updated section about integration with properties

content/complex_integration.html

@@ -100,7 +100,8 @@
             <li><a href="#section1">Contour integral</a></li>
             <li><a href="#section2">Numerical evaluation Exp 1</a></li>
             <li><a href="#section3">Numerical evaluation Exp 2</a></li>
-            <li><a href="#section4">Antiderivatives</a></li>
+            <li><a href="#section4">Properties</a></li>
+            <li><a href="#section5">Antiderivatives</a></li>
             <!-- Add more sections as needed -->
         </ul>
     </div>
@@ -361,6 +362,108 @@ <h3>Exploration 2</h3>
 
             <div id="section4">
 
+                <h2>Properties</h2>
+
+                <p>
+                    All the following familiar properties of integrals can be proved directly 
+                    from the definition given in (\ref{integral-real-cv}). 
+                    If $w(t)= u(t)+iv(t)$ and $s(t)= \upsilon(t)+i \nu(t)$ 
+                    are complex-valued functions of a real variable $t$ continuous 
+                    on an interval $[a,b],$ then
+                </p>
+                <div class="scroll-wrapper">
+                    \begin{eqnarray}
+                    \int_a^b k \, w(t)\, dt &=& k \int_a^b w(t)\,dt, \quad k\text{ is a complex contant,}\\
+                    \int_a^b \big[ w(t) + s(t)\big] \,dt &=&  \int_a^b  w(t)\,dt +  \int_a^b  s(t) \,dt,\\
+                    \int_a^b w(t)  \,dt &=&  \int_a^c  w(t)\,dt +  \int_c^b  w(t) \,dt \quad c\in[a,b], \\
+                    \int_a^b w(t)  \,dt &=&  -\int_b^a  w(t)\,dt.
+                    \end{eqnarray}
+                </div>
+
+                <p>
+                    From definition (\ref{contour-integral}), and the properties just mentioned above, 
+                    it also follows immediatly that
+                </p>
+                <div class="scroll-wrapper">
+                    \begin{eqnarray}
+                    \int_{C} z_0 \, f(z) \, dz &=& z_0 \int_{C} f(z) \, dz, \quad z_0 \text{ is a complex contant,}\\
+                    \int_{C} \big[ f(z) + g(z) \big] \, dz &=& \int_{C} \, f(z) \, dz + \int_{C}  g(z) \, dz \\
+                    \end{eqnarray}
+                </div>
+
+                <p>
+                    Now consider the contour defined for the integral (\ref{contour-integral}).
+                    The contour $-C$ consists of the same set of points but with the order
+                    reversed so that the new contour extends from $z_2$ to $z_1,$ as shown in Figure 3.
+                </p>
+
+                <figure>
+                    <img src="../images/chp04/reversed-contour.svg" alt="Reversed contour" title="Reversed contour" style="width:60%;">
+                        <figcaption>
+                            The contour $-C$  extends from $z_2$ to $z_1.$
+                        </figcaption>
+                </figure>
+
+                <p>
+                    The contour $-C$ has a parametric representation $z=z(-t)$ with $-b\leq t \leq -a.$
+                    Hence
+                </p>
+                <div class="scroll-wrapper">
+                    \begin{eqnarray*}
+                    \int_{-C} f(z) \, dz = \int_{-b}^{-a} f\left[z(-t)\right]\frac{d}{dt}z(-t)\,dt = -\int_{-b}^{-a} f\left[z(-t)\right]z'(-t)\,dt.
+                    \end{eqnarray*}
+                </div>
+                <p>
+                    In we use the substition $\tau = -t$ in the last integral, then
+                </p>
+                <div class="scroll-wrapper">
+                    \begin{eqnarray*}
+                    \int_{-C} f(z) \, dz = -\int_{a}^{b} f\left[z(\tau)\right]z'(\tau)\,dt.
+                    \end{eqnarray*}
+                </div>
+                <p>
+                    Hence, we obtain the following property 
+                </p>
+                <div class="scroll-wrapper">
+                    \begin{eqnarray}
+                    \int_{-C} f(z) \, dz = -\int_{C} f(z)\, dz.
+                    \end{eqnarray}
+                </div>
+
+                <p> 
+                    Finally, consider a path $C,$ paramtrized by $z(t)$ for $t\in[a,b],$
+                    that consists of a contour $C_1$
+                    from $z_1$ to $z_2$ followed by a contour $C_2$ from 
+                    $z_2$ to $z_3,$ the initial point of $C_2$ being the final point of $C_1.$
+                    Then
+                </p>
+                <div class="scroll-wrapper">
+                    \begin{eqnarray}\label{sum-paths}
+                    \int_{C} f(z) \, dz = \int_{C_1} f(z) \, dz  + \int_{C_2} f(z) \, dz.
+                    \end{eqnarray}
+                </div>
+
+                <figure>
+                    <img src="../images/chp04/sum-contours.svg" alt="Sum of contours" title="Sum of contours" style="width:60%;">
+                        <figcaption>
+                            $C= C_1 + C_2.$
+                        </figcaption>
+                </figure>
+
+                <div class="practice">
+                    <p>
+                        <strong>Exercise 4:</strong>
+
+                        Use the properties for integrals of functions $w(t)$ to prove 
+                        (\ref{sum-paths}).
+                    </p>
+                </div>
+            </div>
+
+            <hr>
+
+            <div id="section5">
+
                 <h2>Antiderivatives</h2>
 
                 <p>
@@ -416,14 +519,14 @@ <h2>Antiderivatives</h2>
                 </p>
 
                 <figure>
-                    <img src="../images/chp04/equalpath.jpg" alt="Piecewise smooth curve" title="Piecewise smooth curve" style="width:50%;">
+                    <img src="../images/chp04/path-independence.svg" alt="Path independence" title="Path independence" style="width:60%;">
                     <figcaption>
                         Independent paths forming closed curve
                     </figcaption>
                 </figure>
 
                 <p>
-                    Considering Figure 3, we have
+                    Considering Figure 5, we have
                     $$\int_{C_1} f\left(z\right)dz=\int_{C_2} f\left(z\right)dz$$
                     because
                     </p><div class="w3-responsive">