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...
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    Visual complex analysis problem

    you mean...meet at a given angle c? i am stuck, to be honest^^
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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...?
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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...
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    Visual complex analysis problem

    an ellipse, is my guess?
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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.
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    Construct geometric line with nested root

    nevermind :) but you gave me some new ways to think about this... maybe there are some other viewpoints?
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    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)
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    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.
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    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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    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.
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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..
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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.
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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
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