# 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 $$z$$.