# MATH3401

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 $c_1 \lt c_2 \lt \cdots \lt c_n\in (a,b) \quad \text{ such that}$

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 $$u$$ & $$v$$").

### 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.