ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: /0 on July 24, 2010, 12:29:44 pm

Title: Complex Analysis
Post by: /0 on July 24, 2010, 12:29:44 pm: In the complex derivative,

$f'(z) = \lim_{h \to 0}\frac{f(z+h)-f(z)}{h}$

Can 'h' be anything?

This looks a bit like the directional derivative formula in $\mathbb{R}^2$ :

$D_uf(x) = \lim_{h \to 0} \frac{f(\vec{x}+h\vec{u})-f(\vec{x})}{h}$

But yet there is a 'unique' $f'(z)$ ? I would have thought $f'(z)$ should vary depending on the direction you approach.

(This question of course leads to how the Cauchy-Riemann equations came about)

Thanks
Title: Re: Complex Analysis
Post by: zzdfa on July 24, 2010, 01:03:59 pm: Remember the definitions: a function is differentiable only if that limit exists.
If you get 2 different values depending on which direction you approach, then the limit does not exist and hence the function is not differentiable. if it -is- unique, then we say it is differentiable.

for example f(z)=Re(z)+2Im(z)= is not complex differentiable at 0 because the limit doesn't exist.
Title: Re: Complex Analysis
Post by: /0 on July 24, 2010, 01:58:58 pm: But if you think of the function as a 'contour map' on the x-y plane ( $z=x+yi$ )... if every point in a domain has derivatives from every side equal, then surely the function must be constant?
Title: Re: Complex Analysis
Post by: zzdfa on July 24, 2010, 02:10:54 pm: It is true that if the directional derivatives of f:R^2->R at each point are the same for every direction, then f is constant. (because if directional derivative is the same in all directions it implies the gradient is 0 at that point. if the gradient at 0 at every point then f is constant)

However you can't think of a complex function C->C as a contour map, remember the range of the function the complex numbers, so the graph is a surface in C^2, or R^4.

just try a simple example, check that if f(z)=z^2 then $f'(0)=\lim_{h \to 0} \frac{ (z+h)^2 - z^2}{h} = 0$
Title: Re: Complex Analysis
Post by: /0 on July 25, 2010, 02:12:34 am: Ah, ok thanks zzdfa =)

Another Q, if you approach a point along the imaginary axis then:

$f'(z) = \lim_{k \to 0} \frac{f(z+ik)-f(z)}{ik}=-i\frac{\partial f}{\partial y}$

Is the $-i$ from the $\frac{1}{i}$ in the limit?

Since $\frac{\partial f}{\partial y} = \lim_{h \to 0} \frac{f(x,y+h)-f(x,y)}{h}$

It seems that if you set $h = ik$ , then it essentially becomes the first limit above:
Since $\frac{\partial f}{\partial y} = \lim_{ik \to 0} \frac{f(x,y+ik)-f(x,y)}{ik}$ ( $k \to 0$ is the same as $ik \to 0$ right?)

So is it really ok to take the $\frac{1}{i}$ out?
Title: Re: Complex Analysis
Post by: zzdfa on July 25, 2010, 12:08:41 pm: Quote from: /0 on July 25, 2010, 02:12:34 am

It seems that if you set $h = ik$ , then it essentially becomes the first limit above:
Since $\frac{\partial f}{\partial y} = \lim_{ik \to 0} \frac{f(x,y+ik)-f(x,y)}{ik}$ ( $k \to 0$ is the same as $ik \to 0$ right?)

Yes the -i is from the 1/i, your argument doesn't work because $\lim_{ik \to 0} \frac{f(x,y+ik)-f(x,y)}{ik}$ is actually equal to $\lim_{k \to 0} \frac{f(x-k,y)-f(x,y)}{ik}=i \frac{\partial f}{\partial x}$

because ik->0 is the same as k->0,
and
$f(x,y+ik)=f(x+i(y+ik))=f(x-k+iy)=f(x-k,y)$

I think where you might be getting confused is the association between the function f:C->C and the function f*:R^2->C.

given a function with complex domain, f(z), the associated function f*:R^2->C is defined by f*(x,y)=f(x+yi).

of course in the above we used f to denote both f and f* because you can just look at the number of arguments to figure out which is which.
Title: Re: Complex Analysis
Post by: /0 on July 29, 2010, 10:04:22 pm: Thanks zzdfa xD

When people talk of a function being differentiable, do they mean differentiable in a neighbourhood, or differentiable everywhere?
Usually people only test differentiability at (0,0), is there any reason why?

The lecture notes give examples or functions which are non-differentiable:
$\bar{z}$ , $Re(z)$ , $Im(z)$ , $|z|$ and $Arg(z)$
Does this mean differentiable nowhere, or does it mean not differentiable at a selection of points?

Also, the tute says that $f(z) = |z|^2$ is not analytic at $z=0$ . But $f(z) = |x+iy|^2 = x^2+y^2 = u(x,y)+iv(x,y)$

So $\frac{\partial u}{\partial x} = 2x$ , $\frac{\partial v}{\partial y}=0$ , $\frac{\partial u}{\partial y}=2y$ , $\frac{\partial v}{\partial x}=0$ .

So at $(x,y) = (0,0)$ , the Cauchy Riemann equations are satisfied, so why isn't the function analytic at z = 0?

Thanks
Title: Re: Complex Analysis
Post by: zzdfa on July 29, 2010, 11:51:49 pm: You might see your lecturer use (0,0) alot in examples, when computing derivatives, because it's simplifies the calculations a bit. But of course there is no special significance since you can translate any function by f(z-p).

Generally in complex analysis when we say differentiable we mean differentiable on some open set U.

You can say that your function f is differentiable at (0,0) with derivative 0, but you can't say it's analytic because analyticity at p means 'differentiable in an open set containing p'. so to show that f is analytic around p you have to show that the CR equations are satisfied in an open set around p, which is isn't in your case.
Title: Re: Complex Analysis
Post by: humph on July 30, 2010, 12:45:18 am: Quote from: zzdfa on July 29, 2010, 11:51:49 pm
You might see your lecturer use (0,0) alot in examples, when computing derivatives, because it's simplifies the calculations a bit. But of course there is no special significance since you can translate any function by f(z-p).

Generally in complex analysis when we say differentiable we mean differentiable on some open set U.

You can say that your function f is differentiable at (0,0) with derivative 0, but you can't say it's analytic because analyticity at p means 'differentiable in an open set containing p'. so to show that f is analytic around p you have to show that the CR equations are satisfied in an open set around p, which is isn't in your case.

No, when you say differentiable, you just mean differentiable at a point. When it is holomorphic/analytic at a point, then it must be differentiable in a neighbourhood of that point.
Title: Re: Complex Analysis
Post by: zzdfa on July 30, 2010, 10:48:19 am: ah yeah, I even contradicted myself in the next sentence :-\
Title: Re: Complex Analysis
Post by: humph on July 30, 2010, 11:26:08 am: Quote from: /0 on July 29, 2010, 10:04:22 pm
Thanks zzdfa xD

When people talk of a function being differentiable, do they mean differentiable in a neighbourhood, or differentiable everywhere?
Usually people only test differentiability at (0,0), is there any reason why?

The lecture notes give examples or functions which are non-differentiable:
$\bar{z}$ , $Re(z)$ , $Im(z)$ , $|z|$ and $Arg(z)$
Does this mean differentiable nowhere, or does it mean not differentiable at a selection of points?

Also, the tute says that $f(z) = |z|^2$ is not analytic at $z=0$ . But $f(z) = |x+iy|^2 = x^2+y^2 = u(x,y)+iv(x,y)$

So $\frac{\partial u}{\partial x} = 2x$ , $\frac{\partial v}{\partial y}=0$ , $\frac{\partial u}{\partial y}=2y$ , $\frac{\partial v}{\partial x}=0$ .

So at $(x,y) = (0,0)$ , the Cauchy Riemann equations are satisfied, so why isn't the function analytic at z = 0?

Thanks
Who's your tutor? As was mentioned, it's differentiable at zero, but analytic (well I prefer holomorphic) means it has to be differentiable in a neighbourhood of zero, which isn't the case here.
Title: Re: Complex Analysis
Post by: /0 on July 30, 2010, 01:33:57 pm: My tutor's Davidson Ng, but we haven't had a tutorial yet.

I understand what you mean, that it must be differentiable in a neighbourhood... but the theorem that I used is as follows:

Let $f: D \to \mathbb{C}$ be given by $f(z) = u(x,y)+iv(x,y)$ , for $z=x+iy$ . Assume that all partial derivatives $\frac{\partial u}{\partial x}$ , $\frac{\partial v}{\partial x}$ , $\frac{\partial u}{\partial y}$ , $\frac{\partial v}{\partial y}$ exist, are continuous, and satisfy the Cauchy-Riemann equations at $z=z_0$ . Then $f$ is holomorphic at $z=z_0$ .

It never says anything about being differentiable in a neighbourhood, only about the continuity of the partial derivatives
Title: Re: Complex Analysis
Post by: Ahmad on August 01, 2010, 11:11:59 am: Where did you see that?

$f(x+iy) = x^2+y^2$ satisfies CR equations at z = 0 and nowhere else, the partials are continuous but f is nowhere analytic.
Title: Re: Complex Analysis
Post by: /0 on August 01, 2010, 11:59:49 am: I believe it's called the Looman-Menchoff Theorem

The version I quoted is what's in the lecture notes. Is there a difference?
Title: Re: Complex Analysis
Post by: Ahmad on August 01, 2010, 03:18:44 pm: Yeah. The result you should remember is that if CR equations are satisfied at a point and the partials are continuous at the point then the function is differentiable at the point. Then being holomorphic/analytic in some open set just means differentiable everywhere in the open set, so CR equations are satisfied + continuous partials everywhere in the open set. Analytic at a single point would mean differentiable in some open set containing the point.
Title: Re: Complex Analysis
Post by: /0 on August 01, 2010, 03:20:17 pm: Ooohhhh, I see now, thanks =D
Title: Re: Complex Analysis
Post by: /0 on August 22, 2010, 12:21:19 pm: Suppose you need to integrate a real function such as $\int_{-\infty}^\infty\frac{1}{x^6+1}\,dx$ so you introduce a complex function which has poles at the blue crosses in the picture (and some others in the lower half plane that aren't important). Aren't both of the possible contours just as valid as each other? The only difference is that one of them encloses 4 poles while the other only encloses 2, but once you shrink the small semi-cirlces to zero radius they should have zero contribution anyway..
Title: Re: Complex Analysis
Post by: humph on August 23, 2010, 11:04:49 am: Of course. But in either case the little semicircles may have a non-neglible contribution even as you shrink them smaller and smaller.
Title: Re: Complex Analysis
Post by: /0 on August 25, 2010, 04:06:11 am: Ah thanks humph, I skimmed over that lemma before
Title: Re: Complex Analysis
Post by: /0 on August 27, 2010, 03:20:18 pm: Bit of a dumb question here,

If $z^2 = e^{2(\log{|z|}+i\ arg(z))}$ , then why doesn't $f(z) = z^2$ have a branch cut on the negative real axis (assuming the principal logarithm)?
Title: Re: Complex Analysis
Post by: humph on August 27, 2010, 04:11:19 pm: The branch cut from the negative logarithm is because when you approach it from one side you get a different limit than from the other side. But note that $e^{2(\pi)} = e^{2(-\pi)}$ , so it doesn't make any difference when you take an integer power.
Title: Re: Complex Analysis
Post by: /0 on August 27, 2010, 05:33:47 pm: o right, thanks
Title: Re: Complex Analysis
Post by: /0 on August 29, 2010, 12:20:50 am: Just a few last minute questions xD

1. If you have a function like $\frac{1}{z\sqrt{z^2-1}}=\frac{1}{z}\left(e^{\frac{1}{2}log(z^2-1)}\right)^{-1}=\frac{1}{z}\left(e^{\frac{1}{2}(\log(z+1)+\log(z-1))}\right)^{-1}$ .

Is it ok if for $\log(z+1)$ and $\log(z-1)$ you choose different intervals of existence for $arg(z+1)$ and $arg(z-1)$ ?
i.e. can you rotate each of the branch cuts however you like?

And would this mean you have to deal with each logarithm as a separate entity? (i.e. $arg(z+1)(z-1)\neq arg(z+1)+arg(z-1)$ anymore?)

2. Also, for $\int_{-1}^1 \frac{1}{(x+2)\sqrt{x^2-1}}\,dx$ , we take $f(z) = \frac{1}{(z+2)\sqrt{z^2-1}}$ . The solutions take the left contour, but I'm wonder, why couldn't we just take the right contour and just ignore the pole at $z=-2$ ?

Thanks
Title: Re: Complex Analysis
Post by: /0 on November 17, 2010, 08:45:17 pm: From the PDEs exam:

"

a) Obtain the general solution of the first order PDE:

$yu_x+xu_y-yu=xe^x$

b) If we prescribe $u(x,y)=\phi(x,y)$ on the upper portion of the hyperbola $y^2-x^2=1$ , $y \geq 1$ , show that no solution exists unless $\phi(x,y)$ is of a special form. Find this form and show that in such a case there are infinitely many solutions."

I tried solving this the way it's done in the lecture notes but it didn't work =/

$yu_x+xu_y=yu+xe^x$

$\frac{dx}{ds}=y$ , $\frac{dy}{ds}=x$ , $\frac{du}{ds}=yu+xe^x$

I can't solve any of these straight away so the usual method is already looking shaky

But I notice that,

$\frac{dy}{dx}=\frac{x}{y} \Rightarrow y^2-x^2 = C$

And,

$\frac{du}{ds}=\frac{dx}{ds}u+\frac{dy}{ds}e^x$ , which means $\frac{du}{dx}=u$ , $\frac{du}{dy}=e^x$

Hence, $u=Ae^x$ , but $A$ is a function of $y^2-x^2$ so $u=A(y^2-x^2)e^x$ .

Still, this isn't the solution... and I couldn't do the next part of the question

How can you solve it?
Title: Re: Complex Analysis
Post by: humph on November 17, 2010, 09:32:36 pm: Hah, this was one of my assignment questions a couple of years ago. I'll just post it up in full.

Let us parametrise the characteristic curve in the plane of this partial differential equation by $(x(s),y(s))$ , and let $u(s) = (x(s),y(s))$ . As the chain rule implies that
$ \frac{du}{ds} = \frac{dx}{ds} u_x + \frac{dy}{ds} u_y, $
it follows that we can write
$ \frac{du}{ds} = x(s)e^{x(s)} + y(s)u(s) , \qquad \frac{dx}{ds} = y(s) , \qquad \frac{dy}{ds} = x(s). $
Writing $(x(s),y(s)) = \underline{z}(s)$ , we see that the latter two ordinary differential equations are equivalent to the vector-valued ordinary differential equation
$ \frac{d \underline{z}}{ds} = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \underline{z}(s). $
The matrix above has eigenvalues $\pm 1$ with corresponding eigenvectors $(1 , \pm 1)$ , and so the solution to this ordinary differential equation is given by
$ \underline{z}(s) = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{s} + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-s} , $
where $c_1 , c_2 \in \mathbb{R}$ . Thus we have that $x(s) = c_1 e^{s} + c_2 e^{-s}$ and $y(s) = c_1 e^{s} - c_2 e^{-s}$ . Substituting these values into the earlier expression for $du/ds$ , we have that
$ \frac{du}{ds} = (c_1 e^{s} + c_2 e^{-s})e^{c_1 e^{s} + c_2 e^{-s}} + (c_1 e^{s} - c_2 e^{-s})u(s), $
which, by multiplying through by the integrating factor $e^{-c_1 e^{s} - c_2 e^{-s}}$ , can be rearranged to read
$ \frac{d}{ds}\left(e^{-c_1 e^{s} - c_2 e^{-s}} u\right) = c_1 e^{s} + c_2 e^{-s}, $
and so we can integrate to find that
$ e^{-c_1 e^{s} - c_2 e^{-s}} u(s) = c_1 e^{s} - c_2 e^{-s} + u_0, $
or equivalently, by substituting back in $x = c_1 e^{s} + c_2 e^{-s}$ , $y = c_1 e^{s} - c_2 e^{-s}$ , and then multiplying through by $e^x$ ,
$ u(x,y) = e^{x}\left(y + u_0\right), $
where $u_0$ depends on the chosen characteristic curve. These characteristic curves are the solutions to
$ \frac{dy}{dx} = \frac{dy}{ds} \frac{ds}{dx} = \frac{x}{y} , $
which, by separating variables, we can solve to find that
$ y^2 - x^2 = C, $
where $C \in \mathbb{R}$ . Thus we have that $u_0 = u(0) = f(C)$ for any arbitrary function $f$ , and so
$ u(x,y) = e^{x}\left(y + f\left(y^2 - x^2\right)\right). $

With this solution of the partial differential equation, we therefore have that on the upper portion of the hyperbola $y^2 - x^2 = 1$ , $y \geq 1$ ,
$ u(x,y) = e^{x}\left(y + f(1)\right). $
Thus if $\phi$ is not of the form $\phi(x,y) = e^{x}\left(y + \phi_0\right)$ for some $\phi_0 \in \mathbb{R}$ , then we cannot have that $u(x,y) = \phi(x,y)$ on the upper portion of the hyperbola $y^2 - x^2 = 1$ , $y \geq 1$ . If, on the other hand, $\phi$ is of this form, then there are infinitely many solutions of the partial differential equation satisfying this prescribed condition, as there are infinitely many functions $f$ satisfying $f(1) = \phi_0$ ; for example, we could have $f\left(y^2 - x^2\right) = m\left(y^2 - x^2\right) - m + \phi_0$ for any $m \in \mathbb{R}$ .
Title: Re: Complex Analysis
Post by: /0 on November 17, 2010, 11:31:04 pm: lol wow that's pretty full-on. I don't know why it didn't occur to me to use the theory of DE systems from 2405... i mean, that stuff's made for these PDEs. But even if it did the double exponentials would probably have put me off.
But yeah that definitely makes sense, thanks a bunch humph :)