We need to find the absolute value and the square roots of the following complex numbers: a. $3 + 2i$ b. $15 + 8i$ c. $6 + 2i$

1. Problem Description

We need to find the absolute value and the square roots of the following complex numbers:

a. $3 + 2i$

b. $15 + 8i$

c. $6 + 2i$

2. Solution Steps

a. For $3 + 2i$:

Absolute value:

The absolute value of a complex number $a + bi$ is given by $\sqrt{a^2 + b^2}$.

So, the absolute value of $3 + 2i$ is $\sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}$.

Square roots:

Let $\sqrt{3 + 2i} = x + yi$, where $x$ and $y$ are real numbers.

Squaring both sides, we get $3 + 2i = (x + yi)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi$.

Equating the real and imaginary parts, we have:

$x^2 - y^2 = 3$

$2xy = 2$, so $xy = 1$ and $y = \frac{1}{x}$.

Substituting $y = \frac{1}{x}$ into $x^2 - y^2 = 3$, we get $x^2 - \frac{1}{x^2} = 3$.

Multiplying by $x^2$, we have $x^4 - 1 = 3x^2$, so $x^4 - 3x^2 - 1 = 0$.

Let $u = x^2$. Then $u^2 - 3u - 1 = 0$.

Using the quadratic formula, $u = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$.

Since $x$ is real, $x^2$ must be positive. So we take the positive root, $u = x^2 = \frac{3 + \sqrt{13}}{2}$.

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

If $x = \sqrt{\frac{3 + \sqrt{13}}{2}}$, then $y = \frac{1}{x} = \frac{1}{\sqrt{\frac{3 + \sqrt{13}}{2}}} = \sqrt{\frac{2}{3 + \sqrt{13}}} = \sqrt{\frac{2(3 - \sqrt{13})}{(3 + \sqrt{13})(3 - \sqrt{13})}} = \sqrt{\frac{6 - 2\sqrt{13}}{9 - 13}} = \sqrt{\frac{6 - 2\sqrt{13}}{-4}} = \sqrt{\frac{\sqrt{13} - 3}{2}}$.

If $x = -\sqrt{\frac{3 + \sqrt{13}}{2}}$, then $y = -\sqrt{\frac{\sqrt{13} - 3}{2}}$.

Thus, the square roots are $\pm \left( \sqrt{\frac{3 + \sqrt{13}}{2}} + i\sqrt{\frac{\sqrt{13} - 3}{2}} \right)$.

b. For $15 + 8i$:

Absolute value:

The absolute value of $15 + 8i$ is $\sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$.

Square roots:

Let $\sqrt{15 + 8i} = x + yi$.

Squaring both sides, we get $15 + 8i = (x + yi)^2 = (x^2 - y^2) + 2xyi$.

Equating the real and imaginary parts, we have:

$x^2 - y^2 = 15$

$2xy = 8$, so $xy = 4$ and $y = \frac{4}{x}$.

Substituting $y = \frac{4}{x}$ into $x^2 - y^2 = 15$, we get $x^2 - \frac{16}{x^2} = 15$.

Multiplying by $x^2$, we have $x^4 - 16 = 15x^2$, so $x^4 - 15x^2 - 16 = 0$.

Let $u = x^2$. Then $u^2 - 15u - 16 = 0$.

$(u - 16)(u + 1) = 0$.

Since $x$ is real, $x^2$ must be positive. So we take the positive root, $u = x^2 = 16$.

$x = \pm 4$.

If $x = 4$, then $y = \frac{4}{4} = 1$.

If $x = -4$, then $y = \frac{4}{-4} = -1$.

Thus, the square roots are $\pm (4 + i)$.

c. For $6 + 2i$:

Absolute value:

The absolute value of $6 + 2i$ is $\sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$.

Square roots:

Let $\sqrt{6 + 2i} = x + yi$.

Squaring both sides, we get $6 + 2i = (x + yi)^2 = (x^2 - y^2) + 2xyi$.

Equating the real and imaginary parts, we have:

$x^2 - y^2 = 6$

$2xy = 2$, so $xy = 1$ and $y = \frac{1}{x}$.

Substituting $y = \frac{1}{x}$ into $x^2 - y^2 = 6$, we get $x^2 - \frac{1}{x^2} = 6$.

Multiplying by $x^2$, we have $x^4 - 1 = 6x^2$, so $x^4 - 6x^2 - 1 = 0$.

Let $u = x^2$. Then $u^2 - 6u - 1 = 0$.

Using the quadratic formula, $u = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-1)}}{2(1)} = \frac{6 \pm \sqrt{36 + 4}}{2} = \frac{6 \pm \sqrt{40}}{2} = \frac{6 \pm 2\sqrt{10}}{2} = 3 \pm \sqrt{10}$.

Since $x$ is real, $x^2$ must be positive. So we take the positive root, $u = x^2 = 3 + \sqrt{10}$.

$x = \pm \sqrt{3 + \sqrt{10}}$.

If $x = \sqrt{3 + \sqrt{10}}$, then $y = \frac{1}{x} = \frac{1}{\sqrt{3 + \sqrt{10}}} = \sqrt{\frac{1}{3 + \sqrt{10}}} = \sqrt{\frac{3 - \sqrt{10}}{9 - 10}} = \sqrt{\frac{3 - \sqrt{10}}{-1}} = \sqrt{\sqrt{10} - 3}$.

If $x = -\sqrt{3 + \sqrt{10}}$, then $y = -\sqrt{\sqrt{10} - 3}$.

Thus, the square roots are $\pm (\sqrt{3 + \sqrt{10}} + i\sqrt{\sqrt{10} - 3})$.

3. Final Answer

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

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

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