Suggested exercise Section II subsection 1 on page 483

Mondo

I don't see it like that. As it says on page 483 "You can then quickly get a feel... across the contour"
In the example $H(z)=\overline{z}^2z$ and so $\overline{H}=z^2\overline{z}=(z\overline{z})z=|z|^2z$
So $\overline{H}$ is a vector in the same direction as $z$.
$K$ is the vertical line from $1-i$ to $1+i$ and $W$ is the integral of the imaginary part of $\overline{H}$ along this line and by symmetry it is clearly zero.
$F$ is the integral of the real part of $\overline{H}$ along this same line, a positive number and so the integral of $H$ along this line segment is a positive multiple of $i$.

Vasco

Mondo

No, because $W$ is the integral along $K$ of the tangential component, which is the imaginary part of $\overline{H}$. Similarly for $F$ but for the normal to $K$ which is the real part of $\overline{H}$. This is the most important thing to understand here.

Vasco

Mondo

Yes.
$\overline{H}=|z|^2z$ and $ds=dy$ so
$\displaystyle W[\overline{H},K]=\int_{-1}^{+1} \text{Im}(\overline{H})dy=\int_{-1}^{+1} \alpha ydy=0$ where $\alpha=|z|^2$ and
$\displaystyle F[\overline{H},K]=\int_{-1}^{+1} \text{Re}(\overline{H})dy$
Since the real part of $\overline{H}$ is positive along $K$ then this integral is positive and so we can see that
LHS of (11) $=0+i(\text{positive real number})$

This is an algebraic solution, but you should be able to see this in geometric and physical terms as Needham says just after (11) in the book.

Vasco

Mondo

No, if $\overline{H}=|z|^2$ then the integral is even more obviously zero, since the imaginary part of $\overline{H}$ is zero.
I would advise you to reread this part of chapter 11 of the book to make sure you fully understand it.


No, because $\text{Re}(\bar H)$ is not constant along $K$ because $|z|^2$ is not constant along $K$ and also $\int_{-1}^{+1} dy=2$ not $i$.
Are you sure you understand integration properly?

By physical I mean visualise the Polya vector field graphically in relation to $K$. Draw some pictures either by hand or using a computer.

Vasco

Ok let's clarify it, I prepared a simple graphics that depicts our example:



So we have our curve of integration $K$ in green here. First thing I don't get is why in your integrals you write $\int_{-1}^{1}$ while we integrate from $\int_{1-i}^{1+i}$ which is a entirely different path. Hence:

should rather be $\int_{1-i}^{1+i} dy=2i$. Let's even make a simple simulation base on my plot - we can treat it as four segments, each of length $0.5i$ hence $4+0.5i = 2i$
Next let's proceed to integration.
Here I would also start simple and write equation for work along K for a simple function which Ploya vector is defined as $\bar H = z$. We both agreed that work is a tangent component of our vector hence $W[\bar H, K] = \int_{1-i}^{1+i} \bar H \cdot T dy$. If instead of calculating an integral we write a Riemann's sum of midpoints for these four segments on my plot we get $\sum_{1-i}^{1+i} = (\frac{(1-i)+(1-0.5i)}{2})0.5i + (\frac{(1-0.5)+1}{2})0.5i + (\frac{1+(1+0.5i)}{2})0.5i + (\frac{(1+0.5i)+(1+i)}{2})0.5i = 2i$ So the work is purely imaginary on this path.
Next what if $\bar H = |z|^2$? Continuing my Riemann's sum example we have four sums of a real numbers because $|z|^2 = a^2 + b^2$. Since $|1+i|^2 = |1-i|^2 = 2$ and $|1+0.5|^2 = |1-0.5|^2 = 1.25$ the calculation is simple $(\frac{2+1.25}{2})0.5i + (\frac{1+1.25}{2})0.5i + (\frac{1+1.25}{2})0.5i + (\frac{2+1.25}{2})0.5i = 2.75i$ (Exact result for this integral is 2.66(6) so my four point approximation is not that bad). But the key aspect here is the integral is not 0 as you said here:


Do you agree Vasco?

Thank you Vasco.


I noticed that the book uses $ds$ for some integrals in this chapter but I somehow continued treating it as $dz$. In fact at the beginning of subsection II, where the Polya vector field is introduced there is a relation between $dz$ and $ds$ -> $dz = e^{i\alpha}ds$. So $ds$ is a small, real increment which $dz$ transforms into a complex increment - is that fair to say?
Nonetheless, for our curve $K$, how can we say its length is $2$? It starts at $(1-i)$ and ends at $(1+i)$ so $(1+i) - (1-i) = i + i = 2i$. Which is not surprise since we only move on the imaginary axis

Mondo
It is not quite correct what you say about $dz$ and $ds$. Here $dz$ is not a transformation(you need to read the first few pages of chapter 1 to learn the nomenclature - modulus, argument etc).
$dz$ is a complex number with modulus $ds$ and argument $\alpha$.
The difference between two complex numbers (subtraction) is not a distance but another complex number or vector. Have you read the part of the book which explains about regarding complex numbers as vectors? If you have, you seem to have forgotten it!
The distance between two complex numbers is the modulus of the vector which joins them i.e.
$|(1+i)-(1-i)|=|2i|=2$
In mathematics one has to be precise and words that appear in normal speech often have different more precise meanings in mathematics.
Distance is a real positive or zero quantity in mathematics.
Notice that if we subtract the two numbers the other way round we get $-2i$ but $|-2i|=2$ again. Two vectors pointing in opposite directions but with the same length.

Hmm, it is true that calculation underneath Figure 4 on page 477 use only real numbers. This can also be seen from the last calculation at the bottom of page 482. $Hdz$ contains both work and flux components and they are both real (vector magnitude times appropriate trig function times distance traveled).
I think it may all become clear once we answer the above question why the path of integration has length $2$ instead of $2i$. Additionally why the work integral is calculated from imaginary part of $\bar H$.


I am not referring to the integral on the left of (11) which is a complex integral, but to the integrals which represent $W$ and $F$ which are both real integrals.

And so must be $X \cdot N$ as this is a multiplication by a different unit vector to get a perpendicular component.



Well we already got purely imaginary distances which from physical perspective also shall be real. That is why an imaginary work did not surprised me at first.

Vasco

Mondo

I hope you can see from my diagrams what Needham means when he writes near the top of page 483 just under result (11):

"... draw the Polya vector field of any function you wish to integrate. You can then quickly get a feel for the value of the integral by looking at how much the field flows along and across the contour"

Vasco