w4

1. Given \(z_k \in \mathbb{C}\) such that \(\left| z_k \right| = 2\) for all \(k = 1, 2, 3, 4, 5\) and \(\left| z_1 + z_2 + z_3 + z_4 + z_5 \right| = 8\). Find \(\left| z_1^{-1} + z_2^{-1} + z_3^{-1} + z_4^{-1} + z_5^{-1} \right|.\)

   Proof

   Given \(Z_k = (2, \theta)\) means \(Z_{k}=2\cos\theta+i2\sin\theta\).

   Since \(Z_k = (2, \theta)\), \(Z_{k}^{-1}=\left(\frac12,-\theta\right)\), then

   \(\left|Z_1+Z_2+Z_3+Z_4+Z_5\right|=\left|2\left(\cos\theta_1+\cdots+\cos\theta_5\right)+i2(\sin\theta_1+\cdots+\sin\theta_5)\right|=8\)

   Now,

   \(\left|Z_1^{-1}+Z_2^{-1}+Z_3^{-1}+Z_4^{-1}+Z_5^{-1}\right|=\left|\frac12\left(\cos-\theta_1+\cdots+\cos-\theta_5\right)+i\frac12(\sin-\theta_1+\cdots+\sin-\theta_5)\right|\)

   \(=\left|\frac12(\cos\theta_1+\cdots+\cos\theta_5)-i\frac12(\sin\theta_1+\cdots+\sin\theta_5)\right|\)

   Since \(\cos \theta_1 + \cdots + \cos \theta_5 - i (\sin \theta_1 + \cdots + \sin \theta_5) = \overline{\cos \theta_1 + \cdots + \cos \theta_5 + i (\sin \theta_1 + \cdots + \sin \theta_5)}\),

   Then, \(\left| \cos \theta_1 + \cdots + \cos \theta_5 - i (\sin \theta_1 + \cdots + \sin \theta_5) \right| = \left| \cos \theta_1 + \cdots + \cos \theta_5 + i (\sin \theta_1 + \cdots + \sin \theta_5) \right|\)

   Hence, \(\left|Z_1^{-1}+Z_2^{-1}+Z_3^{-1}+Z_4^{-1}+Z_5^{-1}\right|=\frac12\times\frac82=2\)

2. Let \(p \in \mathbb{R}[X]\) be a polynomial.

   • (a) Prove that, if \(\alpha \in \mathbb{C}\) is a root of \(p\), then \(\overline{\alpha}\) is also a root of \(p\).

   $\begin{aligned} p(\bar{z}) &= \sum_{j=0}^{n} a_j \bar{z}^j = \sum_{j=0}^{n} {a_j} \overline{(z)^j} = \sum_{j=0}^{n} \bar{a_j} \overline{z^j} \quad \text{(since \(a_j \in \mathbb{R} \implies \bar{a_j} = a_j\))} \end{aligned} $

   $\begin{aligned} = \sum_{j=0}^{n}\overline{ a_j z^j} = \overline{\sum_{j=0}^{n} a_j z^j} = \overline{p(z)} = \overline{0} = 0\end{aligned} $ - (b) Deduce from the previous part that if \(\deg(p)\) is odd, then \(p\) has at least one real root.

   If \(\deg(p) = n\), \(p(x)\) has at most \(n\) roots.

   Then \(p(x)\) has odd roots.

   From (a), we know the roots come by pairs.

   Thus, there must leave at least one root \(\beta\) such that \(\overline{\beta} = \beta\).

   Therefore, \(\beta\) is a real root.

   ‍

3. Solve the following equations in \(\mathbb{C}\):

   • (a) \(z^6 - 2z^4 - 4z^2 + 8 = 0\).

   We note that \(z^2 = 2\) is a root.

   \(z^6 - 2z^4 - 4z^2 + 8 = (z^2 - 2)C(x)\).

   By calculating, \(C(x) = z^4 - 4\).

   \(z^4 - 4 = 0\)

   \(z^2 = \pm 2\)

   \(z = \pm \sqrt{2}, \pm \sqrt{2}i\)

   Therefore, \(z = \{ \pm \sqrt{2}, \pm \sqrt{2}i \}\). - (b) \(z^8 - z^2 = 0\).

   \(z^2 \left( z^3 - 1 \right)\left( z^3 + 1 \right) = 0\)

   \(\Rightarrow \begin{cases} z^2 = 0 \\ z^3 - 1 = 0 \\ z^3 + 1 = 0 \end{cases}\)

   Then, we have

   • \(z_0=cos(0)+isin(0)=1\)
   • \(z_1 = \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)
   • \(z_2 = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)
   • \(z_3 = \cos(\pi) + i\sin(\pi) = -1\)
   • \(z_4 = \cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\)
   • \(z_5 = \cos\left(\frac{5\pi}{3}\right) + i\sin\left(\frac{5\pi}{3}\right) = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)
   • \(z_6=0\)​
   • (c) \(z^4 - 3(1 - \sqrt{3}i)\overline{z} = 0\).

   $$ \begin{aligned} & z^4-3\left(1-\sqrt3i\right)=0\ & z^4=3(1-\sqrt3i)\ & (r\text{cis}\ \theta)^4=3\cdot2\text{cis}\left(-\frac{\pi}{3}\right)\cdot r\text{cis}(-\theta)\ & r^4\text{cis}(4\theta)=6r\text{cis}\left(-\frac{\pi}{3}-\theta\right)\ & \implies\begin{cases}r^3=6\implies r=\sqrt[3]{6}\ 4\theta=-\frac{\pi}{3}-\theta+2k\pi\end{cases}\ & \implies5\theta=-\frac{\pi}{3}+2k\pi,\;k\in\mathbb{Z}\ & \theta=-\frac{\pi}{15}+\frac{2k\pi}{5},\;k\in\mathbb{Z}\ & z_{k}=\sqrt[3]{6}\text{cis}\left(-\frac{\pi}{15}+\frac{2k\pi}{5}\right),\;k\in{0,1,2,3,4}\end{aligned} $$

4. Find \(a \in \mathbb{R}\) such that \(z = -i\) is a root of \(z^3 - z^2 + z + 1 + a\). For that \(a\), factorize the polynomial in \(\mathbb{C}\) and in \(\mathbb{R}\).

   Since \(z=-i\) is a root, by theorem we know \(z=i\) has to be a root.

   Thus \(z^2=-1\) is a root, then \(-i+1+i+1+a=0\implies a=-2\)​

   In \(C\), we have \(z^3-z^2+z+1+a=(z+i)(z−i)(z−1)\)

   In R, we have \(z^3-z^2+z+1+a=\left(z^2+1\right)\left(z-1\right)\)

5. Consider the polynomial \(p(X) = X^5 - 3X^4 + 3X^3 - 2X^2 + 2X - 4\).

   • (a) Check that \((1 + i)\) is a root of \(p(X)\).

   Yes.

   $ (1+i)^1=1+i \ (1+i)^2=2i \ (1+i)^3=-2+2i \ (1+i)^4=-4 \ (1+i)^5=-4-4i $​

   Since \(p(1+i)=(-4-4i)-3(-4)+3(-2+2i)-2(2i)+2(1+i)-4=0\), thus \(1+i\) is a root - (b) Fully factorize \(p(X)\) over \(\mathbb{R}\).

   Since \(1+i\) is a root, then \(1-i\) has to be a root, we have \((X-(1+i))(X-(1-i))=X^2-2X+2.\)

   We can calculate \(\frac{p\left(x\right)}{X^2-2X+2}=X^3-X^2-X-2\)​

   Test \(x=2\), we know it is a root.

   Then \(X^3-X^2-X-2=(X-2)(X^2+X+1)\)

   Finally, \(p(X)=(X-2)(X^2-2X+2)(X^2+X+1)\) - (c) Fully factorize \(p(X)\) over \(\mathbb{C}\).

   Since \(\begin{gathered} X^2-2X+2=(X-1+i)(X-1-i) \\ X^2+X+1=\left(X-\left(-\frac12+\frac{\sqrt{3}}2i\right)\right)\left(X-\left(-\frac12-\frac{\sqrt{3}}2i\right)\right) \end{gathered}\)​

   then \(p(X)=(X-2)(X-1-i)(X-1+i)\left(X+\frac12-\frac{\sqrt3}2i\right)\left(X+\frac12+\frac{\sqrt3}2i\right)\)