I'm a bit rusty on my complex numbers, how would you solve the following problem on paper?

Determine and sketch (graph) the set of all complex numbers of form: $$z_n=\frac{2n+1}{n-i},n\in\mathbb R$$

Rationalizing yields $$S=\left\{z_n\in\mathbb C : z_n=\frac{n + 2 n^2}{1 + n^2}+\frac{1 + 2 n}{1 + n^2}i\right\}$$

How do I proceed to sketch (graph) this now on paper? (Wolframalpha yields a circle)

I assume I need to find the center and the radius of that circle which would be enough to sketch the graph, but I can't quite proceed from this point on.

  • $\begingroup$ $n\in\mathbb{R}$ or $n\in\mathbb{Z}$.? $\endgroup$
    – Nosrati
    Nov 16, 2017 at 19:34
  • $\begingroup$ @MyGlasses $n\in\mathbb R$ . $\endgroup$
    – Vepir
    Nov 16, 2017 at 19:35

3 Answers 3


Say $$w= {2n+1\over n-i}\Longrightarrow n ={wi+1\over w-2}$$

Since $\overline{n}=n$ we have:

$$ (\overline{w}-2)(wi+1)= (-i\overline{w}+1)(w-2)$$


$$ |w|^2i-2wi+\overline{w}-2 = -i|w|^2+2i\overline{w}+w-2$$

so $$ |w|^2 = (\overline{w}+w)+{w-\overline{w}\over 2i}$$

Now, since $w=x+yi$ we have

$$ x^2+y^2 = 2x+y$$

So this is circle $$(x-1)^2+(y-{1\over 2})^2 ={5\over 4}$$

  • $\begingroup$ Would you mind noting why is $\overline{x}=x$? $\endgroup$
    – Vepir
    Nov 16, 2017 at 19:54
  • $\begingroup$ Well $x$ is real number, no? $\endgroup$
    – nonuser
    Nov 16, 2017 at 19:55
  • $\begingroup$ Oh right, obviously. Thanks for showing me this approach. $\endgroup$
    – Vepir
    Nov 16, 2017 at 19:59

I'll use $x$ instead of $n$.

Note that$$\frac{2x^2+x}{x^2+1}=1+\frac{x^2+x-1}{x^2+1}=1+\frac{2x^2+2x-2}{2(x^2+1)}$$and that$$\frac{2x+1}{x^2+1}=\frac12+\frac{-x^2+4x+1}{2(x^2+1)}.$$So$$\frac{2x+1}{x-i}=1+\frac i2+\frac{2x^2+2x-2}{2(x^2+1)}+\frac{-x^2+4x+1}{2(x^2+1)}i.$$Now, observe that$$\left(\frac{2x^2+2x-2}{2(x^2+1)}\right)^2+\left(\frac{-x^2+4x+1}{2(x^2+1)}\right)^2=\frac54.$$Therefore, your set is contained in the circle with center $1+\frac i2$ and radius $\frac{\sqrt5}2$.


$\dfrac{1}{\overline{z_n}} = \dfrac{n+i}{2n+1} = \dfrac{n}{2n+1}+\dfrac{i}{2n+1}$

Evidently $\dfrac{1}{\overline{z_n}}$ lies on the line $L: 2x+y=1$

$z_n$ hence lies on the curve obtained by inverting $L$ in the circle $|z|=1$. The result is a circle passing through origin $O$. The radius and centre of this circle are found using properties of inversion as below:

Since $L$ is at a distance of $\dfrac{1}{\sqrt 5}$ from origin, the foot of the perpendicular $P$ from $O$ transforms to a point $P'$ at a distance of $\sqrt 5$ along the line $x-2y=0$ i.e. to $P'(2,1)$. Also $P'$ becomes the other end of the diameter from $O$. Hence the radius of the circle is $\dfrac{\sqrt 5}{2}$ and center is $\left(1, \dfrac{1}{2} \right)$

Hence $z_n$ lies on $\left(x-1\right)^2+\left(y-\dfrac{1}{2}\right)^2 = \dfrac{5}{4}$


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