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Chapter 4 Exercise

1. Empty 2. Solution: False. Consider $1=(z^m+1)+(-z^m)\notin \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$. Note that \[(z^m+1)\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}\]and\[-z^m\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\},\]it follows that $\{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$ is not…

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Chapter 1 Exercise A

1.Solution: Because $(a+bi)(a-bi)=a^2+b^2$, one has\[\frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}.\]Hence\[c=\frac{a}{a^2+b^2},d=-\frac{b}{a^2+b^2}.\] 2. Solution1:From direct computation, we have\[\left(\frac{-1+\sqrt{3}i}{2}\right)^2=\frac{-1-\sqrt{3}i}{2},\]hence \[\left(\frac{-1+\sqrt{3}i}{2}\right)^3=\frac{-1-\sqrt{3}i}{2}\cdot\frac{-1+\sqrt{3}i}{2}=1.\]This means $\dfrac{-1+\sqrt{3}i}{2}$ is a cube root of 1. Solution2: Note that \[(a+bi)+(a-bi)=2a\] and \[(a+bi)(a-bi)=a^2+b^2,\] it follows that $\dfrac{-1+\sqrt{3}i}{2}$…

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