Plot a complex set in the complex plane

My lecturer only explained how to plot complex numbers on the complex plane, but he didn't explain how to plot a set of complex numbers.

I did some research online but I didn't find any clear explanation or method. I have an exercise to practice but I don't know how to even start!. Any help would be really appreciated!

• Are you familiar with Analytic Geometry and the equations of circles, eclipses etc on the plane? – MathematicianByMistake Oct 24 '17 at 15:19
• Yes, how can I use them here? A friend explained me that with the equations(as in a.) I can obtain the radios and centre of a circunference. But I do not know how to do it, and the other cases I do not have any equation – Evoked Oct 24 '17 at 15:24
• Check the answer for a hint and a general approach. Your friend is correct. – MathematicianByMistake Oct 24 '17 at 15:25

HINT

Let's tackle the first one. A similar approach is required for the other two.

For $z=x+iy$, $x,y\in \mathbb{R}$ we get $$|z-1+i|=1\Rightarrow\\|x-1+i(y+1))|=1\Rightarrow\\\sqrt{(x-1)^2+(y+1)^2}=1\Rightarrow\\(x-1)^2+(y+1)^2=1$$

Does this-hopefully it does-remind you a more general equation of a circle?

Let's also look at the third one as well.

We have $$\operatorname{Re}\Big(\frac{z+1}{z-1}\Big)=\Re\Big(\frac{z+1}{z-1}\Big)=\Re\Big[\frac{(z+1)(\bar{z}-1)}{(z-1)(\bar{z}-1)}\Big]=\Re\Big[\frac{(z+1)(\bar{z}-1)}{|z-1|^2}\Big]\gt1$$ Now for $z=x+iy$, $x,y\in \mathbb{R}$ you can substitute on the last relationship and obtain an equation for $x,y$-an inequality to be precise.

• Yes it does! Since in the others there are inequalities, the representation would no longer be a circle? – Evoked Oct 24 '17 at 15:30
• I will help you get an equation for the last one as well, but-and don't take it personally-we can't do your homework for you, there would be no point in that if you don't try yourself. If the answer satisfies you you can upvote/accept.. – MathematicianByMistake Oct 24 '17 at 15:32
• This is not my homework! This is me trying to understand an exercise I searched for in a book! Please don't take in a bad way my words but I'm just trying to understand something, not asking for a particular class free – Evoked Oct 24 '17 at 18:32
• @Evoked No worries! I am only happy that the answer was helpful! Feel free to ask any questions on MSE. It is great for you that you wish to learn more and self-study. :-) – MathematicianByMistake Oct 24 '17 at 18:36