Hey, I'm not sure if this is part of the syllabus but my teacher gave us some extension questions to do (apparently our school doesn't teach harder 3u saying we ought to have already have learned it...) but could someone help me with this:
By considering the cases x ≥ 0, −1 < x < 0 and x ≤ −1 (or otherwise) prove that 1+x+x^2+x^3+x^4 is always positive.
Alrighty... I'll give the explanation a go, but it's been a very long time since I've any complex analysis this difficult. (VCE complex numbers is so much easier lol).
I'm assuming that you've already described the locus. Mainly,
> circle of radius 2 centered at \(2+0i\)
> line \(\text{Im}(z)=0\) (essentially the real axis)
All the points on the circle are fine except one, and we'll see why this is in a second.
Let's consider individually different sections of the line \(\text{Im}(z)=0\). For our convenience, let's write \(z=a+0i\).
If \(a<2\) (\(a-2<0\)), then we have \(\arg(z-2)=\arg(a-2)=\pi\) and \(\arg(z^2)=\arg(a^2)=0\) (since \(a^2>0\)).
Note that obviously we cannot have \(z=2+0i\) since \(\arg(0)\) is undefined.
If \(a>2\) (\(a-2>0\)), then we have \(\arg(z-2)=\arg(a-2)=0\) and \(\arg(z^2)=\arg(a^2)=0\) (since \(a^2>0\)).
Therefore, what is essentially left of the line is the ray given by \(\arg(z-2)=0\).
Altogether, the locus of \(z\) is given by \[\left\{z\mid \arg(z-2)=0\right\}\cup\left\{z\mid\ |z-2|=2\right\}\setminus\{0\},\] which can be read as "the set of values \(z\) that satisfy \(\arg(z-2)=0\) or \(|z-2|=2\), and \(z\neq0\)"