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Let’s consider $w = z + \frac1z$

and I want to find out the effect of this transformation on the circle $| z | = 1$

A method that seems effective

if $z = x + iy$ and $|z| = 1$ then $1/z = z^*$

So $w = z + \frac1z = z + z^* = x + iy + x – iy = 2x$

So the image, $w$, runs along the real axis and further, since $-1 \leq x \leq +1$ then $-2 \leq w \leq +2$. That is the image is constrained to run between $-2$ and $+2$ on the real axis.

The very useful Geogebra visualisation allowed me to input the transformation then drag $z_1$ roughly around the path of the circle $|z| = 1$. The image point $z_2$ in the Geogebra, which plays the role of w, traced a path along the real axis $(-2,+2)$ as the algebra above showed.

Now my challenge is to perform the transformation on $|z| = 2$.

Using the Geogebra visualisation, the image $z_2$ appears as if it could be an ellipse.

If I do similar algebra to above starting with $|z|= 2$ and $z = x + iy$

I get $w = \frac{5}{4}x + i\frac{3}{4}y$

And I can’t work out how to interpret / represent this in coordinate or polar form in order to create an “equation” for the image.

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    you have shown that $r=\frac{5}{4}x$ and $s=\frac{3}{4}y$ where $(r,s)$ is the image of $(x,y)$. Therefore $\left(\frac{4}{5}r\right)^2+\left(\frac{4}{3}s\right)^2=4$ – 2017-02-26
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    Aha! It fits ! Thank-you. – 2017-02-26

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In general, the image of circle $\,|z|=r \gt 0\,$ is an ellipse with foci at $\,\pm2\,$, except for the case $r=1$ when the ellipse degenerates to segment $\,[-2,2]\,$ on the real axis.

Let $u$ be a square root of $z\,$, thus $\,u^2=z\,$ and $\,|u|^2=|z|=r\,$. Then:

$$ w+2=u^2+\frac{1}{u^2}+2=\left(u+\frac{1}{u}\right)^2 $$

Therefore:

$$ |w+2|=\left|u+\frac{1}{u}\right|^2 = \left(u+\frac{1}{u}\right)\left(\bar u+\frac{1}{\bar u}\right) = |u|^2+\frac{1}{|u|^2} + \frac{u}{\bar u} + \frac{\bar u}{u}=r+\frac{1}{r} + \frac{u}{\bar u} + \frac{\bar u}{u} \tag{1} $$

Redoing the same steps for $\,w-2\,$ gives:

$$ |w-2| =r+\frac{1}{r} - \frac{u}{\bar u} - \frac{\bar u}{u} \tag{2} $$

Adding $\,(1)+(2)\,$:

$$ |w+2|+|w-2|=2\left(r+\frac{1}{r}\right) $$

The latter is the equation of an ellipse with foci at $\,\pm 2\,$ and semi-major axis $\,r+\cfrac{1}{r} \ge 2\,$. The equality holds (only) for $r=1\,$, in which case the ellipse degenerates to the real segment $[-2,2]\,$.

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    Thanks for the detailed answer. I learned a lot from it. However, I have difficulty understanding your step $|u + \frac{1}{u}|^2 = (u + \frac{1}{u})(\overline{u} + \frac{1}{\overline{u}})$. I have tried substituting u = x +iy and also u = polar form but the algebra gets very messy and I can't see that they are equivalent. I suppose I am not seeing some obvious equivalence here between the two forms of the expression. – 2017-02-28
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    @CliveLong $|v|^2=v \bar v$ for any $\forall v \in \mathbb{C}\,$ (in fact, that's the definition of the complex magnitude in some contexts). If $v = u+1/u$ then $\bar v = \bar u + 1/\bar u$ which gives the equality you quote. – 2017-02-28
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    Thanks, that makes sense. The final detail for me is this. Is the following correct? To use $|v|^2 = v\overline{v}$ in this case I need to show $\overline{u + \frac{1}{u}} = (\overline{u} + \frac{1}{\overline{u}})$ (i). Which is $\overline{u + \frac{1}{u}} = \frac{\overline{u^2 + 1}}{\overline{u}}$ (ii). But also, $\overline{u} + \frac{1}{\overline{u}} =\frac{(\overline{u})^2 + 1}{\overline{u}}$ (iii). So combining (i) , (ii) and (iii) it reduces to showing $\overline{u^2 + 1} = (\overline{u})^2 + 1$ (iv). – 2017-02-28
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    And then (iv) can be shown be making u = x + iy and expanding it all out and comparing real and imaginary parts? – 2017-02-28
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    @CliveLong $\overline{u+1/u}=\bar u + 1/\bar u$ follows from the fundamental properties of the complex conjugate: $\,\overline{a+b}=\bar a + \bar b\,$ and $\overline{ab}=\bar a\,\bar b\,$, with the latter also implying that $\overline{1/a}=\overline{1} / \bar a = 1 / \bar a\,$. – 2017-02-28
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    Neat.We like neat. – 2017-02-28