$$5x - 3y = 24$ $(ii)$Multiplying $(i)$ by $3$ and $(ii)$ by $5$,we get:$9x + 15y = -18$$25x - 15y = 120$Adding these equations: $34x = 102$,so $x = 3$.Substituting $x=3$ into $(i)$: $3(3) + 5y = -6$ $\Rightarrow 9 + 5y = -6$ $\Rightarrow 5y = -15$ $\Rightarrow y = -3$.Thus,$x=3$ and $y=-3$.

Find the real numbers $x$ and $y$ if $(x-iy)(3+5i)$ is the conjugate of $-6-24i$.

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