# Complex numbers: Show z belongs to R

• Sep 12th 2012, 09:37 AM
theintervurt
Complex numbers: Show z belongs to R
Could you let me know if I proved this correctly ? Hope the text is readable enough

http://img3.imageshack.us/img3/297/35578078.png
• Sep 12th 2012, 09:48 AM
kalyanram
Re: Complex numbers: Show z belongs to R
I did not follow your argument. But you can try $\displaystyle z=x+iy, x,y \in \mathbb{R}$
Given $\displaystyle Im(z)=0 \implies y=0 \implies z=x, x \in \mathbb{R} \therefore z \in \mathbb{R}$
• Sep 12th 2012, 10:05 AM
theintervurt
Re: Complex numbers: Show z belongs to R
z is given to me as (2 - 3i)^ν + (2 + 3i)^ν where ν belongs to Ν*. http://upload.wikimedia.org/math/9/b...a33480a0d9.png is a fact. However I am not sure whether I have "moved down" the conjugation correctly in the third and fourth equations.
• Sep 12th 2012, 10:55 AM
HallsofIvy
Re: Complex numbers: Show z belongs to R
You appear now to be saying that you are given a specific value of z and asked to prove that its imaginary part is real. As kalyanram said, the imaginary part of any complex number is, by definition, real but it appears that, here, you are asked to actually find the real and imaginary parts of z.
• Sep 12th 2012, 11:07 AM
theintervurt
Re: Complex numbers: Show z belongs to R
I am given a specific value of z and I am asked to prove it belongs to R. Sorry for not being more specific, I thought the thread title and what's on the picture was enough.
• Sep 12th 2012, 11:28 AM
Plato
Re: Complex numbers: Show z belongs to R
Quote:

Originally Posted by theintervurt
I am given a specific value of z and I am asked to prove it belongs to R. Sorry for not being more specific, I thought the thread title and what's on the picture was enough.

Written in polar form:
$\displaystyle (a+bi)^v=r^v\exp(vti)~\&~(a-bi)^v=r^v\exp(-\vti)$.
Add those together we get $\displaystyle 2r^v\cos(vt)$ which is a real number.
• Sep 12th 2012, 12:44 PM
johnsomeone
Re: Complex numbers: Show z belongs to R
Yes, your algebraic manipulations are correct. The presentation of your reasoning isn't very transparent, but, still, it's correct; it's just takes a bit to figure out what you're doing. Although this is fine for figuring out why it's true, I'd suggest that on a graded problem, you rewrite your reasoning more carefully. Above all, show respect for the equal sign!!

Also, it's common to deduce that $\displaystyle Im(z) = 0$ by showing that $\displaystyle z = \bar{z}$. It's a worthwhile technique to remember, as the need to show a complex number is

actually real is a fairly common occurance, and it's often easy to see when $\displaystyle w = \bar{w}$, where w is some complex expression in z.

Working it out that way is extremely similar to how you worked out the problem, but, the " by showing $\displaystyle z = \bar{z}$ " approach is particular to the $\displaystyle Im(z) = 0$ case,

whereas the (correct) formula you used, $\displaystyle Im(z) = \frac{z - \bar{z}}{2i}$, holds for any complex number of any imaginary component.

----

Ex: Your problem is a particular case of a general situation. Whenever symmetrically adding functions of $\displaystyle z$ and $\displaystyle \bar{z}$, the result will be real.

Prop: Let $\displaystyle A \subset \mathbb{C}$ satisfy $\displaystyle z \in A \Leftrightarrow \bar{z} \in A$. Let $\displaystyle f : A \rightarrow \mathbb{C}$ satisfy $\displaystyle \overline{f(z)} = f(\bar{z})$.

Then the function $\displaystyle g : A \rightarrow \mathbb{C}$, that's defined by $\displaystyle g(z) = f(z) + f(\bar{z})$, is real valued, i.e. $\displaystyle g(A) \subset \mathbb{R}$.

Proof:
(Note that nothing is being assumed about A being open, compact, whatever, or f being analytic, continuous, etc.. This is purely a consequence of sets,
functions, and f "passing through" the conjugation. Also, the requirement about $\displaystyle z \in A \Leftrightarrow \bar{z} \in A$ is only to guarantee that $\displaystyle f(\bar{z})$ makes sense
when it appears in the definition of g.)

Let $\displaystyle z \in A.$ Then $\displaystyle \overline{g(z)} = \overline{( f(z) + f(\bar{z}) )} = \overline{f(z)} + \overline{f(\bar{z})} = f(\bar{z}) + f( \ \overline{(\bar{z})} \ )$

$\displaystyle = f(\bar{z}) + f(z) = f(z) + f(\bar{z}) = g(z)$. Thus $\displaystyle \overline{g(z)} = g(z) \ \forall z \in A$.

Since $\displaystyle \overline{g(z)} = g(z) \ \forall z \in A$, have that $\displaystyle Im(g(z)) = 0 \ \forall z \in A$. Thus $\displaystyle g(z) \in \mathbb{R} \ \forall z \in A$.

Note: $\displaystyle \overline{f(z)} = f(\bar{z})$ is the common case. It's true for powers of z, exponentials, and hence complex-trig functions. The main theroem guaranteeing it is that is that any holomorphic function h that's real valued for real values (Im(z)=0 implies Im(h(z)) = 0) (a very common occurance) satisfies $\displaystyle \overline{h(z)} = h(\bar{z}) \ \forall z$.

Question: With the same f and A, if instead you had $\displaystyle g(z) = f(z) - f(\bar{z})$, then what could you say about about g? What about if $\displaystyle g(z) = f(z)f(\bar{z})$?