Property of complex numbers.

Question

Let $z \in \mathbb{C}$ such that $Re(z^{n})\geq0, \forall n\in\mathbb{N}$, where $Re(z^{n})$ is the real part of $z^{n}$. Show that $z\in\mathbb{R}^{+}$.

If $z=a+bi$, $a,b\in\mathbb{R}$, then for $n=1\Rightarrow Re(z)\geq0\Rightarrow a\geq0$. So $a\in\mathbb{R}^{+}$, now we only need show that $b=0$.

• $n=2$

  $Re(z^{2})=a^{2}-b^{2}\geq0\Rightarrow a^{2}\geq b^{2}\Rightarrow a\geq|b|.$

• $n=3$

  $Re(z^{3})=a^{3}-3ab^{2}\geq0\Rightarrow\frac{a}{\sqrt{3}}\geq|b|.$

• $n=4$

  $Re(z^{4})=a^{4}-6a^{2}b^{2}+b^{4}\geq0\Rightarrow(a^{2}-3b^{2})^{2}-8b^{4}\geq0\Rightarrow\frac{a}{\sqrt{\sqrt{8}+3}}\geq|b|.$

  $\vdots$

In this way I think we can make a sequence $(a_{n})_{n\in\mathbb{N}}$ such that $a_{n}\geq|b|, \forall n\in\mathbb{N}$ and $a_{n}\longrightarrow0$ when $n\longrightarrow\infty$, so by the Squeeze theorem $|b|=0\Rightarrow b=0$.

Is that right(Can anyone give me some hints on how to find this sequence?)? Or I can show what I want in an easier way?

Thanks!

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Alec:

$z=\rho e^{i\theta}=\rho(\cos(\theta)+i\sin(\theta))\Rightarrow z^{n}=e^{in\theta}=\rho^{n}(\cos(n\theta)+i\sin(n\theta)).$

Suppose the number has ANY imaginary component, i.e., $\sin(\theta)\neq0\Rightarrow\theta\neq k\pi,\forall k\in\mathbb{Z}$.

You are saying that for some $n_{0}\in\mathbb{N}$ we'll have: $\cos(n_{0}\theta)<0$. Like this:

For $0<\theta<\frac{\pi}{2}$ then take $\frac{\pi}{2\theta}<n_{0}<1+\frac{\pi}{2\theta}$, so $0<\theta<\frac{\pi}{2}<n_{0}\theta<\theta+\frac{\pi}{2}<\pi$.

In other words I'm adding $\theta$ until $n_{0}\theta$ lies on 2nd quadrant.

We can do the same way for $\frac{3\pi}{2}<\theta<2\pi.$

Answer 1

Quick answer / hint

Use the polar form, if the number has ANY imaginary component at all eventually $z^n$ will rotate and be in the $<0$ real side.

Requirements:

You must know the $re^{j\theta}=r(\cos(\theta)+j\sin(\theta))$ form of complex numbers.

This ought to be sufficient but if it is not drop me a comment and I'll add more. I've assumed this is a "I haven't seen the method" question, hence my hint.

Comment from: Ofir Schnabel

To add to Alec Teal answer you need to use De Moivre's formula http://en.wikipedia.org/wiki/De_Moivre%27s_formula

Answer 2

Write $z=r.e^{i.\theta}$. Therefore $z^n=r^n.e^{i.n\theta}$. The real part of $z^n$ is $r^n.\cos{n.\theta}$ and the only way to have it positive whatever n is $\theta=0$.