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Complex stuff questions (1)

Markov

Member
Feb 1, 2012
149
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
Jan 26, 2012
1,403
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
Feb 1, 2012
149
Mmmm, I think it is for $|z|=1,$ does make sense?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Well for 1. I think it would be easiest to substitute $ z = x + iy$ and see what happens...
 

Markov

Member
Feb 1, 2012
149
Can anybody check my work for 2), 3) and 4) please?
 
Jan 31, 2012
54
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!