The problem requires us to simplify several complex number expressions, find the quotient of complex numbers in standard form, and find the absolute value and square roots of complex numbers.

1. Problem Description

The problem requires us to simplify several complex number expressions, find the quotient of complex numbers in standard form, and find the absolute value and square roots of complex numbers.

2. Solution Steps

1a. $(5+2i)(8+6i)$

Multiply the two complex numbers using the distributive property (FOIL method):

$(5)(8) + (5)(6i) + (2i)(8) + (2i)(6i) = 40 + 30i + 16i + 12i^2$

Since $i^2 = -1$:

$40 + 46i + 12(-1) = 40 + 46i - 12 = 28 + 46i$

1b. $(\frac{1}{3} + \frac{2}{5}i) + (\frac{1}{2} + \frac{1}{4}i)$

Add the real and imaginary parts separately:

$(\frac{1}{3} + \frac{1}{2}) + (\frac{2}{5} + \frac{1}{4})i = (\frac{2}{6} + \frac{3}{6}) + (\frac{8}{20} + \frac{5}{20})i = \frac{5}{6} + \frac{13}{20}i$

1c. $(-7-3i) + (-4+4i)$

Add the real and imaginary parts separately:

$(-7-4) + (-3+4)i = -11 + i$

1d. $(4+i\sqrt{3}) + (-6-2i\sqrt{3})$

Add the real and imaginary parts separately:

$(4-6) + (\sqrt{3} - 2\sqrt{3})i = -2 - i\sqrt{3}$

1e. $(6-7i) - (7-6i)$

Subtract the real and imaginary parts separately:

$(6-7) + (-7-(-6))i = -1 + (-7+6)i = -1 - i$

1f. $(-1-2i)^2$

Square the complex number:

$(-1-2i)(-1-2i) = (-1)(-1) + (-1)(-2i) + (-2i)(-1) + (-2i)(-2i) = 1 + 2i + 2i + 4i^2 = 1 + 4i + 4(-1) = 1 + 4i - 4 = -3 + 4i$

1g. $(-5i)(8i)$

Multiply the complex numbers:

$(-5)(8)(i)(i) = -40i^2 = -40(-1) = 40$

2a. $\frac{4i}{3-2i}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $3+2i$:

$\frac{4i(3+2i)}{(3-2i)(3+2i)} = \frac{12i + 8i^2}{9 - 4i^2} = \frac{12i - 8}{9 + 4} = \frac{-8 + 12i}{13} = -\frac{8}{13} + \frac{12}{13}i$

2b. $\frac{2+5i}{3+7i}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $3-7i$:

$\frac{(2+5i)(3-7i)}{(3+7i)(3-7i)} = \frac{6 - 14i + 15i - 35i^2}{9 - 49i^2} = \frac{6 + i + 35}{9 + 49} = \frac{41 + i}{58} = \frac{41}{58} + \frac{1}{58}i$

2c. $\frac{4-10i}{-3+7i}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $-3-7i$:

$\frac{(4-10i)(-3-7i)}{(-3+7i)(-3-7i)} = \frac{-12 - 28i + 30i + 70i^2}{9 - 49i^2} = \frac{-12 + 2i - 70}{9 + 49} = \frac{-82 + 2i}{58} = \frac{-41+i}{29} = -\frac{41}{29} + \frac{1}{29}i$

2d. $\frac{-1-i}{-2-3i}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $-2+3i$:

$\frac{(-1-i)(-2+3i)}{(-2-3i)(-2+3i)} = \frac{2 - 3i + 2i - 3i^2}{4 - 9i^2} = \frac{2 - i + 3}{4 + 9} = \frac{5-i}{13} = \frac{5}{13} - \frac{1}{13}i$

2e. $\frac{3+9i}{4-i}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $4+i$:

$\frac{(3+9i)(4+i)}{(4-i)(4+i)} = \frac{12 + 3i + 36i + 9i^2}{16 - i^2} = \frac{12 + 39i - 9}{16 + 1} = \frac{3 + 39i}{17} = \frac{3}{17} + \frac{39}{17}i$

2f. $\frac{-4+9i}{-3-6i}$

Multiply the numerator and denominator by the conjugate of the denominator, which is $-3+6i$:

$\frac{(-4+9i)(-3+6i)}{(-3-6i)(-3+6i)} = \frac{12 - 24i - 27i + 54i^2}{9 - 36i^2} = \frac{12 - 51i - 54}{9 + 36} = \frac{-42 - 51i}{45} = \frac{-14 - 17i}{15} = -\frac{14}{15} - \frac{17}{15}i$

3a. $3+2i$

Absolute value: $|3+2i| = \sqrt{3^2 + 2^2} = \sqrt{9+4} = \sqrt{13}$

To find the square root, let $\sqrt{3+2i} = a+bi$. Squaring both sides:

$3+2i = (a+bi)^2 = a^2 + 2abi - b^2$.

Equating real and imaginary parts: $a^2 - b^2 = 3$ and $2ab = 2$, so $ab=1$. $b = 1/a$.

$a^2 - \frac{1}{a^2} = 3$. $a^4 - 1 = 3a^2$. $a^4 - 3a^2 - 1 = 0$.

$a^2 = \frac{3 \pm \sqrt{9+4}}{2} = \frac{3 \pm \sqrt{13}}{2}$. Since $a$ is real, $a^2 = \frac{3+\sqrt{13}}{2}$.

$a = \pm \sqrt{\frac{3+\sqrt{13}}{2}}$.

If $a = \sqrt{\frac{3+\sqrt{13}}{2}}$, then $b = \frac{1}{a} = \sqrt{\frac{2}{3+\sqrt{13}}} = \sqrt{\frac{2(3-\sqrt{13})}{9-13}} = \sqrt{\frac{2(3-\sqrt{13})}{-4}} = -\sqrt{\frac{-3+\sqrt{13}}{2}} = \sqrt{\frac{\sqrt{13}-3}{2}}$. Since $ab=1$ they must be of the same sign, so $b$ should be positive as well.

Square roots are $\pm(\sqrt{\frac{3+\sqrt{13}}{2}} + i \sqrt{\frac{\sqrt{13}-3}{2}})$.

3b. $15+8i$

Absolute value: $|15+8i| = \sqrt{15^2 + 8^2} = \sqrt{225+64} = \sqrt{289} = 17$

To find the square root, let $\sqrt{15+8i} = a+bi$. Squaring both sides:

$15+8i = (a+bi)^2 = a^2 + 2abi - b^2$.

Equating real and imaginary parts: $a^2 - b^2 = 15$ and $2ab = 8$, so $ab=4$. $b = 4/a$.

$a^2 - \frac{16}{a^2} = 15$. $a^4 - 16 = 15a^2$. $a^4 - 15a^2 - 16 = 0$.

$(a^2 - 16)(a^2+1) = 0$. So $a^2 = 16$ or $a^2 = -1$. Since a is real, $a^2 = 16$.

$a = \pm 4$.

If $a = 4$, then $b = 4/4 = 1$. If $a = -4$, then $b = 4/(-4) = -1$.

Square roots are $\pm(4+i)$.

3c. $6+2i$

Absolute value: $|6+2i| = \sqrt{6^2 + 2^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$

Let $\sqrt{6+2i} = a+bi$.

$6+2i = a^2-b^2 + 2abi$.

$a^2-b^2=6$ and $2ab=2$, so $ab=1$ and $b = 1/a$.

$a^2 - 1/a^2 = 6$. $a^4-1 = 6a^2$, so $a^4-6a^2-1=0$.

$a^2 = \frac{6 \pm \sqrt{36+4}}{2} = \frac{6 \pm \sqrt{40}}{2} = \frac{6 \pm 2\sqrt{10}}{2} = 3 \pm \sqrt{10}$.

Since $a$ is real, $a^2 = 3 + \sqrt{10}$. Thus $a = \pm \sqrt{3+\sqrt{10}}$.

Then $b = 1/a = \pm \sqrt{\frac{1}{3+\sqrt{10}}} = \pm \sqrt{\frac{3-\sqrt{10}}{9-10}} = \pm \sqrt{\sqrt{10}-3}$.

Square roots are $\pm(\sqrt{3+\sqrt{10}} + i \sqrt{\sqrt{10}-3})$.

3. Final Answer

1a. $28+46i$

1b. $\frac{5}{6} + \frac{13}{20}i$

1c. $-11+i$

1d. $-2 - i\sqrt{3}$

1e. $-1-i$

1f. $-3+4i$

1g. $40$

2a. $-\frac{8}{13} + \frac{12}{13}i$

2b. $\frac{41}{58} + \frac{1}{58}i$

2c. $-\frac{41}{29} + \frac{1}{29}i$

2d. $\frac{5}{13} - \frac{1}{13}i$

2e. $\frac{3}{17} + \frac{39}{17}i$

2f. $-\frac{14}{15} - \frac{17}{15}i$

3a. Absolute value: $\sqrt{13}$. Square roots: $\pm(\sqrt{\frac{3+\sqrt{13}}{2}} + i \sqrt{\frac{\sqrt{13}-3}{2}})$

3b. Absolute value: $17$. Square roots: $\pm(4+i)$

3c. Absolute value: $2\sqrt{10}$. Square roots: $\pm(\sqrt{3+\sqrt{10}} + i \sqrt{\sqrt{10}-3})$