MATH3401
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 \(t\in \R\).
Define:
\[w'(t) = u'(t) + i v'(t).\]
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 \[c_1 \lt c_2 \lt \cdots \lt c_n\in (a,b) \quad \text{ such that}\]
- \(w\) is continuous on each of \[(a,c_1), (c_1,c_2), \ldots, (c_n,b)\]
- \(\ds \lim_{t\ra c_j^-} w\left(t\right)\), \(\ds \lim_{t\ra c_j^+} w\left(t\right)\) both exist (they may or may not coincide).
- \(\ds \lim_{t\ra a^+} w\left(t\right)\), \(\ds \lim_{t\ra b^-} w\left(t\right)\) 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 \((4)\) & \((5)\)
\[ \left| \int_a^b w\left(t\right)dt \right| = r \leq \int_a^b \left| w\left(t\right) \right|dt \qquad \square \]
Contours
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.