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 $b=0$.

  • $n=2$

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

  • $n=3$


  • $n=4$



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?



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

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 $\frac{3\pi}{2}<\theta<2\pi.$

  • 3
    $\begingroup$ Do you know how to represent complex numbers in polar/trigonometric form? If yes, it might be simpler that way. $\endgroup$ – quid Dec 31 '14 at 14:49
  • $\begingroup$ Good, I like the edit and you've got the gist, but the proof is a mouthful! Notice that $Re(z^n)\ge0$ so that means for $n=1$, that is $Re(z)\ge 0$ now lets take $z$ and write it as $z=Re(z)+jIm(z)$ from this you find $\theta$, when you multiply complex numbers the radius multiples and the angle adds, so if you have $\theta=0$ you can keep multiplying it by itself but $0+0=0$ always, if $\theta$ is ANYTHING non-zero at all, then $\theta+\theta$ is different to just $\theta$, we're rotating. Eventually this will get into the half where $Re(z^n)<0$ $\endgroup$ – Alec Teal Dec 31 '14 at 17:37
  • $\begingroup$ You do know $ae^{j\phi}be^{j\theta}=abe^{j(\theta+\phi)}$ right? This is the crucial part, then $(re^{j\theta})^n=r^ne^{jn\theta}$ $r^n\ge0$ by definition! As $r\ge 0$ as I said above, so the angle is the important part, $n=1,2,...,3$ is rotating by $\theta$ each time $\endgroup$ – Alec Teal Dec 31 '14 at 17:39

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.


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

  • $\begingroup$ See my editions please. $\endgroup$ – donikvep Dec 31 '14 at 16:37

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 $\theta=0$.


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