9.4

1. Evaluate \((\frac{1}{2}+\frac{1}{2}i)^{10}\)

   Since \(z\)=0.5+0.5i,\(|z|=\sqrt{(0.5^2+0.5^2)}=\frac{\sqrt2}{2}\)

   \(z=\frac{\sqrt2}{2}(cos\theta+isin\theta)\)​

   \(\frac{\sqrt2}{2}cos\theta=0.5\), \(cos\theta=\frac{1}{\sqrt2}\),then \(sin\theta=\frac{1}{\sqrt2}\)

   Hence \(z=(\frac{\sqrt2}{2},\frac{\pi}{4})\)​

   \(z^{10}=((\frac{\sqrt2}{2})^{10},10\cdot \frac{\pi}{4})\)​

   \(=(\frac{1}{32},\frac{\pi}{2})\)​

   Therefore, \(z^{10}=\frac{1}{32}(cos\frac{\pi}{2}+isin\frac{\pi}{2})=\frac{1}{32}i\)​

2. For every \(z\in C\) prove that \(|z|\leq |Re(z)|+|Im(z)|\leq\sqrt2|z|\)​

   Suppose \(z=(r,\theta)\), then \(Re(z)=rcos\theta\), \(Im(z)=rsin\theta\), \(|z|=r\)

   Then \(|Re(z)|+|Im(z)|=r(cos\theta)+r(sin\theta)=r(\sqrt2sin(\theta+\frac{\pi}{4}))=\sqrt2r(sin(\theta+\frac{\pi}{4}))\leq \sqrt2r=\sqrt2|z|\)​

   Then we prove \(|z|\leq |Re(z)|+|Im(z)|\)

   Let \(z=a+bi\), then we need to prove \(\sqrt{a^2+b^2}\leq |a|+|b|\)​

   Equivalently, we prove \(a^2+b^2\leq (a+b)^2=a^2+b^2+2|ab|\)​

   Since \(a,b\geq 0\), clearly, \(\sqrt{a^2+b^2}\leq |a|+|b|\)​

   Therefore \(|z|\leq |Re(z)|+|Im(z)|\leq\sqrt2|z|\)

3. Prove that the sum of the nth roots of unity is 0

   Let \(w=cis(\frac{2\pi}{n})\) then the nth roots of unity are 1,w,\(w^2\),...,\(w^{n-1}\) with \(w\neq 1\)

   Then the sum of the nth roots of unity are \(\frac{1(1-w^n)}{1-w}\), since \(w^n=1\)

   Then the sum is 0

4. Prove \(cos(\frac{2\pi }{n})+cos(\frac{4\pi }{n})+...+cos(\frac{2\pi(n-1) }{n})=-1\)​

   For the angular variable (n\in N~fixed)

   Then \(z_j=cos(\frac{2\pi j}{n})+isin(\frac{2\pi j}{n})\)

   Hence \(z_1+z_2+...+z_n=cos(\frac{0\pi }{n})+cos(\frac{2\pi }{n})+cos(\frac{4\pi }{n})+...+cos(\frac{2\pi(n-1) }{n})+i[sin(\frac{0\pi }{n})+sin(\frac{2\pi }{n})+sin(\frac{4\pi }{n})+...+sin(\frac{2(n-1)\pi }{n})]\)

   Therefore \(cos(\frac{0\pi }{n})+cos(\frac{2\pi }{n})+cos(\frac{4\pi }{n})+...+cos(\frac{2\pi(n-1) }{n})=0\)

   Then \(cos(\frac{2\pi }{n})+cos(\frac{4\pi }{n})+...+cos(\frac{2\pi(n-1) }{n})=-1\)

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