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Let $f(z) = \frac1 2(z + \frac 1 z)$. I want to find $f(B(0, 1))$

My work so far

Let $w \in \mathbb{C}$

$$f(z) = w \Leftrightarrow \frac 1 2 (z + \frac 1 z) = w$$ $$z + \frac 1 z = 2w$$ $$z^2 - 2wz + 1 =0$$ $$(z - w)^2 - w^2 + 1 = 0$$ $$(z - w)^2 = w^2 - 1$$ $$z = \pm \sqrt{w^2 - 1} +w$$

Now because we want to find $f(B(0, 1))$ we assume that $|z| < 1$. Let's consider first case i.e. when $z = \sqrt{w^2 - 1} + w$:

$$|\sqrt{w^2 - 1} + w|<1$$

If I rewrite $w$ by using $w = x + iy$

$$|\sqrt{(x + iy)^2 - 1} + x + iy| < 1$$

My idea was to rewrite left side of the inequality in form of $x_1 + iy_1$ and then calculate its modulus: $\sqrt{x_1^2 + y_1^2}$. However I find it very hard to find this form of the expression on the left side. Could you please give me a hand in doing so?

Lucian
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1 Answers1

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One possible approach is to use polar coordinates. As shown in Maps circles onto ellipses, every circle of radius $0 < r < 1$ is mapped to an ellipse with width and height $$ a = \frac 12 \left( r + \frac 1 r\right) \, , \, b = \frac 12 \left( \frac 1 r - r\right) \, , $$ centered at the origin. With $r$ varying from $1$ to $0$, the width takes all values in $(1, \infty)$, and the height takes all values in $(0, \infty)$.

This shows that the image of the unit disk is $\hat{\Bbb C} \setminus [-1, 1]$, with $z=0$ being mapped to the point at infinity.

Another approach is to write $$ f(z) = \frac 12 \left( z + \frac 1 z\right) = \frac{\left(\frac{z+1}{z-1}\right)^2 +1}{\left(\frac{z+1}{z-1}\right)^2 -1} $$ as the composition of fractional linear mappings (aka Möbius transformations) and a square.

The first transformation $z \to \frac{z+1}{z-1}$ maps the unit disk onto the left halfplane. This is mapped to the “slit plane” $\Bbb C \setminus (-\infty, 0]$ by squaring, and the last transformation maps this to $\hat{\Bbb C} \setminus [-1, 1]$.

Martin R
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