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Exercise 1.1 (Solutions)

15 Hits, Last modified: 17 months ago

|c|c|} \hline + & 0 \\ \hline 0 & 0 \\ \hline \end{array} \] As $0+0=0 \in \{0\}$. This implies $\{... } \hline \times & 0 \\ \hline 0 & 0 \\ \hline \end{array} \] As $0\times 0=0 \in \{0\}$ This impli... |c|c|} \hline + & 1 \\ \hline 1 & 2 \\ \hline \end{array} \] As $1+1=2 \notin \{1\}$. This implies... } \hline \times & 1 \\ \hline 1 & 1 \\ \hline \end{array} \] As $1\times 1=1 \in \{1\}$. This implie

Exercise 1.2 (Solutions)

5 Hits, Last modified: 17 months ago

box{In fact if } z=a+bi, \mbox{ then } -z=-a-bi. \end{array}\nonumber\] - Associative Law for Addition ... -(i^2)^{9} \cdot i \\ & =-(-1)^{9} i=-(-1)i =i. \end{align} **Question 4(iv)** Simplify: $\displays... \frac{i}{i}\\ &=\frac{i}{i^2}=\frac{i}{-1}=-i. \end{align} </panel> <panel> **Question 5(i)** Writ... in \mathbb{R}, \text{ as } x,y \in \mathbb{R}. \end{align} Now \begin{align} \text{Product}&=z\cdot