Recent content by raphael3d
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Visual complex analysis problem
clearly we live in shifted time zones =) well thank you, i will look into that...surely there will be a wiki or something similar. this is a problem of "visual complex analysis" by tristan needham. a wonderful book :) here is it: http://www.mathsisfun.com/geometry/circle-theorems.html...- raphael3d
- Post #10
- Forum: Calculus and Beyond Homework Help
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Visual complex analysis problem
you mean...meet at a given angle c? i am stuck, to be honest^^- raphael3d
- Post #8
- Forum: Calculus and Beyond Homework Help
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Visual complex analysis problem
if a is 1 and b is i on the unit circle, then z lies in the first quadrant? i would guess the angle where a and b meet z doesn't change as long as z lies between them...?- raphael3d
- Post #7
- Forum: Calculus and Beyond Homework Help
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Visual complex analysis problem
two lines from two distinctive points a,b to one point z. whereas those lines form angles with the horizontal and the difference between those angles is constant. all the points lie on a circle... i have drawn the lines and points and angles, but i don't know how to proceed from here... what...- raphael3d
- Post #5
- Forum: Calculus and Beyond Homework Help
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Visual complex analysis problem
an ellipse, is my guess?- raphael3d
- Post #3
- Forum: Calculus and Beyond Homework Help
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Visual complex analysis problem
Homework Statement Explain geometrically why the locus of z such that arg [ (z-a)/(z-b) ] = constant is an arc of a certain circle passing through the fixed points a and b. i tried to visualize the equation in a cartesian co-system but in doing so, i was not very successful.- raphael3d
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- Analysis Complex Complex analysis Visual
- Replies: 10
- Forum: Calculus and Beyond Homework Help
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Undergrad Construct geometric line with nested root
nevermind :) but you gave me some new ways to think about this... maybe there are some other viewpoints?- raphael3d
- Post #5
- Forum: General Math
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Undergrad Construct geometric line with nested root
a prependicular of length 1 on the end of the root2 line means root((root2)^2 + 1^2) = root3 which is not equal to root(root2 +1)- raphael3d
- Post #3
- Forum: General Math
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Undergrad Construct geometric line with nested root
well it is easy to construct sqrt(2) with a triangle with two sides of length 1. but what about sqrt(2 + sqrt(2)) or the iteration sqrt(2 + sqrt(2 + sqrt(2))). the question is how to construct a line with length sqrt(sqrt(2)) i guess(beginning with lines of length 1), but i am not sure.- raphael3d
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- Geometric Line Root
- Replies: 4
- Forum: General Math
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Graduate Proof geometrically that QX and QY are the roots
excuse me sir, here is a the working link: http://img717.imageshack.us/img717/4029/screenshot20110106at123.png thank you -
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Graduate Proof geometrically that QX and QY are the roots
http://img717.imageshack.us/img717/4...10106at123.png its a construction from felix klein, around 1900. it should be made without trigonometric functions or complex algebra etc. the ratios of lengths should be sufficient, but i have only found one similar triangle and then i got stuck... -
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Show that M(z) maps the unit circle to itself.
well after multiplication of M(z)*M(z)', i get 1. i guess that's it.thank you dick.- raphael3d
- Post #20
- Forum: Calculus and Beyond Homework Help
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Show that M(z) maps the unit circle to itself.
but what does the complex conjugate of (z-a)/(a'z-1) looks like? i have a mental blackout as it seems..- raphael3d
- Post #18
- Forum: Calculus and Beyond Homework Help
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Show that M(z) maps the unit circle to itself.
meaning M(z)*M(z)' = |M(z)|2, but how can i evaluate |(z-a)/(a'z-1)|2 ? thank you for your patience.- raphael3d
- Post #15
- Forum: Calculus and Beyond Homework Help
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Show that M(z) maps the unit circle to itself.
you mean M(z)*M(z') of i am not mistaken? assuming M(z)=z => M(z)*M(z') = z*z' = |z|^2=1- raphael3d
- Post #13
- Forum: Calculus and Beyond Homework Help