# Complex Analysis - Cauchy Riemann Equations + Laurent Expansions

• August 4th 2008, 02:41 PM
dougwm
Complex Analysis - Cauchy Riemann Equations + Laurent Expansions
Hi all, i have a couple of questions on my complex analysis course that i cannot do and was wondering if you could help at all.

QUESTION 1

a) Suppose that the functions f(z) = u(x,y) + iv(x,y) and g(x) = v(x,y) + iu(x,y) are analytic in some domain D. Show that both u and v are constant functions

b) Let f be a holomorphic function on the punctured disk D'(0,R) = {z in complex plane : 0 < |z| < R}, where R > 0 is fixed. Write down the formulae for c(subscript)n in the Laurent expansion

f(z) = sum from n = minus infinity to infinity c(subscript)n . z^n (by this i mean cn multiplied by z^n)

Using these formulae, prove that if f is bounded on D'(0,R), it has a removable singularity at 0.

c) FInd the maximal radius R > 0 for which the function f(z) = (sin z)^-1 is holomorphic in D'(0,R), and find the principal part of its Laurent expansion about z(subscript)0 = 0 (centred at 0)
(the z(subscript)0 refers to: sum from k = minus infinity to infinity c(subscript)k(z - z(subscript)0)^k the Laurent expansion)

QUESTION 2

Find the Laurent exansions of the function f(z) = z / (z^2 - 1) valid for:
i) 0 < |z - 1| < 2
ii) |z + 1| > 2
iii) |z| > 1

Thank you
Doug
• August 4th 2008, 07:13 PM
ThePerfectHacker
Quote:

Originally Posted by dougwm
a) Suppose that the functions f(z) = u(x,y) + iv(x,y) and g(x) = v(x,y) + iu(x,y) are analytic in some domain D. Show that both u and v are constant functions

Use the Cauchy-Riemann equations,
$u_x = v_y$, $u_y = -v_x$, $v_x = u_y$, and $v_y = -u_x$.
Thus,
$v_y = - v_y$, $u_x=-u_x$, $v_x = - v_x$, and $u_y = -u_y$.
Thus, $\nabla u = \nabla v = \bold{0}$.
Thus, if $D$ is a region (open connected set) then $u,v$ are constant.
• August 4th 2008, 07:19 PM
ThePerfectHacker
Quote:

Originally Posted by dougwm
b) Let f be a holomorphic function on the punctured disk D'(0,R) = {z in complex plane : 0 < |z| < R}, where R > 0 is fixed. Write down the formulae for c(subscript)n in the Laurent expansion

f(z) = sum from n = minus infinity to infinity c(subscript)n . z^n (by this i mean cn multiplied by z^n)

Using these formulae, prove that if f is bounded on D'(0,R), it has a removable singularity at 0.

The theorem says that we can write,
$f(z) = \sum_{-\infty}^{\infty}a_n z^n$ where $a_n = \frac{1}{2\pi i} \oint \limits_{|z|=R} \frac{f(\mu)}{\mu^{n+1}} d\mu$.
But we do not need this to prove that if $f$ is bounded then it has a removable singularity.
If $|f|\leq A$ then $|zf(z)| \leq A|z|$.
But $\lim_{z\to 0}A|z| = 0$ thus, $\lim_{z\to 0}|zf(z)| = 0$.

Quote:

c) FInd the maximal radius R > 0 for which the function f(z) = (sin z)^-1 is holomorphic in D'(0,R), and find the principal part of its Laurent expansion about z(subscript)0 = 0 (centred at 0)
(the z(subscript)0 refers to: sum from k = minus infinity to infinity c(subscript)k(z - z(subscript)0)^k the Laurent expansion)
If $z\not = 0$ then $(\sin z)^{-1}$ is defined as long as $z\not = \pi k$.
Thus, this radius is $R=\pi$.
• August 5th 2008, 02:10 AM
wingless
Quote:

Originally Posted by ThePerfectHacker
If $|f|\leq A$ then $|zf(z)| \leq A|z|$.
But $\lim_{z\to 0}A|z| = 0$ thus, $\lim_{z\to 0}|zf(z)| = 0$.

But the formula will be used to prove that

"If $\lim_{z\to z_0} (z-z_0)f(z) = 0$ then $z_0$ is a removable singularity."

In order to prove this we have to show that all $a_{-n}$'s are 0 (this is the definition of removable singularity).

$a_n = \frac{1}{2\pi i} \int \limits_{\gamma_r} f(w)(w-z_0)^{-n-1}~dw$

$a_{-n} = \frac{1}{2\pi i} \int \limits_{\gamma_r} f(w)(w-z_0)^{n-1}~dw$

( $\gamma_r$ is a circle of radius r centered at $z_0$)

We know that f is bounded, $|f(w)(w-z_0)| \leq \epsilon$,

$|f(w)(w-z_0)| \leq \epsilon \quad \rightarrow \quad |f(w) \cdot r| \leq \epsilon \quad \rightarrow \quad |f(w)| \leq \frac{\epsilon}{r}$

Now we can use the formula.

$|a_{-n}| = \left |\frac{1}{2\pi i} \int \limits_{\gamma_r} f(w)(w-z_0)^{n-1}~dw \right | \leq (\frac{1}{2\pi})(2\pi r) (\frac{\epsilon}{r}r^{n-1}) = \epsilon r^{n-1}$

And $\epsilon r^{n-1}$ goes to 0 for smaller circles.
• August 5th 2008, 03:35 AM
dougwm
Thank you both very much for your quick replies and great responses. They have helped alot.

For the question 2, about the Laurent expansions I have:

z/(z^2 - 1)

by partial fractions this becomes:

1/2(z - 1) + 1/2(z + 1)

so for:

0 < |z - 1| < 2

1/2(z + 1) = 1/2(2 + z - 1) = 1/4(1 + (z - 1)/2)

= 1/4 x (sum from n to infinity) (-1)^n . [|z - 1|/2]^n

this proves |z - 1| < 2

not too sure how to prove that |z - 1| > 0 though.

for:

|z + 1| > 2 i have:

1/2(z - 1) = 1/2(-2 + z + 1) = 1/2 . 1/(z + 1) . 1/(1 - 2/(z + 1))

= 1/2 . 1/(z + 1) . (sum from n to infinity) [ 2/|z + 1|]^n

which proves |z + 1| > 2

That is what i have so far, i am not sure how to do the other half of the partial differential equation in both cases. To prove it totally, also a little unsure on the third part.

Thank you once again
Doug
• August 5th 2008, 08:29 AM
Opalg
Quote:

Originally Posted by dougwm
For the question 2, about the Laurent expansions I have:

z/(z^2 - 1)

by partial fractions this becomes:

1/2(z - 1) + 1/2(z + 1)

so for:

0 < |z - 1| < 2

1/2(z + 1) = 1/2(2 + z - 1) = 1/4(1 + (z - 1)/2)

= 1/4 x (sum from n to infinity) (-1)^n . [|z - 1|/2]^n

this proves |z - 1| < 2

not too sure how to prove that |z - 1| > 0 though.

That's correct. You have found the series expansion of 1/(2(z+1)) (in powers of z-1) which converges when |z - 1| < 2. But don't forget to include the other partial fraction 1/(2(z-1)). This is a negative power of z-1, so it is only defined when z≠1, or in other words when |z - 1| > 0.

Quote:

Originally Posted by dougwm
for:

|z + 1| > 2 i have:

1/2(z - 1) = 1/2(-2 + z + 1) = 1/2 . 1/(z + 1) . 1/(1 - 2/(z + 1))

= 1/2 . 1/(z + 1) . (sum from n to infinity) [ 2/|z + 1|]^n

which proves |z + 1| > 2

Again, the calculation is correct, but it's not appropriate to say that it "proves" |z + 1| > 2. What you should asy is that this series converges when |z + 1| > 2. (Here too, don't forget to include the other part of the partial fraction expression.)

For the third part of the problem, in the region |z| > 1, you need to get series expansions of both parts of the partial fraction expression:

$\frac1{2(z - 1)} + \frac1{2(z + 1)} = \frac1{2z}\left(\frac1{1-\frac1z} + \frac1{1+\frac1z}\right) = \ldots.$
• August 5th 2008, 10:24 AM
ThePerfectHacker
Quote:

Originally Posted by wingless
But the formula will be used to prove that

"If $\lim_{z\to z_0} (z-z_0)f(z) = 0$ then $z_0$ is a removable singularity."

I think your approach was better than mine because it is like the poster asked.

However, the proof of the theorem (which is quoted) does not need to proven with approximating the integral. Remember a point $z_0$ is a removable singularity if there exists a function $g$ in the punctured neighborhood of $z_0$ such that $f(z) = g(z)$ for all $0<|z-z_0| and $g$ is analytic at $z_0$. Now to prove this define the function $h(z) = (z-z_0)f(z)$ for $0<|z-z_0| and $h(z_0)=0$. Then clearly $h$ is analytic on the punctured neighrhood and continous at $z_0$. It follows that $h$ is analytic on the entire disk. Finally, define $g(z)=h(z)/(z-z_0)$ for $z\not = 0$ and $g(z_0)=h'(z_0)$. Then $g$ is analytic on the entire disk and $g(z) = h(z)$ for all $0<|z-z_0|. Thus, $f$ has removable singularity.