Simplify: a) (5+4i)(3+7i)(2-3i) b) (2-3i)(3+2i)/(4-3i)

12. Simplify:

• a) $(5+4i)(3+7i)(2-3i)$

  Step 1: Multiply the first two complex numbers.

  $$(5+4i)(3+7i) = 5(3) + 5(7i) + 4i(3) + 4i(7i)$$ $$= 15 + 35i + 12i + 28i^2$$

  Since $i^2 = -1$:

  $$= 15 + 47i - 28$$ $$= -13 + 47i$$

  Step 2: Multiply the result by the third complex number.

  $$(-13+47i)(2-3i) = -13(2) - 13(-3i) + 47i(2) + 47i(-3i)$$ $$= -26 + 39i + 94i - 141i^2$$

  Since $i^2 = -1$:

  $$= -26 + 133i + 141$$ $$= 115 + 133i$$

  The simplified expression is $\boxed{115 + 133i}$.

• b) $\frac{(2-3i)(3+2i)}{(4-3i)}$

  Step 1: Multiply the complex numbers in the numerator.

  $$(2-3i)(3+2i) = 2(3) + 2(2i) - 3i(3) - 3i(2i)$$ $$= 6 + 4i - 9i - 6i^2$$

  Since $i^2 = -1$:

  $$= 6 - 5i + 6$$ $$= 12 - 5i$$

  Step 2: Divide the result by the denominator.

  $$\frac{12-5i}{4-3i}$$

  Multiply the numerator and denominator by the conjugate of the denominator, which is $(4+3i)$.

  $$\frac{(12-5i)(4+3i)}{(4-3i)(4+3i)}$$

  Step 3: Expand the numerator.

  $$(12-5i)(4+3i) = 12(4) + 12(3i) - 5i(4) - 5i(3i)$$ $$= 48 + 36i - 20i - 15i^2$$ $$= 48 + 16i + 15$$ $$= 63 + 16i$$

  Step 4: Expand the denominator.

  $$(4-3i)(4+3i) = 4^2 - (3i)^2$$ $$= 16 - 9i^2$$ $$= 16 + 9$$ $$= 25$$

  Step 5: Combine the numerator and denominator and express in $a+bi$ form.

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

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

• c) $\frac{\cos3x+i\sin3x}{\cos x+i\sin x}$

  Step 1: Use Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta$.

  $$\frac{\cos3x+i\sin3x}{\cos x+i\sin x} = \frac{e^{i3x}}{e^{ix}}$$

  Step 2: Apply the rules of exponents for division.

  $$e^{i3x - ix} = e^{i(3x-x)} = e^{i2x}$$

  Step 3: Convert back to trigonometric form using Euler's formula.

  $$e^{i2x} = \cos(2x) + i\sin(2x)$$

  The simplified expression is $\boxed{\cos(2x) + i\sin(2x)}$.

13. Express the following in the form $a+ib$.

• a) $\frac{2+3i}{i(4-5i)} + \frac{2}{i}$

  Step 1: Simplify the first term, $\frac{2+3i}{i(4-5i)}$.
  First, simplify the denominator: $i(4-5i) = 4i - 5i^2 = 4i - 5(-1) = 5+4i$.
  So the term becomes $\frac{2+3i}{5+4i}$.
  Multiply the numerator and denominator by the conjugate of the denominator, $(5-4i)$:

  $$\frac{(2+3i)(5-4i)}{(5+4i)(5-4i)} = \frac{2(5) + 2(-4i) + 3i(5) + 3i(-4i)}{5^2 - (4i)^2}$$ $$= \frac{10 - 8i + 15i - 12i^2}{25 - 16i^2} = \frac{10 + 7i + 12}{25 + 16} = \frac{22+7i}{41}$$ $$= \frac{22}{41} + \frac{7}{41}i$$

  Step 2: Simplify the second term, $\frac{2}{i}$.
  Multiply the numerator and denominator by $-i$:

  $$\frac{2}{i} \cdot \frac{-i}{-i} = \frac{-2i}{-i^2} = \frac{-2i}{-(-1)} = \frac{-2i}{1} = -2i$$

  Step 3: Add the simplified terms.

  $$\left(\frac{22}{41} + \frac{7}{41}i\right) + (-2i)$$ $$= \frac{22}{41} + \left(\frac{7}{41} - 2\right)i$$ $$= \frac{22}{41} + \left(\frac{7}{41} - \frac{82}{41}\right)i$$ $$= \frac{22}{41} - \frac{75}{41}i$$

  The expression in the form $a+ib$ is $\boxed{\frac{22}{41} - \frac{75}{41}i}$.

• b) $\frac{1}{2+3i} + \frac{1}{1-2i}$

  Step 1: Simplify the first term, $\frac{1}{2+3i}$.
  Multiply the numerator and denominator by the conjugate of the denominator, $(2-3i)$:

  $$\frac{1}{2+3i} \cdot \frac{2-3i}{2-3i} = \frac{2-3i}{2^2 - (3i)^2} = \frac{2-3i}{4 - 9i^2} = \frac{2-3i}{4+9} = \frac{2-3i}{13}$$ $$= \frac{2}{13} - \frac{3}{13}i$$

  Step 2: Simplify the second term, $\frac{1}{1-2i}$.
  Multiply the numerator and denominator by the conjugate of the denominator, $(1+2i)$:

  $$\frac{1}{1-2i} \cdot \frac{1+2i}{1+2i} = \frac{1+2i}{1^2 - (2i)^2} = \frac{1+2i}{1 - 4i^2} = \frac{1+2i}{1+4} = \frac{1+2i}{5}$$ $$= \frac{1}{5} + \frac{2}{5}i$$

  Step 3: Add the simplified terms.

  $$\left(\frac{2}{13} - \frac{3}{13}i\right) + \left(\frac{1}{5} + \frac{2}{5}i\right)$$

  Group the real and imaginary parts:

  $$$$