\( \mathbb{R}^n \) and \(\mathbb{C}^n \)

1. add Suppose \(a\) and \(b\) are real numbers, not both 0. Find real numbers \(c\) and \(d\) such that \[1/(a+bi) = c+di \]

   We can multiple both sides by \(a-bi\). \[ \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} = \frac{a-bi}{a^2+b^2}\] Equating real and imaginary parts, we get \[ c = \frac{a}{a^2+b^2} \text{ and } d = \frac{-b}{a^2+b^2} \]

2. add Show that \[ \frac{-1 + \sqrt{3}i}{2} \] is a cube root of 1 (meaning that its cube equals 1).

   \[ \Bigg(\frac{-1 + \sqrt{3} i}{2} \Bigg)^3 = \frac{-2 -2i\sqrt{3}}{4} \cdot \frac{-1 + \sqrt{3}i}{2} = \frac{8}{8} = 1\]

3. add Find two distinct square roots of \(i\).

   Suppose \(a, b \in \mathbb{R} \) such that \( (a+bi)^2 = i \). Then, \[ (a+bi)^2 = a^2 +2abi -b^2 = i\] Equating real and imaginary parts, we get \[ a^2 - b^2 = 0 \tag{1} \] \[ 2ab = 1 \tag{2} \] Equation (1) implies \(a = b \) (\(a = -b \) is impossible). Plugging \(a=b\) into (2) we get \( b = \pm \frac{\sqrt{2}}{2}\). Hence, our two roots are \[ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \text{ and } \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\]

4. add Show that \(\alpha + \beta = \beta + \alpha \) for all \(\alpha, \beta \in \mathbb{C}\).

   Let \(\alpha, \beta \in \mathbb{C} \). We can write \[\alpha = a+ bi \text{ and } \beta = c + di \] for \( a,b,c,d \in \mathbb{R}. \) \begin{align*} \alpha + \beta &= (a+bi) + (c+di) \\ & = (a+c) + (b+d)i \\ & = (c+a) + (d+b)i \\ & = (c+di) + (a+bi) \\ & = \beta + \alpha \end{align*}

5. add Show that \( (\alpha + \beta) + \gamma = \alpha + (\beta + \gamma) \) for all \( \alpha, \beta, \gamma \in \mathbb{C}. \)

   We can write \[\alpha = a+ bi \] \[ \beta = c+di\] \[ \gamma = e + fi \] for \( a,b,c,d,e,f \in \mathbb{R}. \) \begin{align*} (\alpha + \beta) + \gamma &= \Big[ (a+bi) + (c+di)\Big] + (e + fi) \\ & = (a+c) + (b+d)i + (e+fi) \\ & = a + c + bi + di + e + fi \\ & = a+bi + c + e + di + fi \\ & = (a+bi) + (c+e) + (d+f)i \\ & = (a+bi) + \Big[(c+di) + (e+fi)\Big] \\ & = \alpha + (\beta + \gamma) \end{align*}

6. add Show that \( (\alpha \beta)\gamma = \alpha(\beta \gamma) \) for all \( \alpha, \beta, \gamma \in \mathbb{C} \)

   We can write \[\alpha = a+ bi \] \[ \beta = c+di\] \[ \gamma = e + fi \] for \( a,b,c,d,e,f \in \mathbb{R}. \) \begin{align*} (\alpha \beta)\gamma &= \big[(a+bi)(c+di)\big](e+fi) \\ & = (ac + dai + bci -bd)(e+fi) \\ & = ac(e+fi) - bd(e+fi) +(e+fi)(da+bc)i \\ & = ace + acfi - bde -bdfi + edai + ebci - fda - fbc \\ & = ace + ebci - fda - bdfi + edai + acfi - bde - fbc \\ & = ce(a+bi) - fd(a+bi) + (a+bi)(de+fc)i \\ & = (a+bi)(ce + fci + dei -fd)\\ & = (a+bi)\big[(c+di)(e+fi)\big] \\ & = \alpha(\beta \gamma) \end{align*}

7. add Show that for every \(\alpha \in \mathbb{C}\), there exists a unique \(\beta \in \mathbb{C}\) such that \[ \alpha + \beta = 0\]

   We will first prove existence. Let \(\alpha = a + bi\) for \(a, b \in \mathbb{R}\). Now, define \(\beta = -a - bi\). Then, \begin{align*} \alpha + \beta &= (a+bi) + (-a - bi) \\ &= (a-a) + (b-b)i \\ &= 0 + 0i\\ &= 0 \end{align*} To prove uniqueness, let \(\gamma \in \mathbb{C}\) such that \(\alpha + \gamma = 0\). Then, \[\gamma = \gamma + (\alpha + \beta) = (\gamma + \alpha) + \beta = 0 + \beta = \beta\] Hence, \(\beta\) is unique.