# Complex Analysis - proof problem

• Oct 13th 2013, 03:40 PM
Naranja
Complex Analysis - proof problem
Hi,

I am asked to tackle the following problem:

"Prove that if $z \in \mathbb_{C}$ and Re $(z^k)$ $\geq$ 0 for all k = 1, 2, 3... then $z \in [0, \infty)$"

My first issue is one of interpretation. What am I looking to show? That if z belongs to [0, $\infty$), then z must be purely real? I don't know too much about the where complex numbers fit into the typical set notation. How would I attempt this problem and what should a starting point be?

What I have attempted so far is to multiply out a few of the lower powers of z and collect real and imaginary parts, but that doesn't seem to get me anywhere of use. The real part of z must be nonnegative from the definition, taking k=1. Then, from k=2 we can see that the real part of z must be greater than the imaginary part, then with k=3, the real part is greater than 3 times the imaginary part. By trial and error, I can keep increasing k and see how this is going but its not rigorous. Is this even the right approach, and if so should I be using a proof by induction?
• Oct 13th 2013, 04:09 PM
Plato
Re: Complex Analysis - proof problem
Quote:

Originally Posted by Naranja
"Prove that if and Re $(z^k)$ $\geq$ 0 for all k = 1, 2, 3... then $z \in [0, \infty)$"

First, I must say that I find this a lame question. But I think you are reading it correctly.

If $z \in \mathbb_{C}$ the we can write it as $z=r\exp(i\theta)$ where $r=|z|~\&~-\pi<\text{Arg}(z)=\theta\le\pi$.

Therefore, $\forall k\in\mathbb{Z}^+~\text{Re}(z^k)=r^k\cos(k\theta)$. So in order for that to a non-negative number, what can you say about $\theta~?$
• Oct 13th 2013, 09:31 PM
Naranja
Re: Complex Analysis - proof problem
Hi Plato,

If I am understanding correctly, I think this implies $\theta$ has to be zero.

I think I am getting the picture now. I can see why z has to be purely real, I can convince myself; but I don't know how I would prove it from this.
• Oct 14th 2013, 12:35 AM
Prove It
Re: Complex Analysis - proof problem
Sorry but this question makes no sense. Because there is no ordering of complex numbers, it's impossible to state what it is greater than...
• Oct 14th 2013, 12:48 AM
SlipEternal
Re: Complex Analysis - proof problem
Quote:

Originally Posted by Prove It
Sorry but this question makes no sense. Because there is no ordering of complex numbers, it's impossible to state what it is greater than...

While the complex numbers may not be ordered, the real numbers are. Since the inequality given is comparing the Real component of $z^k$ to 0, this does make sense.

Quote:

Originally Posted by Naranja
but I don't know how I would prove it from this.

Starting with what Plato wrote, you have $\forall k \in \mathbb{Z}^+, r^k\cos(k\theta) \ge 0$. From $k=1$, you have $r\cos(\theta)\ge 0$. This means $r=0$ or $\theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$. So, suppose $\theta \neq 0$. Show that $\exists k \in \mathbb{Z}^+$ such that $k\theta \in [-\pi,\pi] \setminus \left[-\dfrac{\pi}{2},\dfrac{\pi}{2} \right]$.