Geometric proof that $\left|\frac{i-z}{i+z}\right|<1$ for $z\in \mathbb{H}$

I just read a proof in a complex analysis book that says there exists a biholomorphic (conformal) map between $\mathbb{H}$ and $\mathbb{D}$ and I don't understand one step in the proof, namely the norm of the map (in the title) is less than $1$, using a purely "geometric" argument (i.e. a picture was in the book). I don't understand "geometric division", so I don't immediately see the proof "geometrically", please I do NOT want an algebraic answer (this is fairly straight forward anyway).

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5 Answers

Note that $|i-z|$ is the distance between $z$ and $i$, and that $|i+z|$ is the distance between $z$ and $-i$. The inequality says that $z$ lies closer to $i$ than to $-i$, which is true for points in the upper halfplane.

To make this even more clear, write $\bigl|{i-z \over i+z}\bigr|={|i-z|\over |i+z|}<1$, and multiply through by the denominator $|i+z|$.

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The real axis is in the middle between $i$ and $-i$, hence is the locus of all points having the same distance from $i$ and $-i$. The rest of $\mathbb C$ splits into those points closer to $i$ than to $-i$ (the upper half plane) and those closer to $-i$ (lower half plane).

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Basically you want to prove that any point in $\mathbb{H}$ is closer to $i$ than to $-i$. Draw the triangle with vertices at $i, -i, z$ and observe that the angle at $-i$ is smaller than the angle at $i$. he easiest way to prove this is by extending the line through $i z$ till it intersects the real axix at $y$ and compare the angles in the triangle $i,-i,z$ with the angles in $i,-i,y$.

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The norms of $i - z$ and $i+z$ are the squares of the distances between $z$ and $\pm i$. A point $z$ lies in the upper half plane if and only if it is strictly closer to $i$ than to $-i$, so that this ratio has norm strictly less than $1$ and so lies in the unit disk.

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|z-i|=|z+i| is the set of points equidistant from $i$ and $-i$, the real line. |z-i|<|z+i| is the set of points closer to $i$ than $-i$, the upper half plane.

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