Exercise 7 chapter 7

Gary

I have made a very small modification to the very last line of my solution and re-published.

Vasco

Gary

In part (ii) I think $v=-x$ not $x$ and your definition of $J$ seems to be wrong.
The line of critical points is $y=1/2$. You have made a typo here and written $y=2$.

In (iii) I think $\nu(0)=-1$ and $\nu(i)=+1$.

In (iv) $h$ winds zero times round the origin - and in my opinion that's the point!

Do you disagree with my answers to parts (v) and (vi)?

Vasco

Gary

The last line of my document in part (vi) contains a typo and an error, it should read:

"...we find that $\nu[h(C),0]=0=\nu(i/2)$

I realised that $i/2$ in the square brackets was a typo and should have been zero, and after reading your posts and document I concluded that $\nu(i/2)=0$ not $-1$. I have updated my document to reflect these changes.

Part (iv)

Since the exercise asks us to find the image curve traced by $h(z)$ as $z=2e^{i\theta}$ traverses the circle $|z|=2$, I think that finding three loops overcomplicates the situation unnecessarily.

The image curve $h$ is just the circle centred at $4$ with radius $2$, the right hand circle in your figure 1, and you can see that it does not encircle the origin and so $\nu[h(C),0]=0$, which is confirmed by the TAP for $z=0$ and $z=i$ inside $\Gamma$.

Part (v)

In the analysis of [8] in the book $h(z)$ is written as $ABC$ and in our case $A=\bar{z}=(\bar{z}-0)$ and $B=(z-i)$ and $C=1$ and so we have to consider what $A$ and $B$ do as $z$ goes round $\Gamma$ and then add their windings together to find the winding of $h(\Gamma)$ round $0$.

Part (vi)

In this part we do exactly the same as in part (v), but use $\Gamma$ as the small circle (radius $\epsilon$) centred at $z=i/2$. So as $z$ traverses $\epsilon e^{i\theta}$ we must find what $A=\bar{z}$ does and what $B=(z-i)$ does and add their windings together to get the total winding of $h(\Gamma)$ about the origin.

Vasco