Complex Analysis

Lecture 17

Derivatives of functions \(w(t)\)

Consider a \(\C\)-valued functions of a real variable: \[w(t) = u(t) + i v(t).\] with \(t\in \R\).


\[w'(t) = u'(t) + i v'(t).\]

Derivatives of functions \(w(t)\)

Standard differential laws for functions of a real variable apply:

\((*)\) \(\left( c \cdot w \right)^{\prime} = c \cdot w^{\prime}\),   \(c\in\C\);

\(\quad\) \(\left( w_1 \pm w_2 \right)^{\prime} = w_1^{\prime} \pm w_2^{\prime}\);

\(\quad\) \(\ds \frac{d}{dt}e^{ct} = ce^{ct}\)     ... etc.

\((*)\) Product rule, quotient rule, etc.

Integration of functions \(w(t)\)

Definite and indefinite integrals of such functions:

\((*)\) \(\ds \int_a^b w\left(t\right) dt = \int_a^b u\left(t\right)dt + i \int_a^b v\left(t\right)dt, \quad a,b\in \R\).     (🪁)

i. e. \[\Re\left( \int_a^b w\left(t\right) dt \right) = \int_a^b \Re\left( w(t) \right) dt\qquad (1)\]

\[\Im\left( \int_a^b w\left(t\right) dt \right) = \int_a^b \Im\left( w(t) \right) dt\qquad (2)\]

Integration of functions \(w(t)\)

\(\int_0^{\infty} w\left(t\right)dt\) is defined analogously.

(🪁) certanly makes sense if \(w\) is continuous,

i. e. \[w\in \mathcal C^0\big([a,b]\big).\]

Piece-wise functions

Indeed, ok for so-called piece-wise continuous functions on \([a,b]\), i.e. \(w\) such that there exist \[c_1 \lt c_2 \lt \cdots \lt c_n\in (a,b) \quad \text{ such that}\]

  1. \(w\) is continuous on each of \[(a,c_1), (c_1,c_2), \ldots, (c_n,b)\]
  2. \(\ds \lim_{t\ra c_j^-} w\left(t\right)\), \(\ds \lim_{t\ra c_j^+} w\left(t\right)\) both exist (they may or may not coincide).
  3. \(\ds \lim_{t\ra a^+} w\left(t\right)\), \(\ds \lim_{t\ra b^-} w\left(t\right)\) both exist (Here "exist" means "exist for \(u\) & \(v\)").

Graph of a piece-wise continuous function

Integration of functions \(w(t)\)

Suppose there exists \(W(t) = U(t) + i V(t)\) such that \(W' = w\) on \([a,b]\).

Then the Fundamental Theorem of Calculus holds in the form \[ \int_a^b w\left(t\right)dt = W(b)-W(a). \]

Integral estimate

Next estimate is crucial: Suppose \(w = u+iv \) is
piece-wise continuous (pwc) on \([a,b]\)

Then \[ \left| \int_a^b w\left(t\right)dt \right| \leq \int_a^b \left| w\left(t\right) \right|dt \qquad (3) \]

Integral estimate

Proof: If \(\int_a^b w\left(t\right)dt = 0\),

\(\text{LHS}\) of \((3) = 0\) and \(\text{RHS}\geq 0 \),

so done.

Integral estimate

Proof (cont): Otherwise, there exists \(r>0\) and \(\theta_0\in \R\) such that \[ \int_a^b w\left(t\right)dt = re^{i\theta_0} \qquad (3)' \] \[ \Rightarrow \left| \int_a^b w\left(t\right)dt \right| = r \qquad (4) \]

Integral estimate

Proof (cont): \((3)'\cdot e^{-i\theta_0}\)

\(\Rightarrow r = \ds \int_a^b e^{-i\theta_0} w\left(t\right)dt \)

\( \qquad = \Re\left( \ds \int_a^b e^{-i\theta_0} w\left(t\right)dt \right)\) (since \(r\in \R\))

\( \qquad = \ds \int_a^b \Re\left( \ds e^{-i\theta_0} w\left(t\right) \right) dt\) \(\quad (5)\)

Integral estimate

Proof (cont): \[ \Re\left( \ds e^{-i\theta_0} w\left(t\right) \right) \leq \left| e^{-i\theta_0} w\left(t\right) \right| = |w(t)| \]

Combine this with \((4)\) & \((5)\)

\[ \left| \int_a^b w\left(t\right)dt \right| = r \leq \int_a^b \left| w\left(t\right) \right|dt \qquad \square \]


A contour is a parametrised curve in \(\C\).

Given \(x(t)\), \(y(t)\) continuous \([a,b]\ra \R\),

\[ z(t) = x(t) +i y(t), \quad a\leq t \leq b \] defines an arc in \(\C\).

This is both a set of points (the image \(z\left([a,b]\right)\)), called the trace of the arc, and also a recipe for drawing it/parametrisation.