# 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 $$z^n=w$$.
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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

2.3 Prove that if a limit exists, then it is unique.
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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 $$-\frac{1}{z^2}$$.