The problem asks to express the given complex numbers in the form $a + bi$, where $a$ and $b$ are rational numbers. We have six expressions: (a) $\frac{2}{1-i}$ (b) $\frac{3+i}{4-3i}$ (c) $\frac{3+i}{i}$ (d) $\frac{1+i\sqrt{3}}{\sqrt{3}-2i}$ (e) $\frac{x+yi}{x-yi}$ (f) $\frac{-2+3i}{-i}$

1. Problem Description

The problem asks to express the given complex numbers in the form $a + bi$, where $a$ and $b$ are rational numbers. We have six expressions:

(a) $\frac{2}{1-i}$

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

(c) $\frac{3+i}{i}$

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

(e) $\frac{x+yi}{x-yi}$

(f) $\frac{-2+3i}{-i}$

2. Solution Steps

(a) $\frac{2}{1-i}$

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $1+i$.

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

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

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

$\frac{3+i}{4-3i} = \frac{(3+i)(4+3i)}{(4-3i)(4+3i)} = \frac{12 + 9i + 4i + 3i^2}{16 - 9i^2} = \frac{12 + 13i - 3}{16 + 9} = \frac{9+13i}{25} = \frac{9}{25} + \frac{13}{25}i$

(c) $\frac{3+i}{i}$

To rationalize the denominator, we multiply the numerator and denominator by $-i$ (or the conjugate $-i$).

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

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

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3}+2i$.

$\frac{1+i\sqrt{3}}{\sqrt{3}-2i} = \frac{(1+i\sqrt{3})(\sqrt{3}+2i)}{(\sqrt{3}-2i)(\sqrt{3}+2i)} = \frac{\sqrt{3} + 2i + 3i + 2i^2\sqrt{3}}{3 - 4i^2} = \frac{\sqrt{3} + 5i - 2\sqrt{3}}{3 + 4} = \frac{-\sqrt{3} + 5i}{7} = -\frac{\sqrt{3}}{7} + \frac{5}{7}i$

(e) $\frac{x+yi}{x-yi}$

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $x+yi$.

$\frac{x+yi}{x-yi} = \frac{(x+yi)(x+yi)}{(x-yi)(x+yi)} = \frac{x^2 + 2xyi + y^2i^2}{x^2 - y^2i^2} = \frac{x^2 - y^2 + 2xyi}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2} + \frac{2xy}{x^2 + y^2}i$

(f) $\frac{-2+3i}{-i}$

To rationalize the denominator, we multiply the numerator and denominator by $i$.

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

3. Final Answer

(a) $1 + i$

(b) $\frac{9}{25} + \frac{13}{25}i$

(c) $1 - 3i$

(d) $-\frac{\sqrt{3}}{7} + \frac{5}{7}i$

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

(f) $-3 - 2i$