Complex Analysis

Riemann Sphere

The Point to Infinity

For some purposes it is convenient to introduce the point to 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.

Sorry, the applet is not supported for small screens. Rotate your device to landscape. Or resize your window so it's more wide than tall.

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.

Now we can introduce the following limit concepts:

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

Example 1: If $f(z)=1/z^2$, 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 $f(z)=1/(z-3)$, 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: For $f(z)=e^z$, defined on $D=\{z\,:\,\textbf{Re}(z)>0\}$, we have $$\lim_{z\rightarrow \infty}f(z)=\infty.$$ In fact, for any given $M>0$ we have $$\left|f(z)\right|=\left|e^z\right|=e^x>M$$ whenever $x>\ln M.$ Hence, it suffices to take $R>\max\{0,\ln R\}$.

It is worth to mention that for $f(z)=e^z$ defined on $D_1=\{z\,:\,\textbf{Re} (z)=0\}$, i.e., for $f(iy)=e^{iy}$, there is no limit as $z=iy\rightarrow \infty $, because along the imaginary axis, $e^{iy} = \cos y+i\sin y$ is periodic and not constant.

Finally, for $f(z)=e^z$ defined on $D_2=\{z\,:\,\textbf{Re}(z)<0\}$ we have that $$\lim_{z\rightarrow \infty}f(z)=0.$$

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

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

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