Problem 1.10

[1.10] Fix an \(a\in\mathbb{C}\) and \(b\in\mathbb{R}\). Show that the equation
\(|z^{2}|+Re(az)+b=0\) has a solution if and only if \(|a^2|\geq4b\).

Let \(a\in\mathbb{C}\),\(b\in\mathbb{R}\), and \(z\in\mathbb{C}\).
Then \(a=d+ei\) and \(z=f+gi\) for some \(d,e,f,g\in\mathbb{R}\).
Note that: \[|z^{2}|=z\overline z=f^{2}+g^{2}\]
and \[Re(az)=fd-eg\]

So \(|z^{2}|+Re(az)+b=0\) has solutions \[\begin{aligned} &\iff (f^{2}+g^{2})+(fd-eg)+b=0\\ &\iff f^{2}+fd+g^{2}-eg+b=0\\ &\iff (f+\frac{1}{2}d)^{2}-\frac{1}{4}d^{2}+(g-\frac{1}{2})^{2}-\frac{1}{4}e^{2}+b=0 \text{ (By completing the square)}\\ &\iff (f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}d^{2}+\frac{1}{4}e^{2}-b\\ &\iff(f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}(d^{2}+e^{2})-b\\\end{aligned}\] The above equation is a circle with radius \(r=\sqrt{\frac{1}{4}(d^{2}+e^{2})-b}\), so \(r^{2}\) must be greater than or equal to zero in order for a solution to exist. Thus, \[\begin{aligned} \frac{1}{4}(d^{2}+e^{2})-b\geq0 &\iff(d^{2}+e^{2})\geq4b\end{aligned}\] And since \(a=d+ei\), \(|a^{2}|=a\overline a=d^{2}+e^{2}\). Thus a solution to \(|z^{2}|+Re(az)+b=0\) exists if and only if \(|a^2|\geq4b\).

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