Let's solve these one by one.

(a) Simplify $(5 + j4)(3 + j7)(2 - j3)$

We will simplify this expression in two stages. First, we'll multiply the first two complex numbers, and then multiply the result by the third.

1. Multiply $(5 + j4)$ and $(3 + j7)$: $(5 + j4)(3 + j7) = 5(3) + 5(j7) + j4(3) + j4(j7)$ Simplifying each term: $= 15 + j35 + j12 + j^2(28)$ Recall that $j^2 = -1$, so: $= 15 + j35 + j12 - 28$ Combine the real and imaginary parts: $= (15 - 28) + j(35 + 12) = -13 + j47$

2. Now multiply $(-13 + j47)$ by $(2 - j3)$: $(-13 + j47)(2 - j3) = (-13)(2) + (-13)(-j3) + (j47)(2) + (j47)(-j3)$ Simplifying each term: $= -26 + j39 + j94 + j^2(-141)$ Again, using $j^2 = -1$: $= -26 + j39 + j94 + 141$ Combine the real and imaginary parts: $= (-26 + 141) + j(39 + 94) = 115 + j133$ So, the simplified result of part (a) is: $115 + j133$

(b) Simplify $\frac{(2 - j3)(3 + j2)}{(4 - j3)}$

1. Multiply the numerator $(2 - j3)$ and $(3 + j2)$: $(2 - j3)(3 + j2) = 2(3) + 2(j2) + (-j3)(3) + (-j3)(j2)$ Simplifying each term: $= 6 + j4 - j9 + j^2(-6)$ Using $j^2 = -1$: $= 6 + j4 - j9 + 6 = (6 + 6) + j(4 - 9) = 12 - j5$

2. Now, simplify the denominator $(4 - j3)$: To simplify the division, multiply the numerator and denominator by the conjugate of the denominator, $(4 + j3)$: $\frac{12 - j5}{4 - j3} \times \frac{4 + j3}{4 + j3} = \frac{(12 - j5)(4 + j3)}{(4 - j3)(4 + j3)}$

3. Simplify the denominator $(4 - j3)(4 + j3)$: This is a difference of squares: $(4 - j3)(4 + j3) = 4^2 - (j3)^2 = 16 - (-9) = 25$

4. Simplify the numerator $(12 - j5)(4 + j3)$: $(12 - j5)(4 + j3) = 12(4) + 12(j3) + (-j5)(4) + (-j5)(j3)$ Simplifying each term: $= 48 + j36 - j20 + j^2(-15)$ Again, using $j^2 = -1$: $= 48 + j36 - j20 + 15 = (48 + 15) + j(36 - 20) = 63 + j16$

5. Final simplification: $\frac{63 + j16}{25} = \frac{63}{25} + j\frac{16}{25}$ So the simplified result is: $2.52 + j0.64$

(c) Simplify $\frac{\cos 3x + j \sin 3x}{\cos x + j \sin x}$

This can be recognized as a quotient of complex numbers in polar form, which is represented as: $e^{j\theta} = \cos \theta + j \sin \theta$ So, the expression can be rewritten as: $\frac{e^{j3x}}{e^{jx}} = e^{j(3x - x)} = e^{j2x} = \cos 2x + j \sin 2x$ Thus, the simplified form is: $\cos 2x + j \sin 2x$

Let me know if you'd like more details on any part of this solution or if you have further questions.

Follow-up questions:

1. How do you multiply two complex numbers using the distributive property?
2. What is the significance of multiplying by the conjugate when simplifying complex fractions?
3. How does Euler's formula help simplify expressions involving trigonometric functions and imaginary numbers?
4. Can you demonstrate how to simplify another complex number division?
5. What is the importance of $j^2 = -1$ when simplifying complex expressions?

Tip:

When multiplying complex numbers, always remember to apply the distributive property carefully, and never forget that $j^2 = -1$, which helps in simplifying terms involving imaginary units.