- Thread starter
- #1

Denote $D=\{z\in\mathbb C:|z|<1\}$

1) Show that if $z+\dfrac1z$ is a real number, then $\text{Im}(z)=0$ or $z\in\partial D.$

2) Let $z,w\in\mathbb C$ so that $z\in\partial D.$ Show that $\left| {\dfrac{{z + w}}{{\overline z w + 1}}} \right| = 1.$

3) Let $z_1,z_2,z_3\in\mathbb C$ so that $z_1,z_2,z_3\in\partial D$ and $z_1+z_2+z_3=0.$ Prove that $\dfrac{1}{{{z_1}}} + \dfrac{1}{{{z_2}}} + \dfrac{1}{{{z_3}}} = 0.$

4) Let $A = {\left( {\dfrac{{ - 1 + \sqrt 3 i}}{2}} \right)^n} + {\left( {\dfrac{{ - 1 - \sqrt 3 i}}{2}} \right)^n}.$ Prove that $A=\left\{\begin{array}{rl}2,&\text{if }n\text{ is multiple of }3.\\-1,&\text{Otherwise.}\end{array}\right.$

__Attempts__:

1) if $z+\dfrac1z$ is a real number, then $z + \dfrac{1}{z} = \overline {z + \dfrac{1}{z}} ,$ does this hold? Would it work if I do $z=x+iy$ ? But I don't see how to show that $z\in\partial D.$

2) It's simple algebra, I just put $z=a+bi$ and $w=c+di$ and show the modulus is 1.

3) I don't know if there's a faster way here, I mean by not setting $z_1=a+bi,$ and so on, is there a faster way to proceed?

4) I have $A = 2\cos \left( {\dfrac{{2\pi n}}{3}} \right)$ so if $n$ is a multiple of 3 the result follows, but I don't see how to conclude the second part.