# Complex stuff questions (1)

#### Markov

##### Member
I'll be posting problems and my ideas to solve them, probably if no idea exists, I'll need more help please!

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.

#### Prove It

##### Well-known member
MHB Math Helper
I'll be posting problems and my ideas to solve them, probably if no idea exists, I'll need more help please!

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.
What are you using $\partial D$ to represent?

#### Markov

##### Member
Mmmm, I think it is for $|z|=1,$ does make sense?

#### Prove It

##### Well-known member
MHB Math Helper
Well for 1. I think it would be easiest to substitute $z = x + iy$ and see what happens...

#### Markov

##### Member
Can anybody check my work for 2), 3) and 4) please?

#### Also sprach Zarathustra

##### Member
4.

Perhaps we should find a Polynomial with roots $cos(\frac{2n\pi}{3})$

f(x)=x^4-\frac{5}{4}x^2+\frac{1}{4}

If we factorize it, we will get:

$(x-\frac{1}{2})(x+\frac{1}{2})(x-1)(x+1)$

You can prove that, when 3|n then A=1 or -1 and the rest is follows, I think... hmm!

---------- Post added at 10:29 AM ---------- Previous post was at 09:56 AM ----------

3)

A huge hint:

$z_1,z_2,z_3$ are points of equilateral triangle(...that question from other post).

---------- Post added at 11:02 AM ---------- Previous post was at 10:29 AM ----------

Some better solution for 4:

Put n=3k+1 and n=3k+2 and see what happens!