Points of Analyticity and Singularity

[Bashyboy]

Homework Statement

Determine the points for which $f(z) = \frac{z^2 + 3}{(z-3)^2(z^2 - 4z + 5)}$ is analytic and singular.

Homework Equations

Theorem 133:

Suppose c is a complex constant and suppose the derivatives of the complex
functions f and g exist at z. Then

1. Sums

$\frac{d}{dz} \left[f(z) + g(z) \right] = f'(z) + g'(z)$​

2. Product rule

$\frac{d}{dz} \left[f(z)g(z) \right] = f'(z) g(z) + f(z) g'(z)$​

3. Quotient rule. For those values of $z$ where $g(z) \ne 0$,

$\frac{d}{dz} \left[\frac{f(z)}{g(z)} \right] = \frac{f(z) g'(z) - f'(z) g(z)}{[g(z)]^2}$​

The Attempt at a Solution

Because the question of analyticity is related to a functions differentiability, before we determine at which points it is analytic and singular, let us first ascertain its differentiability.

The function $f(z)$ is a ratio of the two functions $g(z) = z^2 + 3$ and $h(z) = (z-3)^2(z^2 - 4z + 5)$. Theorem 133, part 3, states that, if the functions $f(z)$ and $g(z)$ are differentiable at the point $z$, then their quotient $\frac{f(z)}{g(z)}$ is differentiable at the same point $z$, provided that $g(z)$ is not zero at this point; if it is, then the quotient is not differentiable at that point. We are therefore urged to find at the points for which $h(z)$ is zero:

$(z-3)^2(z^2 - 4z + 5) = 0 \implies$

$(z - 3)^2 = 0 \implies z = 3$

and

$(z^2 - 4z + 5) = 0 \implies (z-5)(z+1) = 0 \implies z=-1 ~~~\mbox{or}~~~ z = 5$

Therefore, the rational function $f(z)$ will not be differentiable at these three points.

Are there any other points at which $h(z)$ is not differentiable? From the same theorem alluded to above, the product rule tells us a similar thing which the quotient rule tells us. The function that is a product of two function is differentiable at every point that the functions in the product are. Sure enough, from problem 130 we know that $z$ and $z^2$ are differentiable functions; and from theorem 131 we know that a function which is the sum of two differentiable functions is differentiable everywhere the two functions are. For instance, because we know that $z$ is differentiable everywhere, as well as any constant function, we know that $z-3$ is also a differentiable function. Therefore, $h(z)$ is differentiable everywhere.

How about the function $g(z)$; at which points is it differentiable? The same observations will reveal that $g(z) = z^2 + 3$ is differentiable everywhere. Therefore, the function is not differentiable at the points $\{-1,3,5\}$, meaning that they are points of singularity. The function is analytic everywhere else.

Pardon me if this sounds rather dull and too formal (this is for homework I have to hand). I just curious to know if my reasoning is sound. I would appreciate it if someone could check my logic.

 

[RUber]

You should double check your roots to the polynomial $z^2 -4z+5$.

 

[RUber]

Your logic is good. Polynomial functions are some of the nicest to work with--always differentiable. Their compositions and products also. As you noticed, their quotients are not always analytic due to singularities where the denominator is equal to zero. But these singularities are always removeable, so even that is nice.

 

[Bashyboy]

Thank you very much for your prompt answers. I have another question relating to analyticity and singularities. I have the function $F(z) = e^x e^{-iy}$, and I have to determine its singularities and analyticities (?). Here is what I believe to be a rather clever answer, but I suppose that you will be the judge:

$F(z) = e^x e^{-iy} = e^{x - iy} = e^{\overline{z}}$. Let $g(z) = e^z$ and $f(z) = \overline{z}$. By the chain rule, we get

$F'(z) = g'(f(z)) f'(z)$

Now, although I don't know what the derivative of $e^{\overline{z}}$, nor do I even know if it exists (we haven't proven this, yet), I do know that the function $f(z) = \overline{z}$ is not differentiable anywhere (I proved this in one of the earlier problems). Because the derivative of the function $F(z)$ involves $f'(z)$, which does not exist, by the product rule $F'(z)$ does not exist either, it is not differentiable anywhere.

Thus, every point would be a singularity.

At first this sounded pretty good; but the more deliberation I give it, the more unsure am I of it...I would like it to be true!

 

[RUber]

I feel like you have complicated that process too much.
Question 1: for what values of x is $e^x$ analytic?
Question 2: for what values of y is $e^{-iy}$ analytic?
Question 3: for what values would their product be analytic?

 

[Bashyboy]

Hmm, I am not sure, because I don't know what the complex derivative of $e^z$ is.

 

[RUber]

Do you know what the derivative of $\cos y - i \sin y$ is? With you x,y notation the assumption is that x and y are both real, right?

 

[Bashyboy]

Does differentiating it require that you differentiate each part, the real and imaginary function? If so, I don't believe that this has been demonstrated to us yet. I believe the easiest thing would be to use the chain rule, as I don't want to use the Cauchy-Riemann equations.

 

[RUber]

I have been having a tough time accepting that this was not analytic. Using the Cauchy Riemann equations, it became quickly apparent that there is a sign problem caused by the conjugate. Probably the exact reason you said $f(z) = \overline z$ is not differentiable.
Nevermind, I concur with your original logic.

 

[Bashyboy]

And with which logic do you concur? The logic I gave in post #4?

 

[RUber]

Yes, becuase you should also have proved that $g(z)=e^z$ is analytic everywhere. So for the same reasons, $f(z)=\overline z$ is not differentiable, any function of the conjugate will run into the same problems. As you said, $g(f(z))=g'(f(z))*f'(z)=g(f(z))*f'(z)$.