Hand in problems

1.10

1.10
Fix \(a \in \mathbb{C}\) and \(b \in \mathbb{R}\). Show that the equation \(|z^2|+Re(az)+b=0\) has a solution if an only if \(|a^2|\geq 4b\). When solutions exist, show the solution set is a circle.

Fix \(a \in \mathbb{C}\) and \(b \in \mathbb{R}\). Suppose that the equation \(|z^2|+Re(az)+b=0\) has a solution. Let \(z=x+iy\) and \(a=c+id\)
So: \[\begin{aligned} |z^2|+Re(az)+b&=0\\ x^2+y^2+cx-dy+b&=0\\ x^2+cx+y^2-dy&=-b\\ x^2+cx+\frac{1}{4}c^2+y^2-dy+\frac{1}{4}d^2&=-b+\frac{1}{4}c^2+\frac{1}{4}d^2\\ (x+\frac{1}{2}c)^2+(y+\frac{1}{2}d)^2&=-b+\frac{1}{4}(c^2+d^2)\\ (x+\frac{1}{2}c)^2+(y+\frac{1}{2}d)^2&=-b+\frac{1}{4}|a^2|\end{aligned}\] Thus the solution set is a circle centered at \(-\frac{1}{2}+\frac{1}{2}i\) with radius \(\sqrt{-b+\frac{1}{4}|a^2|}\). By checking the left hand side, \((x+\frac{1}{2}c)^2+(y+\frac{1}{2}d)^2\geq0\). So the right hand side is: \[\begin{aligned} -b+\frac{1}{4}|a^2|&\geq0\\ \frac{1}{4}|a^2|&\geq b\\ |a^2|\geq4b\end{aligned}\] Meanwhile, assume \(\mid a^2 \mid \geq 4b\). So: \[\begin{aligned} \mid a^2 \mid &\geq 4b\\ a\overline{a}&\geq 4b &\text{ by proposition 1.10f}\\ c^2+d^2&\geq 4b\\ \frac{c^2+d^2}{4}&\geq b & \text{ by algebra}\\ \frac{c^2+d^2}{4}-b &\geq 0\\ \frac{c^2}{4}+\frac{d^2}{4}-b &\geq 0 &\text{ equation of a circle}\end{aligned}\] Since the solution set is a circle, then \(|z^2|+Re(az)+b=0\) has a solution when \(\mid a^2 \mid \geq 4b\).
Therefore \(|z^2|+Re(az)+b=0\) has a solution if and only if \(|a^2| \geq 4b\), as was to be shown.

1.19

1.19
Fix a positive integer \(n\) and a complex number \(w\). Find all solutions to \(z^n=w\).
—————————————————————————————————————–
Let \(n \in \mathbb{Z}\) and \(z,w \in \mathbb{C}\). In polar coordinates, \(w=re^{i\theta}\) so: \(z^n=re^{i\theta}\). Let \(z=me^{i\phi}\). The modulus of \(z\) is \(m^n=r\) ie \(m=\sqrt[n]{r}\). So for \(\theta\) of \(z\), we have \(n\theta=\theta+2\pi k\) for some \(k \in \mathbb{Z}\) and \(\theta=\frac{\theta+2 \pi k}{n}\). So \(z=\sqrt[n]{r}*e*\frac{i(\theta+2 \pi k)}{n}\). By the fundamental theorem of algebra we know that there are \(n\) solutions to this equation. Restricting \(k\) so \(0 \leq k \leq n\), then there are \(n\) unique solutions for \(z\).