Solving the Branch Cut Question on f(z) = arccot(z)

[FelixHelix]

Just covered branch cuts in my undergraduate course but stuck on one of the questions...

Find the domain on which f(z) = arccot(z) is single valued and analytic.

Now, we've looked at ln(z) in class and I understand the principal of limiting the domain but I'm not having much success and see no examples anywhere on the internet.

Some pointers would be great or an explanation of how to tackle this kind of function.

How I thought I may start is:

Let w = cot(z) and in exponential form i* \frac{\exp{(iz)}+\exp{(-iz)}}{\exp{(iz)}-\exp{(-iz)}}

Then let y = exp(iz) so you get:

w = i* \frac{y+y^-1}{y-y^-1}

I though i might be able to solve the last equation but I'm not sure now.

anyway, if you could point me in the right direction I'd be most thankful...

Felix

 

[Mugged]

Have you tried to just graph this function? If you're talking about real-valued f(z) then you could see the domain of z just from the graph.

If you begin dissecting trigonometric functions into their exponential definitions from the Euler equation, then you can run into problems of having imaginary domains.

For example take f(x) = sin(x).
The domain is all real numbers and the range is [-1,1] right?
But what if i set sin(x) = 2? Is there a solution to this? In fact there is but the answer for x is imaginary. So why don't we say that the domain is larger than the real numbers for sin(x) and the range is larger than -1 to 1? Well because we're making some assumptions here and probably in your homework problem too...we have to restrict ourselves to real domains and ranges. So don't bother trying to solve that function with y and y^-1 because it's not the correct direction here.

So for your arccot(z) function you are on the right track when you first look at cot(z). The range of cot(z) would have to be the domain of arccot(z) right? So from the graph of cot(z) you can see this goes from -inf to +inf periodically. You then have to conclude that the domain of arccot is all real numbers.

Hope that helps

 

[FelixHelix]

Hi - Thanks for the post. I believe I am to consider the complex domain as well. My answersheet gives the cut to be (-i,i) but I don't see how you get there or limit it to be multivalued... All examples we've covered have solved for y after some re-expressing hence why I thought that might be the right approach.

Any ideas?

 

[jasonRF]

You should be able to write arccot in terms of logarithms by solving the following equation for z as a function of y:
<br /> \cot z = y = i \frac{e^{i z} + e^{-i z}}{e^{i z} - e^{-i z}}<br />

Then you have already (I Hope!) worked with branch cuts of log.

jason

 

[FelixHelix]

You should be able to write arccot in terms of logarithms by solving the following equation for z as a function of y:
<br /> \cot z = y = i \frac{e^{i z} + e^{-i z}}{e^{i z} - e^{-i z}}<br />

Then you have already (I Hope!) worked with branch cuts of log.

jason

OK... So:

Let w = cot(z) and in exponential form i* \frac{\exp{(iz)}+\exp{(-iz)}}{\exp{(iz)}-\exp{(-iz)}}

Then let y = exp(iz) so you get:

y = i* \frac{w+w^{-1}}{w-w^{-1}}

When I rearrange the equation for y I then solve a quadratic in w so:

So the two answers to w are \pm \frac{(i+y)^ \frac{1}{2}} {(i-y)^ \frac{1}{2}}

and therefore using the positive solution:

arccot(z) = \frac{1}{2} \ln{(i+z)} - \frac{1}{2} \ln{(i-z)}

This identifies that the function is non analytic at i and -i.Is this correct?

Next, do I have to show that the branch cut could be either (-infinity, -i), (-i,i), (i, infinity) and check them some how?

Also, I can't seem to calculate how to make it single valued?