Simplify the following complex number expressions: a) (5+j4)(3+j7)(2-j3) b) ((2-j3)(3+j2))/((4-j3))

$$(5+j4)(3+j7) = 5(3) + 5(j7) + j4(3) + j4(j7)$$ $$= 15 + j35 + j12 + j^2 28$$

Since $j^2 = -1$:

$$= 15 + j47 - 28$$ $$= -13 + j47$$

$$(-13+j47)(2-j3) = -13(2) -13(-j3) + j47(2) + j47(-j3)$$ $$= -26 + j39 + j94 - j^2 141$$

Since $j^2 = -1$:

$$= -26 + j133 + 141$$ $$= 115 + j133$$

The simplified expression is $\boxed{\text{115 + j133}}$.

$$(2-j3)(3+j2) = 2(3) + 2(j2) - j3(3) - j3(j2)$$ $$= 6 + j4 - j9 - j^2 6$$

Since $j^2 = -1$:

$$= 6 - j5 + 6$$ $$= 12 - j5$$

To do this, multiply the numerator and denominator by the conjugate of the denominator $(4+j3)$.

$$\frac{12-j5}{4-j3} = \frac{(12-j5)(4+j3)}{(4-j3)(4+j3)}$$

$$(12-j5)(4+j3) = 12(4) + 12(j3) - j5(4) - j5(j3)$$ $$= 48 + j36 - j20 - j^2 15$$

Since $j^2 = -1$:

$$= 48 + j16 + 15$$ $$= 63 + j16$$

$$(4-j3)(4+j3) = 4^2 - (j3)^2$$ $$= 16 - j^2 9$$

Since $j^2 = -1$:

$$= 16 - (-9)$$ $$= 16 + 9$$ $$= 25$$

$$\frac{63+j16}{25} = \frac{63}{25} + j\frac{16}{25}$$

The simplified expression is $\boxed{\frac{63}{25} + j\frac{16}{25}}$.

$$\cos 3x + j\sin 3x = e^{j3x}$$ $$\cos x + j\sin x = e^{jx}$$

$$e^{j2x} = \cos 2x + j\sin 2x$$

The simplified expression is $\boxed{\cos 2x + j\sin 2x}$.