Functions and limits · Definition
Complex limit
Before you start you need to know
It is the tool that gives meaning to «getting close» in the plane, and complex continuity and the complex derivative are born from it.
Definition
The limit \(L\) of a function \(f\) at \(z_0\) is the complex number that \(f(z)\) gets as close to as desired (\(\varepsilon\)) when \(z\) comes within distance \(\delta\) of \(z_0\), from any direction.
The definition asks that \(|f(z)-L|\) be small whenever \(z\) lies in a small disk around \(z_0\), with \(z\neq z_0\). Since a disk contains points in every direction, the limit demands a single value regardless of path: in \(\mathbb{R}\) there are only two sides, in \(\mathbb{C}\) there are infinitely many directions. The value \(f(z_0)\) plays no role, and \(z_0\) must be an accumulation point of the domain.
The rules are the familiar ones: the limit of a sum, product and quotient (with nonzero limit in the denominator) is the sum, product and quotient of the limits. Moreover \(f=u+iv\) has limit \(L=a+ib\) if and only if \(u\to a\) and \(v\to b\) in the two-real-variable sense, which reduces the computation to limits in \(\mathbb{R}^2\).
To compute, simplify first: \((z^2+4)/(z-2i)=z+2i\) when \(z\neq 2i\), so the limit at \(2i\) is \(4i\). To show that a limit does not exist it suffices to give two paths with different values, as with \(\bar z/z\) at \(0\), which is \(1\) on the real axis and \(-1\) on the imaginary axis. The converse fails: agreeing along several paths does not prove existence.
A useful criterion: \(f(z)\to L\) is equivalent to \(|f(z)-L|\to 0\), and if \(|f(z)-L|\le C|z-z_0|\) the limit is \(L\). With the limit in hand, continuity is stated by saying that the limit equals the value of the function.
Think about it
Why does \(\lim_{z\to 0}\dfrac{z}{|z|}\) not exist?
On the positive real axis it equals \(1\) and on the negative one \(-1\); also on \(z=it\) (\(t>0\)) it equals \(i\). There are paths with different values.
What is \(\lim_{z\to i}(z^2+z)\)?
It is a polynomial, so substitute: \(i^2+i=-1+i\).
Exercises 10
Compute the following limits
- \(\displaystyle \lim_{z\to 1+i}(z^2+1)\)
- \(\displaystyle \lim_{z\to 2i}\dfrac{z^2+4}{z-2i}\)
- \(\displaystyle \lim_{z\to i}\dfrac{z^4-1}{z-i}\)
- \(\displaystyle \lim_{z\to 0}\dfrac{z^2}{|z|}\)
Problems
- If \(\lim_{z\to z_0}f(z)=1+i\) and \(\lim_{z\to z_0}g(z)=2-i\), what is \(\lim_{z\to z_0}f(z)g(z)\)?
- About \(\lim_{z\to 0}\dfrac{\bar z}{z}\) it is claimed that:
- Let \(f(z)=\dfrac{x^2-y^2}{x^2+y^2}\) with \(z=x+iy\). Which is correct about \(\lim_{z\to 0}f(z)\)?
- The value \(f(z_0)\) of a function at the point where the limit is taken:
- If \(|f(z)-(2+i)|\le 3|z|\) for all \(z\), what follows about \(\lim_{z\to 0}f(z)\)?
- Two paths toward \(z_0\) give the same limiting value \(L\). Then:
What comes next
References
- Elias M. Stein, Rami Shakarchi, Complex Analysis, 1st edition, Princeton University Press, 2003. Cap. 1, §2.1
- James Ward Brown, Ruel V. Churchill, Complex Variables and Applications, 9th edition, McGraw-Hill, 2014. §15–16
- Lars V. Ahlfors, Complex Analysis, 3rd edition, McGraw-Hill, 1979. Cap. 2, §1.1