Hand in problems


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.


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 \(z\).


2.3 Prove that if a limit exists, then it is unique.
Proof: Contradiction
Assume two limits exist, call them \(lim_{z \rightarrow z_0}f(z)=a\) and \(lim_{z \rightarrow z_0}f(z)=b\), such that \(a \neq b\). Choose \(\epsilon =\frac{|a-b|}{2}\). Then by definition of limit, there exists \(\delta_a\) such that \(|z-z_0| < \delta_a \Rightarrow |f(z)-a|< \epsilon\), and there exists \(\delta_b\) such that \(|z-z_0| < \delta_b \Rightarrow |f(z)-a|< \epsilon\). Now choose \(\delta=min\lbrace \delta_a,\delta_b \rbrace\), thus both \(|z-z_0| < \delta_a\) and \(|z-z_0| < \delta_b\) are true. Now consider \(|a-b|\): \[\begin{aligned} |a-b|&=|a-f(z)+f(z)-b| &\text{ by algebra}\\ &\leq|a-f(z)|+|f(z)-b| &\text{ by triangle inequality}\end{aligned}\] From above, we know \(|f(z)-a|< \epsilon\) and \(|f(z)-a|< \epsilon\), thus \(|a-b|<2\epsilon\), ie \(\epsilon<\frac{|a-b|}{2}\); which is a contradiction since we chose \(\epsilon =\frac{|a-b|}{2}\). Thus \(a=b\) which implies \(lim_{z \rightarrow z_0}f(z)=a\) is equal to \(lim_{z \rightarrow z_0}f(z)=b\), therefore, if a limit exists, then it is unique.


2.10 Consider the function \(f:\mathbb{C} \backslash \{ 0 \} \rightarrow \mathbb{C}\) given by \(f(z)=\frac{1}{z}\). Apply the definition of the derivative to give a direct proof that \(f'(z)=-\frac{1}{z^2}\). ———————————————————————————————–
Let the function \(f:\mathbb{C} \backslash \{ 0 \} \rightarrow \mathbb{C}\) be given by \(f(z)=\frac{1}{z}\) and let \(z \in \mathbb{C} \backslash \lbrace 0 \rbrace\). To find the derivative, we apply the difference quotient limit: \[\begin{aligned} f'(z)&=\lim\limits_{z \rightarrow z_0} \frac{f(z+h)-f(z)}{h}\\ &=\lim\limits_{z \to x_0} \frac{\frac{1}{(z+h)}-\frac{1}{z}}{h}\\ &=\lim\limits_{z \to z_0} \frac{\frac{z}{z(z+h)}-\frac{z+h}{z(z+h)}}{h}\\ &=\lim\limits_{z \to z_0} \frac{\frac{z-z-h}{z^2+hz}}{h}\\ &=\lim\limits_{z \to z_0} \frac{\frac{-h}{z^2+hz}}{h}\\ &=\lim\limits_{z \to z_0} \frac{-h}{h(z^2+hz)}\\ &=\lim\limits_{z \to z_0} \frac{-1}{(z^2+hz)}\\ &=\frac{-1}{z^2+0z} &\text{apply the limit}\\ &=\frac{-1}{z^2}\end{aligned}\] Thus the derivative of \(f:\mathbb{C} \backslash \{ 0 \} \rightarrow \mathbb{C}\) given by \(f(z)=\frac{1}{z}\) is \(-\frac{1}{z^2}\).


2.15 Prove that if \(f(z)\) is given by a polynomial in \(z\), then \(f(z)\) is entire.

Let \(z_0 \in \mathbb{C}\) and \(f(z)=a_0+a_1z+a_2z^2+...+a_nz^n\) be an arbitrary polynomial in \(z\). Let \(f_i(z)\) be a function of \(z\) given by \(f_i(z)=a_iz^i\) for some \(i \in \lbrace0,1,2,...,n\rbrace\). Then, \(f(z)=f_0(z)+f_1(z)+f_2(z)+...+f_n(z).\) Then by Proposition 2.15 on page 30,