Riemann Sphere

The Point at infinity

For some purposes it is convenient to introduce the point at infinity, denoted by $\infty,$ in addition to the points $z\in \mathbb C.$ We must be careful in doing so, because it can lead to confusion and abuse of the symbol $\infty.$ However, with care it can be useful, if we want to be able to talk about infinite limits and limits at infinity.

In contrast to the real line, to which $+\infty$ and $-\infty$ can be added, we have only one $\infty$ for $\mathbb C.$ The reason is that $\mathbb C$ has no natural ordering as $\mathbb R$ does. Formally we add a symbol $\infty$ to $\mathbb C$ to obtain the extended complex plane, denoted by $\mathbb C^*=\mathbb C \cup \{\infty\},$ and define operations with $\infty$ by the rules

\begin{eqnarray*} z+\infty&=&\infty\\ z\cdot \infty&=&\infty \quad \quad \text{provided } z\neq 0\\ \infty+\infty&=&\infty\\ \infty\cdot\infty&=& \infty\\ \frac{z}{\infty}&=&0 \end{eqnarray*}

for $z\in \mathbb C.$ Notice that some operations are not defined:

$$\frac{\infty}{\infty}\,,\quad 0\cdot \infty\,,\quad \infty-\infty\,,$$

and so forth are for the same reasons that they are in the calculus of real numbers.

The extended complex plane can be mapped onto the surface of a sphere whose south pole corresponds to the origin and whose north pole to the point $\infty.$ All other points of the complex plane can be mapped in a one-to-one fashion to points on the surface of the sphere by using the following construction. Connect the point $z$ in the plane with the north pole using a straight line. This line intersects the sphere at the point $P(z).$ In this way each point $z= x+iy$ on the complex plane corresponds uniquely to a point $P(z)$ on the surface of the sphere. This construction is called the stereographic projection and is illustrated in the following applet.

In the following applet we can observe the unit sphere whose south pole corresponds to the origin of the $z$ plane. Drag the point defined on the $z$ plane, or the sliders, to explore the behaviour of the point $P(z)$ on the sphere.

The extended complex plane is sometimes referred to as the compactified (closed) complex plane. It is often useful to view the complex plane in this way, and knowledge of the construction of the stereographic projection is valuable in certain advanced treatments.

Infinite Limits

Now we can introduce the following limit concepts:

$\displaystyle \lim_{z\rightarrow \infty}f(z)=z_0$ means: For any $\varepsilon>0,$ there is an $R>0$ such that $$|z|> R \quad \Rightarrow \quad \left|\,f(z)-z_0\right| < \varepsilon.$$
$\displaystyle \lim_{z\rightarrow z_0}f(z)=\infty$ means: For any $R>0,$ there is a $\delta>0$ such that $$0 \lt |z-z_0|< \delta \quad \Rightarrow \quad |\,f(z)|>R.$$
$\displaystyle \lim_{z\rightarrow \infty}f(z)=\infty$ means: For any $M>0,$ there is an $R>0$ such that $$|z|> R \quad \Rightarrow \quad |\,f(z)|>M .$$

Example 1: If $f(z)=\dfrac{1}{z^2},$ for $z\neq 0,$ then $$\lim_{z\rightarrow \infty}f(z)=0.$$ In fact, given $\varepsilon>0$ we have $$\left|\frac{1}{z^2}-0\right|=\frac{1}{\left|z^2\right|}=\frac{1}{\left|z\right|^2} <\varepsilon$$ by taking $$\left|z\right|>\frac{1}{\sqrt{\varepsilon}}=R.$$

Example 2: Let $f(z)=\dfrac{1}{z-3},$ for $z\neq 3.$ Then $$\lim_{z\rightarrow 3}f(z)=\infty.$$ In fact, for any given $R>0$ the inequality $$\frac{1}{\left|z-3\right|}>R$$ holds whenever $$0<\left|z-3\right| < \frac{1}{R}=\delta.$$

Example 3: Now let $f(z) = \dfrac{2z^3-1}{z^2+1},$ for $z\neq -1.$ Then \[ \lim_{z\to \infty} f(x) = \infty. \] There are different ways to prove this limit. A first attempt is to find the right inequality. First, let's begin assuming that $|z|\gt 1.$ This implies the following inequalities:

\begin{eqnarray} \left|2z^3+1\right|&\geq & 2\left|z^3\right|-1\gt 0\label{property}\\ \left|z^2\right|+\left|z^2\right|&\geq& \left|z^2\right|+1.\label{property2} \end{eqnarray}

Notice also that

$$ \left|z^2+1\right|\leq \left|z\right|^2+1\;\; \implies\;\; \frac{1}{\left|z\right|^2+1}\leq \frac{1}{\left|z^2+1\right|}. $$

Using inequality (\ref{property}) we have

$$ \frac{2\left|z^3\right|-1}{\left|z^2+1\right|} \geq \frac{2\left|z^3\right|-1}{\left|z\right|^2+1}. $$

And from inequality (\ref{property2}) we obtain

\[ \left|\dfrac{2z^3-1}{z^2+1}\right|\geq \frac{2\left|z\right|^3-1}{\left|z\right|^2+1}\geq \frac{\left|z\right|^3}{\left|z\right|^2+ \left|z\right|^2}= \frac{|z|}{2}. \]

Hence, for any $M>0,$ we can choose $R = 1+ 2M$ to conclude that

\[ \left|\dfrac{2z^3-1}{z^2+1}\right| > M \;\text{ whenever } \;|z|>R. \]

Now we can write the proof in short.

Proof: For any $M\gt0,$ choose $R = 1+ 2M$ such that $|z|\gt R.$ Then

\begin{eqnarray*} \left|\dfrac{2z^3-1}{z^2+1}\right|&\geq& \frac{2\left|z\right|^3-1}{\left|z\right|^2+1} &=& \frac{|z|}{2}\gt \frac{1+2M}{2}\gt M. \quad \blacksquare \end{eqnarray*}

Remark: By the way the choice $R=1+2M$ is obviously not unique. We just need $R\geq 1$ to ensure the second inequality in the proof above, and $R\geq 2M$ to ensure the final inequality. Even if you end up with $|f(z)|>M/2$ at the end, that is an equivalent definition since we have a universal quantifier “for any $M\gt 0$”.

There is an easier way to calculate the limits from Examples 1-3. The following theorem provides a very useful method.

If $z_0$ and $w_0 $ are points in the $z$ and $w$ planes, respectively, then

\begin{eqnarray*} \lim_{z\rightarrow z_0}f(z)=\infty \quad\text{if and only if}\quad \lim_{z\rightarrow z_0}\frac{1}{f(z)}=0,\label{limiinfty01}\\ \lim_{z\rightarrow \infty }f(z)=w_0 \quad\text{if and only if}\quad \lim_{z\rightarrow 0}f\left(\frac{1}{z}\right)=w_0,\label{limiinfty02}\\ \lim_{z\rightarrow \infty }f(z)=\infty \quad\text{if and only if}\quad \lim_{z\rightarrow 0}\frac{1}{f(1/z)}=0.\label{limiinfty03} \end{eqnarray*}

Using this result, we can easily find that

$$\lim_{z\rightarrow -1}\frac{iz+3}{z+1}=\infty \quad \text{since}\quad \lim_{z\rightarrow -1}\frac{z+1}{iz+3}=0$$

and

$$\lim_{z\rightarrow \infty}\frac{2z+i}{z+1}=2 \quad \text{since}\quad \lim_{z\rightarrow 0}\frac{(2/z)+i}{(1/z)+1}=\frac{2+iz}{1+z}=2.$$

Furthermore,

$$\lim_{z\rightarrow \infty}\frac{2z^3-1}{z^2+1}=\infty \quad \text{since}\quad \lim_{z\rightarrow 0}\frac{(1/z^2)+1}{(2/z^3)-1}=\frac{z+z^3}{2-z^3}=0.$$

Although we have successfully used techniques from calculus to compute complex-valued limits, real-variable intuition may not always apply. For example, $\lim_{z\to \infty} e^{-z}$ is not 0; in fact, this limit does not exist.