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May 30

Complex analysis questions

Are all bijective holomorphic functions taken over the entire complex plane linear? What about over the entire Riemann sphere? Also, does anyone know how this image was made? 74.14.109.234 (talk) 00:47, 30 May 2010 (UTC).[reply]

Yes and yes. Algebraist 00:50, 30 May 2010 (UTC)[reply]

Thank you. 74.14.109.234 (talk) 02:21, 30 May 2010 (UTC)[reply]

Bijective meromorphic functions on the Riemann sphere are linear fractional transformations. That's the more interesting fact. Then requiring them to be holomorphic merely says the denominator cannot vanish at any finite point, so the denominator is constant; hence the function is affine. Michael Hardy (talk) 02:55, 30 May 2010 (UTC)[reply]

What do you mean by how the image was made? It is a plot of

\exp(1/z)

, with a color function described in the image's page. It was created by a software that can create plots of functions (possibly Matlab). If you want more specific details you'll have to ask the person who created it, Functor Salad. -- Meni Rosenfeld (talk) 06:41, 30 May 2010 (UTC)[reply]

It has already been asked on his talk page and there is no answer. I don't need to know which program was used for this specific image, just how this type of plot could be done. 76.67.75.115 (talk) 16:34, 30 May 2010 (UTC)[reply]

In the highest generality, to make such a plot you go over each pixel, calculate the corresponding value of the function, calculate from it the required values of hue, saturation and luminance, and color the pixel accordingly. Some software will have functionality to facilitate this process, the syntax of which will depend on the software. Are you asking for an example for such a software, and the syntax for this software? -- Meni Rosenfeld (talk) 17:56, 30 May 2010 (UTC)[reply]

I was asking for a piece of software that could do this, but I found Sage. 76.67.79.184 (talk) 00:52, 31 May 2010 (UTC)[reply]

Algebraic expressions for the standard trigonometric functions

I noticed in the article for the hyperbolic functions that each function has an equivalent algebraic expression (such as the one for the hyperbolic sine, $\sinh(x)={\frac {e^{x}-e^{-x}}{2}}$ ). Do expressions like these exist for the regular trig functions (sine, cosine, tangent, etc.)? I realize that they can be represented by their respective Taylor series, but I'm looking to see if there exists some sort of a finite formula, like the aforementioned. — Trevor K. — 01:12, 30 May 2010 (UTC) —Preceding unsigned comment added by Yakeyglee (talk • contribs)

\cos x={\frac {e^{ix}+e^{-ix}}{2}}

\sin x={\frac {e^{ix}-e^{-ix}}{2i}}

Michael Hardy (talk) 02:53, 30 May 2010 (UTC)[reply]

If you looked at the definition of e^ix you would find that it is

e^{ix}=cis(x)=\cos(x)+i\sin(x)

then if you replace x with -x

e^{-ix}=cis(-x)=\cos(-x)+i\sin(-x)

e^{-ix}=cis(-x)=\cos(x)-i\sin(x)

Postscript: cis(x) is the notation I used while I was doing electrical engineering. It allows me to record on paper the lecture notes very quickly. 139.130.1.226 (talk) 20:52, 1 June 2010 (UTC)[reply]

Exam question thats stumping me...

Hi everyone, I was in an exam, and I came across this question that I wrote down because I had no idea how to solve it:

6sin(x) - 8cos(x) = 5

Having received my marks back from the exam, I know the answer is x = 180n + (-1)^n 30 degrees + 53 degrees 8 minutes. I have no idea why! Can anyone help? 110.175.208.144 (talk) 05:41, 30 May 2010 (UTC)[reply]

Hint: Rewrite the equation as

10(0.6\sin x-0.8\cos x)=5

. -- Meni Rosenfeld (talk) 06:33, 30 May 2010 (UTC)[reply]

Still not following, I'm afraid :( I don't really see how that helps me... 110.175.208.144 (talk) 06:52, 30 May 2010 (UTC)[reply]

Can you verify that the stated numbers are solutions to the equation? Can you draw the graph of f(x) = 6sin(x) - 8cos(x) - 5 ? Can you by brute force crudely estimate a solution to f(x) = 0 ? Perhaps Meni's hint shows a shortcut. (Sorry, we are not going to give the detailed answer away). Bo Jacoby (talk) 07:06, 30 May 2010 (UTC).[reply]

Somewhat more direct hint: The fact that the exam is asking this question implies that the syllabus presumably included List of trigonometric identities#Angle sum and difference identities. Qwfp (talk) 07:09, 30 May 2010 (UTC)[reply]

I can verify that the stated numbers in the answer are correct, as it is in both the textbook and the teacher's marking notes. And I know that you're not going to give the detailed answer :) I just need to get past this first step, methinks... And if it helps, Qwfp, the general formula we are given in class was "In general, the solution for sin(x) = sin(y) is given by x = 180 x n + (-1)^n x y where n is an integer". What's messing with my head is where they got to the 30 degrees and 53 degrees 8 minutes from... 110.175.208.144 (talk) 07:21, 30 May 2010 (UTC)[reply]

You can write

0.6\sin x-0.8\cos x

as

\sin(x-\theta )

where

\sin \theta

is .8; I assume that's where the 53° come in. The angle whose sine is 5/10 = .5 is easy to find. You then have the sines of two expressions being equal. The expressions themselves being equal is one solution but others are obtained from identities which I'm assuming were covered in your class.--RDBury (talk) 07:27, 30 May 2010 (UTC)[reply]

So after RDBury's step I'll end up with sin(x - theta) = sin 30? EDIT: Never mind, I solved it :) Thanks heaps, guys :) 110.175.208.144 (talk) 07:38, 30 May 2010 (UTC)[reply]

Prove

Hello. Is there an easy way to prove: ( ${\overrightarrow {u}}$ × ${\overrightarrow {v}}$ ) ⋅ (( ${\overrightarrow {v}}$ × ${\overrightarrow {w}}$ ) × ( ${\overrightarrow {w}}$ × ${\overrightarrow {u}}$ )) = ( ${\overrightarrow {u}}$ ⋅ ( ${\overrightarrow {v}}$ × ${\overrightarrow {w}}$ ))²? Using algebraic vectors would be a nightmare. Is there any property of dot or cross products that can simplify this identity? Thanks in advance. --Mayfare (talk) 12:42, 30 May 2010 (UTC)[reply]

According to Triple product, you have

a\cdot (b\times c)=b\cdot (c\times a)

and

(a\times b)\times (a\times c)=(a\cdot (b\times c))a

. Using these, the derivation is fairly straightforward. -- Meni Rosenfeld (talk) 13:09, 30 May 2010 (UTC)[reply]

Matrices to the power n in an n-dimensional complex vector space

Hi everybody,

Is it necessarily the case that if A and B are complex n x n matrices and (AB)^n = 0, then (BA)^n must also be 0?

I feel like I should be able to come up with a counterexample but having looked at the case n=2 I can't spot anything - this is just one of 5 parts of a question intended to take half an hour or less on an exam, so it shouldn't take long ("state a counterexample or provide a proof") - I tried looking at the rank briefly to produce a proof but I'm not convinced it's true, so I'd appreciate any suggestions you could make.

Thanks a lot, 131.111.185.68 (talk) 16:39, 30 May 2010 (UTC)[reply]

This is true. Proof outline: The only eigenvalue of AB is 0; the only eigenvalue of BA is 0; Cayley–Hamilton theorem. -- Meni Rosenfeld (talk) 18:07, 30 May 2010 (UTC)[reply]

Thankyou Meni, once again and indeed as always you've been a great help :) 82.6.96.22 (talk) 18:17, 30 May 2010 (UTC)[reply]

You're welcome :) -- Meni Rosenfeld (talk) 18:40, 30 May 2010 (UTC)[reply]

Have also a look to Characteristic polynomial#Characteristic polynomial of a product of two matrices --pm a 21:00, 30 May 2010 (UTC)[reply]