#math #flashcards/math **Problem 1** Supposed $a$ and $b$ are real numbers, not both 0, find real numbers $c$ and $d$ such that $ \frac{1}{a+bi} = c + di $ $ \begin{align} \frac{1}{a+bi} &= \frac{1}{a+bi} \cdot{} \frac{a-bi}{a-bi} \\ &= \frac{a-bi}{a^2 - abi + abi + b^2i^2} \\ &= \frac{a-bi}{a^2 + b^2} \end{align} $ Thus $ c = \frac{a}{a^2+b^2} \text{ and } d = \frac{-b}{a^2+b^2} $ Can validate and check $\frac{1}{\alpha} \cdot \alpha = 1 : \alpha = \frac{1}{a+bi}$. *** **Problem 2** Show that $ \frac{-1 + \sqrt{3}i}{2} $ is a cube root of 1. $ \begin{align} \left(\frac{-1 + \sqrt{3}i}{2} \right)^3 &= \left( \frac{1}{8} \right) \left(-1+\sqrt{3}i\right)^3 \\ &= \left( \frac{1}{8} \right) \left( -1 + \sqrt{3}i\right) \left( -1 + \sqrt{3}i \right) \left( -1 + \sqrt{3}i \right) \\ &= \left( \frac{1}{8} \right) \left( 1 + 3i^2 - 2\sqrt{3}i \right) \left( -1 + \sqrt{3}i \right) \\ &= \left( \frac{1}{8} \right) \left( -1 -3i^2 + 2\sqrt{3}i + \sqrt{3}i + 3\sqrt{3}i^3 -6i^2 \right) \\ &= \left( \frac{1}{8} \right) \left( -1 + 3 + 3\sqrt{3}i - 3\sqrt{3}i + 6 \right) \\ &= \left( \frac{1}{8} \right) \left( 8 \right) \\ \\ &= 1 \end{align} $ *** **Problem 3** Find two distinct square roots of $i$. Let $x^2 = i$ where $x = (a+bi)\in{} \mathbb{C}$ . $ \begin{align} x^2 = (a^2 - b^2 + 2abi) &= i \\ (a^2 - b^2 + 2abi) & = (0 + 1i) \end{align} $ $ \begin{align} a^2-b^2 = 0\\ 2abi = 1i \end{align} $ $ a = \pm b $ So, $ \begin{align} 2(\pm b)bi & = i \\ 2\pm{}b^2i = i \\ b = (\pm)\sqrt{\frac{1}{2}}, (\pm)\sqrt{\frac{1}{2} }i \end{align} $ $b \in{} \mathbb{R}$, so the second term cannot be true, and $a\neq-b$ Limiting us to $a = +b$ from $a = \pm b$. $ \begin{align} a & = + b \\ & = (\pm)\sqrt{\left( \frac{1}{2} \right)} \end{align} $ Thus, $x=\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$ and $x=-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i$ *** **Problem 4** Show that $\alpha + \beta = \beta + \alpha$ for all $\alpha, \beta \in{} \mathbb{C}$. $\begin{align} \alpha + \beta &= (a + c) + (b + d)i && \text{(addition def)} \\ &= (c + a) + (d + b)i && \text{(commutativity of reals)} \\ &= (c + di) + (a + bi) && \text{(additive definition backwards)} \\ &= \beta + \alpha \\ \end{align}$ **Problem 5** Show that $(\alpha + \beta) + \lambda = \alpha + (\beta + \lambda)$ for all $\alpha, \beta, \lambda \in{} \mathbb{C}$. Let $ \begin{align} \alpha &= a + bi \text{}\\ \beta &= c + di \\ \lambda &= e + fi \end{align} $ $\begin{align} (\alpha + \beta) + \lambda &= ((a + c) + (b + d)i) + (e + fi) \\ &= ((a + c) + e) + ((b + d) + f)i \\ &= (a + (c + e)) + (b + (d + f))i && \text{(associativity of reals)} \\ &= \alpha + ((c + e) + (d + f)i) && \text{(definition of addition)} \\ &= \alpha + (\beta + \lambda) && \text{(definition of addition)} \end{align}$ *** **Problem 6** Show that $(\alpha \beta )\lambda = \alpha(\beta \lambda)$ for all $\alpha,\beta,\lambda \in{} \mathbb{C}$. $\begin{align} (\alpha\beta)\lambda &= ((a + bi)(c + di))(e + fi) \\ &= ((ac - bd) + (ad + bc)i)(e + fi) \\ &= (ac - bd)e - (ad + bc)f + ((ac - bd)f + (ad + bc)e)i \\ &= (ace - bde - adf - bcf) + (acf - bdf + ade + bce)i \end{align}$ $\begin{align} \alpha(\beta\lambda) &= (a + bi)((c + di)(e + fi)) \\ &= (a + bi)((ce - df) + (cf + de)i) \\ &= a(ce - df) - b(cf + de) + (a(cf + de) + b(ce - df))i \\ &= (ace - adf - bcf - bde) + (acf + ade + bce - bdf)i \end{align}$ Thus, $(\alpha\beta)\lambda = \alpha(\beta\lambda)$. *** **Problem 7** Show that for every $\alpha \in{} \mathbb{C}$, there exists a unique $\beta \in{} \mathbb{C}$ such that $\alpha + \beta = 0.$ Let $\alpha = a + bi$, $\beta = -a - bi$. Then $\alpha + \beta = (a - a) + (b - b)i = 0.$ To show uniqueness, lets assume towards a contradiction there exists a $\lambda : \alpha - \lambda = 0, \lambda \neq \beta$ $ \begin{align} \lambda & = \lambda + 0 \\ & = \lambda + (\alpha + \beta) &&\text{ (by def) } \\ & = (\lambda + \alpha) + \beta &&\text{ (by associativity) }\\ & =0 + \beta &&\text{ (by def) }\\ & = \beta \end{align} $ Thus, $\lambda = \beta$, which is a contradiction since we stated $\lambda \neq \beta$. Thus is it must be the case that there only exists a unique $\beta \in{} \mathbb{C}$ such that $\alpha + \beta = 0$. *** **Problem 8** Show that for every $\alpha \in{} \mathbb{C}$ with $\alpha \neq 0$, there exists a unique $\beta \in{} \mathbb{C}$ such that $\alpha \beta=1$. Let $\alpha = a + bi$ where $a \text{ and } b$ are not both zero. Let $\beta = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i$ . $ \begin{align} \frac{1}{a+bi} &= \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} \\ & = \frac{a-bi}{a^2 + b^2} \end{align} $ $\beta$ is the multiplicative inverse of $\alpha$: $ \begin{align} (a+bi) \cdot \frac{a-bi}{a^2+b^2} & = \frac{a^2 - abi + abi +b^2}{a^2+b^2} \\ & = \frac{a^2 + b^2}{a^2 + b^2} \\ & = 1 \end{align} $ To show uniqueness, let $\lambda$ be any complex number such that $\alpha \lambda = 1$. $ \begin{align} \lambda & = \lambda \cdot 1 \\ & =\lambda \cdot (\alpha \cdot \beta) & \text{Definition of beta}\\ & = (\lambda \cdot \alpha) \cdot \beta & \text{Associativity of complex numbers} \\ & = 1 \cdot \beta \\ & = \beta \end{align} $ Thus $\lambda = \beta$, so it must be that $\beta$, the multiplicative inverse of $\alpha$, is unique. *** **Problem 9** Show that $\lambda(\alpha + \beta) =\lambda \alpha + \lambda \beta \text{ for all } \lambda, \alpha, \beta \in{} \mathbb{C}$. Let $\alpha = a + bi, \beta = c + di, \lambda = e + fi$. $ \begin{align} \lambda(\alpha + \beta) & = (e+fi)\cdot ((a+bi)+(c+di)) \\ & = (e+fi)\cdot ((a+c)+(b + d)i) \\ & = e(a+c) + e(b+d)i + f(a+c)i + f(b + d)i^2 \\ & = ea + ec + ebi + edi + fai + fci -fb - fd \\ & = (ea+ ebi + fai - fb) + (ec + edi + fci -fd) \\ & = (e + fi)(a+bi) + (e + fi)(c+di) \\ & = \lambda \alpha + \lambda \beta \end{align} $ **Problem 10** Find $x \in{} \mathbb{R}^4$ such that $ (4,-3,1,7) + 2x = (5,9,-6,8). $ $ \begin{align} (4,-3,1,7) + 2x &= (5,9,-6,8) \\ 2x &= (5,9,-6,8) - (4,-3,1,7) \\ x &= \frac{1}{2}(5-4, 9+3, -6-1, 8-7) \\ x &= \frac{1}{2}(1, 12, -7, 1) \\ x &= \left( \frac{1}{2}, 6, \frac{-7}{2}, \frac{1}{2} \right) \\ \end{align} $ **** **Problem 11** Explain why there does not exist $\lambda \in{} \mathbb{C}$ such that $ \lambda(2-3i, 5 + 4i, -6 + 7i) = (12-5i, 7+22i, -32-9i). $ It would require that there exists a $\lambda \in{} \mathbb{C}$ such that $ \begin{align} \lambda(2-3i) &= 12-5i \\ \lambda(5+4i) & = 7 + 22i \\ \lambda(-6+7i) & = -32-9i \\ \end{align} $ For the first equation $ \begin{align} \lambda & = \left( \frac{12-5i}{2-3i} \right) \\ & = \left( \frac{12-5i}{2-3i} \right) \left( \frac{2+3i}{2+3i} \right) \\ & = \frac{24+36i-10i + 15}{4+9} \\ & = \frac{39+26i}{13} \\ & = 3 + 2i \end{align} $ From the second equation $ \begin{align} \lambda &= \left( \frac{5+4i}{7+22i} \right) \\ &=\left( \frac{5+4i}{7+22i} \right) \left( \frac{7-22i}{7-22i} \right) \\ & = \left( \frac{35-110i+28i+88}{49 + 484} \right) \\ & = \frac{123 - 82i}{533} \end{align} $ Thus there does not exist a single $\lambda \in{} \mathbb{C}$ such that the original equation is satisfied. **** **Problem 12** Show that $(x+y) + z = x + (y+z)$ for all $x,y,z \in{} F^n$. Let $ \begin{align} x & = (x_{1} + \dots + x_{n}) \\ y & = (y_{1} + \dots + y_{n}) \\ z & = (z_{1} + \dots + z_{n}) \\ \end{align} $ Let $((x+y)+z)_{i}$ denote the $i^{th}$ element of the list, $(x+y)+z$. Then for $i=1, \dots, n$ $ \begin{align} ((x+y) + z)_{i}& =(x_{i}+y_{i}) + z_{i} \\ &= x_{i} + (y_{i} + z_{i}) & \text{Associativity in F} \\ &= (x + (y + z))_{i} \end{align} $ Thus $(x+y)+z = x + (y+z)$. **** **Problem 13** Show that $(ab)x = a(bx)$ for all $x \in{} F^n$ and all $a,b \in{} F$. Let $(ab)x = \left((ab)x_{1}, \dots, (ab)x_{n})\right)$, $a(bx) = \left(a(bx_{1}), \dots, a(bx_{n})\right)$. $ \begin{align} ((ab)x)_{i} &= (ab)x_{i} \\ & =a(bx_{i}) &\text{Associativity in F} \\ & = (a(bx))_{i} & \text{for i = 1, ..., n} \end{align} $ So, $(ab)x = a(bx)$. **** **Problem 14** Show that $1x = x$ for all $x \in{} F^n$ $ \begin{align} 1x &= 1(x_{1}, \dots, x_{n}) \\ & =(1x_{1}, \dots, 1x_{n}) \\ & =(x_{1}, \dots, x_{n}) & \text{Identity in F}\\ & = x \end{align} $ **** **Problem 15** Show that $\lambda(x+y) = \lambda x + \lambda y$ for all $\lambda \in{} F$ and all $x,y \in{} F^n$ $ \begin{align} \lambda (x+y) & = \lambda((x_{1} + y_{1}), \dots, (x_{n}+y_{n})) \\ & = (\lambda(x_{1} + y_{1}), \dots, \lambda(x_{n}+y_{n})) \\ & = ((\lambda{}x_{1} + \lambda{}y_{1}), \dots, (\lambda x_{n}+\lambda y_{n})) & \text{Distributive in F}\\ & = \lambda x + \lambda y \end{align} $ **Problem 16** Show that $(a+b)x = ax + bx$ for all $a,b \in{} F$ and all $x \in{} F^n$. $ \begin{align} (a+b)x &= ((a+b)x_{1}, \dots, (a+b)x_{n}) \\ & = ((ax_{1} + bx_{1}), \dots, (ax_{n} + bx_{n})) & \text{Distributive in F} \\ & = ax + bx \end{align} $