1. ## Complex Analysis

On my final today, there was the question that no nonreal complex number has a nth real root.

I know this is true but I wasn't able to prove it.

Another question was, Let S be a finite set $\{z_1, z_2, z_3,\cdots z_n\}$ where z=a+bi. Prove S is bounded.

For this one, I said since S is finite, S is can be put in a 1-1 correspondence with $\mathbb{N}$.

Also, I was allowed to assume $S\subset\mathbb{C}$.

I then said $S\subseteq\mathbb{N}$.

Now, we have $S\subseteq\mathbb{N}\subset\mathbb{C}$.

So S must be bounded. I know it is wrong but how should it be done or what could I have added or altered to make it correct?

2. Originally Posted by dwsmith
On my final today, there was the question that no nonreal complex has a real root.

I know this is true but I wasn't able to prove it.
What kind of root? $n$-th root, where $n$ is a positive integer?

Another question was, Let S be a finite set $\{z_1, z_2, z_3,\cdots z_n\}$ where z=a+bi. Prove S is bounded.

For this one, I said since S is finite, S is can be put in a 1-1 correspondence with $\mathbb{N}$.

Also, I was allowed to assume $S\subset\mathbb{C}$.

I then said $S\subseteq\mathbb{N}$.

Now, we have $S\subseteq\mathbb{N}\subset\mathbb{C}$.

So S must be bounded. I know it is wrong but how should it be done or what could I have added or altered to make it correct?
Many things are wrong with this. First, a finite set can't be put in bijection with an infinite set! Two sets can be put in bijection with each other only if they have the same cardinality.

Second, even supposing you had established a bijection, that doesn't mean $S \subset \mathbb{N}$! I can put my fingers in bijection with $\{1, 2, \dots, 10\}$, but that doesn't mean my fingers are themselves integers between 1 and 10.

Finally, even if $S$ were a subset of $\mathbb{N}$, that wouldn't mean that it's bounded. For instance $\mathbb{N}$ itself isn't bounded (as a subset of $\mathbb{C}$).

What you should have said is that the set $\{|z_1|, \dots, |z_n|\}$ consisting of the moduli of the elements of $S$ is a finite set of real numbers. A finite set of real numbers always has a finite upper bound - you should be able to prove this! This upper bound is a number $M$ such that $|z|\leq M$ for every $z \in S$, i.e. it's a bound for $S$ as a subset of $\mathbb{C}$.

3. Yes real nth root.

4. A real raised to an integer power is always real.
Doesn't that show it by contradiction?

Apologies if I overlooked something.

5. You are correct. It would have been easier to try a proof by contradiction.

I was trying to to do it straight forward and wasn't getting anywhere.

6. Originally Posted by dwsmith
Yes real nth root.
Are you satisfied with what I wrote regarding your other problem?

7. I still don't know how to do it but it is fine since the class is over.

8. Originally Posted by dwsmith
I still don't know how to do it but it is fine since the class is over.
Why did you ask the question, then?

I gave the full answer above. What don't you understand?

9. I asked to the question to see if it was something easily understood by me. I don't know how to prove S has an upper bound M. M should be a fraction though correct?

10. Originally Posted by dwsmith
I asked to the question to see if it was something easily understood by me. I don't know how to prove S has an upper bound M. M should be a fraction though correct?
If you can't prove that a finite set of real numbers has a finite upper bound, how could you possibly ever take this "complex analysis" course?
Did you ever take an introductory real analysis course?

11. Haven't taken Real yet. Real isn't a pre-requisite for Complex Analysis.