# Complex numbers

• June 1st 2013, 03:57 AM
Yeah01
Complex numbers
Hi guys. I need some help with I guess a relatively simple question. Could someone please show me with workings the following ?
If the modulus of z+(1/z)=2. What is z?
• June 1st 2013, 04:12 AM
chiro
Re: Complex numbers
Hey Yeah01.

Hint: Can you calculate an expression for the modulus in terms of x and y (for x + iy) or in terms of r and theta (polar co-ordinates)?

(In other words, if you have z + 1/z, what is the modulus in terms of x and y if z = x + iy)?
• June 1st 2013, 04:17 AM
Yeah01
Re: Complex numbers
Hi chiro,
Hope your having a good night thanks for your response. I've tried that still must be doing something wrong. :(
• June 1st 2013, 04:26 AM
chiro
Re: Complex numbers
Hint: the modulus of z is SQRT(x^2 + y^2). If you are dealing with 1/z then the modulus is 1/SQRT(x^2 + y^2) since if z = r*e(i*theta) then 1/z = 1/r * e^(-i*theta) and 1/r is the modulus of 1/z.
• June 1st 2013, 04:38 AM
Yeah01
Re: Complex numbers
Thanks chiro. Let me give it a go
• June 1st 2013, 04:43 AM
Yeah01
Re: Complex numbers
Still no luck. Don't know what I'm doing wrong
• June 1st 2013, 05:07 AM
Yeah01
Re: Complex numbers
Do u think u can help me out chiro?
• June 1st 2013, 03:32 PM
Yeah01
Re: Complex numbers
Hi I got x = 1/4-y and y=1/4-x. What now? Sm I on the right track?
• June 1st 2013, 03:41 PM
Yeah01
Re: Complex numbers
Hi I got x = 1/4-y and y=1/4-x. What now? Sm I on the right track?
• June 1st 2013, 04:03 PM
Plato
Re: Complex numbers
Quote:

Originally Posted by Yeah01
Hi I got x = 1/4-y and y=1/4-x. What now? Sm I on the right track?

I don't think at you are.

As you can see here, this problem has a rather nice answer.

But when I tried it, the algebra is really messy. From that link, I tried those solutions and they work.
• June 1st 2013, 05:40 PM
Yeah01
Re: Complex numbers
Thanks. What did yr working look like. The answer is good but really want work out how to do it.
• June 1st 2013, 06:40 PM
Plato
Re: Complex numbers
Quote:

Originally Posted by Yeah01
Thanks. What did yr working look like. The answer is good but really want work out how to do it.

Well I really have absolutely no idea how to work it out algebraically.
I taught complex variables on and off for over thirty years. But I have never seen a more difficult problem in this category of problems than this one is.

Clearly $z=\pm 1$ are solutions simply by inspection.

But how one gets the other eight solution remains a puzzle to me.

There may well be a very clever trick that I have not thought of.
• June 1st 2013, 07:35 PM
Gusbob
Re: Complex numbers
I believe the most intuitive way to do this is to look at the pre image of the Joukovski's transformation $z\mapsto z+ \frac{1}{z}$. This is a conformal map, and there are some nice ways you can deal with this. I also believe it can be cranked out algebraically in an acceptably nice way. I'm working on that now.
• June 1st 2013, 08:38 PM
Yeah01
Re: Complex numbers
Wow. Thanks. Looking forward to seeing if someone can do the math.
• June 1st 2013, 08:58 PM
Gusbob
Re: Complex numbers
Let $w=\frac{1}{2}\left(z+\frac{1}{z}\right)$. Then the condition in the question is equivalent to saying $|w|=1$. If $w=\pm 1$, you'll recover the trivial answers $z=\pm 1$

Exercise: If $|w|=1$ and $w\not= \pm 1$, show that $\arg\left(\frac{w-1}{w+1}\right)$ is $\frac{\pi}{2}$ or $\frac{3\pi}{2}$. That is, $\frac{w-1}{w+1}$ is purely imaginary.

HINT: Writing $w=x+iy$, you should find that $\frac{w-1}{w+1}=\frac{2iy}{(x+1)^2+y^2}$

Fun facts: $w-1=\frac{1}{2z}(z-1)^2$ and $w+1=\frac{1}{2z}(z+1)^2$

Hence $\frac{w-1}{w+1}=\left(\frac{z-1}{z+1}\right)^2 =ai$ for some real number a, with some restrictions on $a$.

Now you need some bounds on $|a|= \left|\frac{w-1}{w+1}\right|=\left|\frac{2iy}{(x+1)^2+y^2}\right |$ since you had the restriction $|w|=\sqrt{x^2+y^2}=1$, $y\not=0$

I didn't have time to finish calculating the bounds, but note that $|a|=\left|\frac{\sqrt{2}-1}{\sqrt{2}+1}\right|$ and $|a|= \left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right|$ gives the easier solutions in Plato's wolfram alpha link. It comes out by solving $\left(\frac{z-1}{z+1}\right)^2 =ai$.

Good luck with that. I'll work on the bounds later if I have time.