Post by Admin on Jun 21, 2017 18:25:59 GMT
Gary
When producing my solution to exercise 11, I came across a possible error in the book which at the time I documented in my solution. So that you have access to it without looking at my solution I reproduce it below:
Although subsection 1 Integration along a Circular Arc, of section VI Power Functions on page 395 of the book claims to be a generalisation to a general integer power $w=z^m$, it seems to me that it needs a slight modification to the equations on page 396 to accommodate values of $m<-1$.
If we define $n\equiv m+1$ as in subsection 2 Complex Inversion as a Limiting Case*, then when $n<0~(m<-1)$ we need to rewrite the line below figure 17 on page 396, which in the book (year 2000 printing) is written thus
$\rho\widetilde{\phi}=|\widetilde{\Delta}|~~~~~\implies~~~~~\rho=\dfrac{A^{m+1}}{m+1}$.
This is because when $n=m+1<0$, $\widetilde{\phi}=(m+1)\phi=n\phi<0$ and so we should write
$\rho|\widetilde{\phi}|=|\widetilde{\Delta}|~~~\implies~~~~\rho=\dfrac{A^{m+1}}{|m+1|}$
and the turning angle for $\widetilde{\Delta}$ is $\widetilde{\phi}=n\phi$, positive when $n$ is positive and negative when $n$ is negative.
Vasco
When producing my solution to exercise 11, I came across a possible error in the book which at the time I documented in my solution. So that you have access to it without looking at my solution I reproduce it below:
Although subsection 1 Integration along a Circular Arc, of section VI Power Functions on page 395 of the book claims to be a generalisation to a general integer power $w=z^m$, it seems to me that it needs a slight modification to the equations on page 396 to accommodate values of $m<-1$.
If we define $n\equiv m+1$ as in subsection 2 Complex Inversion as a Limiting Case*, then when $n<0~(m<-1)$ we need to rewrite the line below figure 17 on page 396, which in the book (year 2000 printing) is written thus
$\rho\widetilde{\phi}=|\widetilde{\Delta}|~~~~~\implies~~~~~\rho=\dfrac{A^{m+1}}{m+1}$.
This is because when $n=m+1<0$, $\widetilde{\phi}=(m+1)\phi=n\phi<0$ and so we should write
$\rho|\widetilde{\phi}|=|\widetilde{\Delta}|~~~\implies~~~~\rho=\dfrac{A^{m+1}}{|m+1|}$
and the turning angle for $\widetilde{\Delta}$ is $\widetilde{\phi}=n\phi$, positive when $n$ is positive and negative when $n$ is negative.
Vasco
