4.2 Limits of Complex Functions

Definition 4.8. Let \(D\subseteq \mathbb {C}\). Let \(f: D\longrightarrow \mathbb {C}\) be a function and let \(z_0\) be a limit point of \(D\). A complex number \(L\) is said to be a limit of the function \(f\) as \(z\) approaches \(z_0\) if \(\forall \varepsilon > 0 \,\exists \, \delta > 0 \ni \left |f(z) - L\right | < \varepsilon \) whenever \(z\in D\) and \(0 < \left |z - z_0\right | < \delta \).

If \(\forall \varepsilon > 0 \,\exists \, \delta > 0 \ni \) whenever \(z \in D \cap B'(z_0,\delta ), \, f(z) \in B(L;\varepsilon )\) \[\lim \limits _{z \rightarrow z_0} f(z) = L\]

Example 4.9.

\(f(z) = \begin {cases} 3z^2, & \left |z\right | < 1\\\\ 3, & \left |z\right |= 1\\ \end {cases}\)

\(f: \overline {B}(0;1) \longrightarrow \mathbb {C}\)

\(\lim \limits _{z \rightarrow 1} f(z) = \mathbb {C}\)

\(\lim \limits _{z \rightarrow z_0} f(z)\) does not exist for any \(z_0\) with \(\left |z_0\right | = 1\) and \(z_0 \neq \pm 1\).

Example 4.10. Show that \(\lim \limits _{z \rightarrow 0}\frac {z}{\overline {z}}\) does not exist.

Solution
We show that this limit does not exist by finding two different ways of letting \(z\) approach 0 that yields different values of \(\lim \limits _{z \rightarrow 0}\frac {z}{\overline {z}}\).
First let \(z\) approach 0 along the real axis, that is \(z = x + 0i\). Then for these points, we have \[\lim \limits _{z \rightarrow 0}\frac {z}{\overline {z}}= \lim \limits _{z \rightarrow 0}\frac {x + i0}{x - i0} = \lim \limits _{z \rightarrow 0} 1 = 1\] On the other hand, if we let \(z\) approach 0 along the imaginary axis then \(z = 0 + iy\). And for this approach, we have \[\lim \limits _{z \rightarrow 0}\frac {z}{\overline {z}} = \lim \limits _{z \rightarrow 0}\frac {0 + iy}{0 - iy} = \lim \limits _{z \rightarrow 0} (-1) = -1\] Hence the two limits are different and we conclude that \(\lim \limits _{z \rightarrow 0}\frac {z}{\overline {z}}\) does not exist.

Example 4.11. Use the \(\varepsilon - \delta \) definition to show that \(\lim \limits _{z \rightarrow (1 + i)}(2 + i) z = 1 + 3i\).

Solution
According to the definition, \(\lim \limits _{z \rightarrow (1 + i)}(2 + i) z = 1 + 3i\) if for every \(\varepsilon > 0\), there exist \(\delta > 0\) such that \(\left |(2 + i)z - (1 + 3i)\right | < \varepsilon \) whenever \(0 < \left |z - (1 + i)\right | < \delta \). To do this, we need to find the value of \(\delta \) for any given \(\varepsilon \), that is given any value of \(\varepsilon \), we must find a real number \(\delta > 0\) such that if \(0 < \left |z - (1 + i)\right | < \delta \), then \(\left |(2 + i) z - (1 + 3i)\right | < \varepsilon \).

\begin {align*} \text {Now}\quad \left |f(z) - L\right | & = \left |(2 + i)z - (1 + 3i)\right | = \left |(2 + i) -\Big (z - \frac {1 + 3i}{2 + i}\Big )\right |\\ & = \left |2 + i\right |\,\left |z - \frac {1 + 3i}{2 + i}\right | \end {align*}

Now \(\left |2 + i\right | = \sqrt {5}\) and \(\frac {1 + 3i}{2 + i} = 1 + i\). So we have

\begin {align*} \left |f(z) - L\right | & = \sqrt {5}\,\left |z - (1 + i)\right | < \varepsilon \\ & \implies \left |z - (1 + i)\right | < \frac {\varepsilon }{\sqrt {5}} \end {align*}

Then we can choose \(\delta = \frac {\varepsilon }{\sqrt {5}}\)

Theorem 4.12. Suppose that \(f(z) = u(x,y) + i v(x,y), \quad z_0 = x_0 + i y_0\) and \(L = u_0 + i v_0\). Then \(\lim \limits _{(x,y) \rightarrow z_0}f(z) = L\) if and only if \(\lim \limits _{(x,y)\rightarrow (x_0, y_0)} u(x,y) = u_0\) and \(\lim \limits _{(x,y)\rightarrow (x_0, y_0)} v(x,y) = v_0\)

Conversely, suppose both real limits hold. Given \(\varepsilon >0\) choose \(\delta _1,\delta _2\) making \(\left |u-u_0\right |<\frac {\varepsilon }{2}\) and \(\left |v-v_0\right |<\frac {\varepsilon }{2}\), and let \(\delta =\min (\delta _1,\delta _2)\). The third inequality then gives \[\left |f(z)-L\right |\leq \left |u-u_0\right |+\left |v-v_0\right |<\varepsilon .\] So the complex limit holds.

The theorem is what licenses every later computation of a complex limit by treating the real and imaginary parts separately. □

Example 4.13. Use the theorem to compute \(\lim \limits _{z \rightarrow (1 + i) }(z^2 + i)\)

Solution
Since \(f(z) = (z^2 + i) = (x + iy)^2 + i = x^2 - y^2 + (2xy + 1)i\). So let \(u(x,y) = x^2 - y^2\) and \(v(x,y) = 2xy + 1\) and \(z_0 = 1 + i\), identifying \(x_0 = 1;\, y_0 = 1\)
We find \(u_0\) and \(v_0\) by computing these two real parts. \(u_0 = \lim \limits _{(x,y) \rightarrow (1,1)} (x^2 - y^2 )\) and
\(v_0 = \lim \limits _{(x,y) \rightarrow (1,1)} (2xy + 1)\). Hence \begin {align*} u_0 & = \lim \limits _{(x,y) \rightarrow (1,1)} (x^2 - y^2 ) = 1^2 - 1^2 = 0\\\\ v_0 & = \lim \limits _{(x,y) \rightarrow (1,1)} (2xy + 1) = 2 + 1 = 3 \end {align*}

So \(L = u_0 + iv_0 = 0 + 3i = 3i\)

Theorem 4.14. Suppose that \(f\) and \(g\) are complex functions. If \(\lim \limits _{z\rightarrow z_0} f(z) = L\) and \(\lim \limits _{z\rightarrow z_0} g(z) = M\), then

1.: \(\displaystyle {\lim \limits _{z\rightarrow z_0}C( f(z)) = CL}\,,\,\) a complex number.
2.: \(\displaystyle {\lim \limits _{z\rightarrow z_0} (f(z)\pm g(z)) = L\pm M}\)
3.: \(\displaystyle {\lim \limits _{z\rightarrow z_0} (f(z)\cdot g(z) )= L\cdot M}\)
4.: \(\displaystyle {\lim \limits _{z\rightarrow z_0} \frac {f(z)}{g(z)} = \frac {L}{M}}\)

Proof. The definition of limit is word for word the real one with modulus in place of absolute value, so the real proofs transfer unchanged. We give (3) and (4), since those are where the work is.

(3) The product

Write \[f g-LM=f(g-M)+M(f-L).\] Because \(f\) has a limit it is bounded near \(z_0\): taking \(\varepsilon =1\) in the definition gives \(\left |f(z)\right |<\left |L\right |+1\) on some punctured disc. Hence \[\left |fg-LM\right |\leq \left (\left |L\right |+1\right )\left |g-M\right | +\left |M\right |\left |f-L\right | ,\] and both terms can be made smaller than \(\frac {\varepsilon }{2}\) by taking \(z\) close enough to \(z_0\).

(4) The quotient

It suffices to prove \(\dfrac {1}{g}\rightarrow \dfrac {1}{M}\) and then apply (3). Since \(M\neq 0\), taking \(\varepsilon =\frac {\left |M\right |}{2}\) gives a punctured disc on which \(\left |g\right |>\frac {\left |M\right |}{2}\), so \(g\) does not vanish there and \[\left |\frac {1}{g}-\frac {1}{M}\right |=\frac {\left |M-g\right |} {\left |g\right |\left |M\right |}\leq \frac {2}{\left |M\right |^{2}}\left |g-M\right |,\] which tends to \(0\). Parts (1) and (2) are the same argument with less bookkeeping.

Note that (4) needs \(M\neq 0\); nothing here says what happens when the limit of the denominator is zero, and that case is exactly where poles come from. □

Example 4.15. Compute the following limits.

1.: \(\displaystyle {\lim \limits _{z \rightarrow i}\frac {(3 + i)z^4 - z^2 + 2z}{z + 1}}\)
2.: \(\displaystyle {\lim \limits _{z \rightarrow (1 + \sqrt {3}i)} \frac {z^2 - 2z + 4}{z - 1 -\sqrt {3}i}}\)

Solution.

1.: The denominator does not vanish at \(z=i\), so the quotient is continuous there and the limit is found by substitution. Using \(i^2=-1\) and \(i^4=1\), \[\frac {(3+i)(1)-(-1)+2i}{i+1}=\frac {4+3i}{1+i} =\frac {(4+3i)(1-i)}{(1+i)(1-i)}=\frac {4-4i+3i+3}{2}=\frac {7}{2}-\frac {1}{2}i .\]
2.: Here the denominator does vanish, so the numerator must be factored. Its roots are \[z=\frac {2\pm \sqrt {4-16}}{2}=1\pm \sqrt {3}i ,\] so \(z^2-2z+4=\big (z-1-\sqrt {3}i\big )\big (z-1+\sqrt {3}i\big )\). Cancelling the common factor — legitimate because \(z\neq 1+\sqrt 3 i\) while the limit is taken — leaves \[\lim _{z\rightarrow 1+\sqrt {3}i}\big (z-1+\sqrt {3}i\big )=2\sqrt {3}i .\]

Limits Involving Infinity
We can allow \(L\) to be \(\infty \).

Definition 4.16. Let \(D\subseteq \mathbb {C}\) and let \(f: D\longrightarrow \mathbb {C}\) and let \(z_0\) be a limit point of \(D\). We say that \(\,\displaystyle {\lim \limits _{z\rightarrow z_0} f(z) = \infty }\,\) if given \(M> 0 \,\exists \,\delta > 0 \ni \forall z \in D\) and \(0 < \left |z - z_0\right | < \delta ,\, \left |f(z)\right |> M\) ( is arbitrary large).

We say \(\displaystyle {\lim \limits _{z\rightarrow \infty } f(z) = L}\) if given \(\varepsilon > 0\,\exists \, M> 0 \ni \left |f(z) - L\right | < \varepsilon \) whenever
\(z \in \big \{W:\,\left |W\right |> M\big \}\,\left |z\right | > M\).

Example 4.17. Show that \(\displaystyle {\lim \limits _{z\rightarrow \infty } \frac {3z^2}{(1 + i) z^2 - z + 2} = \frac {3}{1 + i}}\,,\,\) for \(\left |z\right | > M\quad \) (\(M\) is large).

\begin {align*} \left |\frac {3z^2}{(1 + i)z^2 - z + 2} - \frac {3}{1 + i}\right | & = \left |\frac {3z - 6}{(1 + i) ((1 + i)z^2 - z + 2)}\right |\\\\ & = \left |\frac {3/z - 6/z^2}{(1 + i)\Big ((1 + i)- \frac {1}{z} + \frac {2}{z^2}\Big )}\right |\quad \cdots \quad (*) \end {align*}

\(\implies \,\displaystyle {\left |(1 + i) - \frac {1}{z} + \frac {2}{z^2}\right |\geq \sqrt {2}- \left |\frac {1}{z} - \frac {2}{z^2}\right |\,}\) when \(\left |z\right |\geq M\) then \(\frac {1}{\left |z\right |} \leq \frac {1}{M}\)

\begin {align*} \left |\frac {1}{z} - \frac {2}{z^2}\right | & \leq \frac {1}{M} + \frac {2}{M^2}\\\\ \implies \, -\left |\frac {1}{z} - \frac {2}{z^2}\right | & \geq -\frac {1}{M} - \frac {2}{M^2}\\\\ \implies \, \left |(1 + i)- \frac {1}{z} + \frac {2}{z^2}\right | & \geq \sqrt {2} + \frac {1}{M} + \frac {2}{M^2} \end {align*}

\begin {align*} (*)& \leq \frac {3/M + 6/M^2}{\sqrt {2}\Big (\sqrt {2} + 1/M + 2/M^2\Big )} = \frac {3}{\sqrt {2}}\Bigg (\frac {M + 2}{\sqrt {2}M^2 + M + 2}\Bigg )\\ &\leq \frac {3}{\sqrt {2}}\cdot \frac {2M}{2\sqrt {2}M^2} = \frac {3}{2M}\qquad (2M \leq \sqrt {2}M^3) \end {align*}

for large \(M \, (*)\) is arbitrary small. Thus the limit.

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