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?
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)?
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 $\displaystyle z=\pm 1$ 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.
I believe the most intuitive way to do this is to look at the pre image of the Joukovski's transformation $\displaystyle z\mapsto z+ \frac{1}{z}$. 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.
Let $\displaystyle w=\frac{1}{2}\left(z+\frac{1}{z}\right)$. Then the condition in the question is equivalent to saying $\displaystyle |w|=1$. If $\displaystyle w=\pm 1$, you'll recover the trivial answers $\displaystyle z=\pm 1$
Exercise: If $\displaystyle |w|=1$ and $\displaystyle w\not= \pm 1$, show that $\displaystyle \arg\left(\frac{w-1}{w+1}\right)$ is $\displaystyle \frac{\pi}{2}$ or $\displaystyle \frac{3\pi}{2}$. That is, $\displaystyle \frac{w-1}{w+1}$ is purely imaginary.
HINT: Writing $\displaystyle w=x+iy$, you should find that $\displaystyle \frac{w-1}{w+1}=\frac{2iy}{(x+1)^2+y^2}$
Fun facts: $\displaystyle w-1=\frac{1}{2z}(z-1)^2$ and $\displaystyle w+1=\frac{1}{2z}(z+1)^2$
Hence $\displaystyle \frac{w-1}{w+1}=\left(\frac{z-1}{z+1}\right)^2 =ai$ for some real number a, with some restrictions on $\displaystyle a$.
Now you need some bounds on $\displaystyle |a|= \left|\frac{w-1}{w+1}\right|=\left|\frac{2iy}{(x+1)^2+y^2}\right |$ since you had the restriction $\displaystyle |w|=\sqrt{x^2+y^2}=1$, $\displaystyle y\not=0$
I didn't have time to finish calculating the bounds, but note that $\displaystyle |a|=\left|\frac{\sqrt{2}-1}{\sqrt{2}+1}\right| $ and $\displaystyle |a|= \left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right|$ gives the easier solutions in Plato's wolfram alpha link. It comes out by solving $\displaystyle \left(\frac{z-1}{z+1}\right)^2 =ai$.
Good luck with that. I'll work on the bounds later if I have time.