We are given several complex number problems. (2a) Express $\frac{1}{z}$ in the form $a + ib$, where $z = x + iy$ is a nonzero complex number. (2b) Given $z + \frac{1}{z} = k$, where $k$ is a real number, prove that $z$ is real or $|z| = 1$. (23a) Express $\frac{1}{i^3}$ in the form $a+ib$. (23b) Express $i^{15}$ in the form $a+ib$. (23c) Express $i^{1002}$ in the form $a+ib$. (24a) Express $\frac{\sqrt{3}+1}{\sqrt{3}-1} + \sqrt{3} - 1$ in the form $a+b\sqrt{3}$, where $a$ and $b$ are rational numbers.

1. Problem Description

We are given several complex number problems.

(2a) Express $\frac{1}{z}$ in the form $a + ib$, where $z = x + iy$ is a nonzero complex number.

(2b) Given $z + \frac{1}{z} = k$, where $k$ is a real number, prove that $z$ is real or $|z| = 1$.

(23a) Express $\frac{1}{i^3}$ in the form $a+ib$.

(23b) Express $i^{15}$ in the form $a+ib$.

(23c) Express $i^{1002}$ in the form $a+ib$.

(24a) Express $\frac{\sqrt{3}+1}{\sqrt{3}-1} + \sqrt{3} - 1$ in the form $a+b\sqrt{3}$, where $a$ and $b$ are rational numbers.

2. Solution Steps

(2a)

We have $z = x + iy$, so $\frac{1}{z} = \frac{1}{x + iy}$.

To express this in the form $a + ib$, we multiply the numerator and denominator by the conjugate of the denominator:

$\frac{1}{x + iy} = \frac{1}{x + iy} \cdot \frac{x - iy}{x - iy} = \frac{x - iy}{x^2 + y^2} = \frac{x}{x^2 + y^2} - i \frac{y}{x^2 + y^2}$.

Therefore, $a = \frac{x}{x^2 + y^2}$ and $b = -\frac{y}{x^2 + y^2}$.

(2b)

Given $z + \frac{1}{z} = k$, where $k$ is real.

Let $z = x + iy$. Then $x + iy + \frac{1}{x + iy} = k$.

$x + iy + \frac{x - iy}{x^2 + y^2} = k$.

Separating real and imaginary parts:

$x + \frac{x}{x^2 + y^2} = k$ and $y - \frac{y}{x^2 + y^2} = 0$.

From the imaginary part, $y(1 - \frac{1}{x^2 + y^2}) = 0$, so either $y = 0$ or $1 - \frac{1}{x^2 + y^2} = 0$.

If $y = 0$, then $z = x$, which means $z$ is real.

If $1 - \frac{1}{x^2 + y^2} = 0$, then $x^2 + y^2 = 1$, so $|z|^2 = 1$, and $|z| = 1$.

Thus, $z$ is real or $|z| = 1$.

(23a)

$\frac{1}{i^3} = \frac{1}{i^2 \cdot i} = \frac{1}{-i} = \frac{1}{-i} \cdot \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = i = 0 + 1i$.

(23b)

$i^{15} = i^{12} \cdot i^3 = (i^4)^3 \cdot i^3 = (1)^3 \cdot i^3 = i^3 = i^2 \cdot i = -i = 0 - 1i$.

(23c)

$i^{1002} = i^{1000} \cdot i^2 = (i^4)^{250} \cdot i^2 = (1)^{250} \cdot (-1) = -1 = -1 + 0i$.

(24a)

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

So $a = 1$ and $b = 2$.

3. Final Answer

(2a) $\frac{x}{x^2 + y^2} - i \frac{y}{x^2 + y^2}$

(2b) Proof completed.

(23a) $0 + 1i$

(23b) $0 - 1i$

(23c) $-1 + 0i$

(24a) $1 + 2\sqrt{3}$