Complex Equations: A = {z complex, bar{z}^5 - |z|^5 = 0}, B = {z complex, (z-2)/(bar{z}+4) - (z+4)/(bar{z}-2) = 4}, C =...

[EmanDroid]

Hi, good morning to all. I would have some help about these three exercises.

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[math]A\, =\, \{z\, \in\, \mathbb{C}\, :\, \bar{z}^5\, -\, |z|^5\, =\, 0\}[/math]
[math]B\, =\, \left\{z\, \in\, \mathbb{C}\, :\, \dfrac{z\, -\, 2}{\bar{z}\, +\, 4}\, -\, \dfrac{z\, +\, 4}{\bar{z}\, -\, 2}\, =\, 4\right\}[/math]
[math]C\, =\, \{z\, \in\, \mathbb{C}\, :\, z^3\, +\, 9iz^2\, -\, 27z\, -\, 30i\, =\, 0\}[/math]

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I don't know how to solve them. Thank you for your support and help

 

[Dr.Peterson]

These aren't exercises; they are definitions of sets. What is it that you want to do with them?

Please read our submission guidelines and follow them: state the entire problem, tell us the context, and show what you DO know about the problem.

 

[EmanDroid]

These aren't exercises; they are definitions of sets. What is it that you want to do with them?

Please read our submission guidelines and follow them: state the entire problem, tell us the context, and show what you DO know about the problem.

Sorry, I didn't express well my problem. These are sets, and the problem is to find elements of these sets by solving these equations. Thank you.

 

[Dr.Peterson]

So, what methods do you know? What have you tried? Where did you get stuck?

I can see the answer to the first just by thinking about which terms are real, or by replacing z with x+iy.

 

[pka]

1) \(\displaystyle \overline{z}^5-|z|^5=0\)
You should know that \(\displaystyle |z|\) is a real number so what about \(\displaystyle \overline{z}^5-|z|^5~?\).

2) use the well known fact that \(\displaystyle \frac{1}{z}=\frac{\overline{z}}{|z|^2}\) so that \(\displaystyle \frac{z-2}{\overline{z}+4}=\frac{(z-2)(z+4)}{|z+4|^2}\)

3) Note that \(\displaystyle z^3+9iz^2-27z-30i=(z+3i)^3-3i=0\)

 

[HallsofIvy]

For the first one if we write \(\displaystyle z= re^{i\theta}\) then \(\displaystyle |z|^5= r^5\) and \(\displaystyle \overline{z}^5= r^5e^{-5i\theta}\). So \(\displaystyle \overline{z}^5- |z|^5= r^5(e^{-5i\theta}- 1)= 0\). So either r= 0 so z= 0 or \(\displaystyle e^{-5i\theta}= 1\) so \(\displaystyle -5\theta= 2n\pi\) for some integer n. In that case, \(\displaystyle \theta= -n\frac{2\pi}{5}\).