Working in complex number field

Question

I have to draw the graphic of "group of points given by the equation $$(|z^2|-3|z|+2)(z^4+4)=0$$ I solved the first part by factoring and obtaining $|z|=1$ and $|z|=2$ so in the graphic I have the points that stay on two circumferences with the respective radii. For the second part: $$z^4=-4 \to z=\sqrt2*i^{(1/4)}$$ so I'm sure I have 4 points that "form a square" but I don't know how to proceed correctly. I saw a question similar to mine but the answers were given using exponential forms that I don't know and that are not in my program. I need a guideline that show me how to proceed when I have to solve this kind of equation or also the simpler equation form $z=v^{n/n}$, I need to have clear ideas and I'll be grateful if you can help me. Thank you in advance!

Answer 1

From $z^4=-4$ you get $$z^2=\pm 2i$$ There are two main non-exponential ways to proceed from here. The first is polar form $$z^2=r(\cos \theta + i \sin \theta)$$ Then from de Moivre's theorem you get $$z=\pm \sqrt r \left(\cos \frac{\theta}2+\sin \frac{\theta}2\right)$$

See if you can do that for $z^2=2i$ and for $z^2=-2i$.

The other way is to let $z=a+bi$ for real $a$ and $b$, so (for example) $$z^2=2i$$ becomes $$(a^2-b^2)+(2ab)i=2i$$ so that $$a^2-b^2=0$$ and $$2ab=2$$ That is two equations with two unknowns, which is solvable, giving two solutions for $a$ and $b$. Then do the same with $z^2=-2i$.

Either method give you four solutions for $z^4=-4$ (either four $\theta$'s or four pairs $(a,b)$). Both ways the solutions are in a square in the complex plane around $0$.

Do you need more help?