We are asked to solve two complex number problems: (e) Simplify $\frac{1}{(2+i)(\sqrt{3}-2i)}$ (f) Simplify $5-4i + \frac{5}{3-4i}$

1. Problem Description

We are asked to solve two complex number problems:

(e) Simplify $\frac{1}{(2+i)(\sqrt{3}-2i)}$

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

2. Solution Steps

(e) Simplify $\frac{1}{(2+i)(\sqrt{3}-2i)}$

First, expand the denominator:

$(2+i)(\sqrt{3}-2i) = 2\sqrt{3} - 4i + i\sqrt{3} - 2i^2 = 2\sqrt{3} - 4i + i\sqrt{3} + 2 = (2\sqrt{3}+2) + i(\sqrt{3}-4)$

Then, we have:

$\frac{1}{(2\sqrt{3}+2) + i(\sqrt{3}-4)}$

To rationalize the denominator, multiply both numerator and denominator by the conjugate of the denominator, which is $(2\sqrt{3}+2) - i(\sqrt{3}-4)$.

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

The denominator becomes:

$(2\sqrt{3}+2)^2 + (\sqrt{3}-4)^2 = (4(3) + 8\sqrt{3} + 4) + (3 - 8\sqrt{3} + 16) = (12 + 8\sqrt{3} + 4) + (19 - 8\sqrt{3}) = 16 + 19 + 8\sqrt{3} - 8\sqrt{3} = 35$

So the expression becomes:

$\frac{(2\sqrt{3}+2) - i(\sqrt{3}-4)}{35} = \frac{2\sqrt{3}+2}{35} + i\frac{4-\sqrt{3}}{35}$

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

To rationalize the denominator of the fraction, multiply both numerator and denominator by the conjugate of the denominator, which is $3+4i$.

$\frac{5}{3-4i} \cdot \frac{3+4i}{3+4i} = \frac{5(3+4i)}{(3-4i)(3+4i)} = \frac{15+20i}{3^2 + 4^2} = \frac{15+20i}{9+16} = \frac{15+20i}{25} = \frac{15}{25} + \frac{20}{25}i = \frac{3}{5} + \frac{4}{5}i$

So the expression becomes:

$5-4i + \frac{3}{5} + \frac{4}{5}i = 5+\frac{3}{5} - 4i + \frac{4}{5}i = \frac{25}{5}+\frac{3}{5} - \frac{20}{5}i + \frac{4}{5}i = \frac{28}{5} - \frac{16}{5}i$

3. Final Answer

(e) $\frac{2\sqrt{3}+2}{35} + i\frac{4-\sqrt{3}}{35}$

(f) $\frac{28}{5} - \frac{16}{5}i$