Recent content by Tranquillity

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    MHB K5 Graph: Hamiltonian Circuits & Analysis

    I have never done anything related with Graph Theory but with a google search for your problem I found this pdf, where it states that the number of Hamilton circuits in K(n) is (n-1)! [factorial] Hope the link will help you with the knowledge you have on graph theory...
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    MHB Fluid Dynamics in Action: (Mechanics)

    If I am not mistaken, you are looking for plane Couette flow! Have a look at http://www.maths.manchester.ac.uk/~dabrahams/MATH45111/files/lecture11.pdf (first three pages) Note that in your problem, the moving plate and the one at rest, are the other way around! Hope this helps! Kindest regards
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    MHB A second complex analysis question

    Thanks for that! :)
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    MHB A second complex analysis question

    Shouldn't be z= (pi/2) + 2*pi*k - ln (2+/- sqrt(5))*i, where k in Z ? Since following Prove it's way, we have for e^(iz) = i(2+sqrt(5)) that y=-ln(2+sqrt(5)) and that e^(ix) = e^(i*pi/2) which holds if and only if ix=i*pi/2 + 2*pi*k Similarly for e^(iz) = i(2-sqrt(5)) that y=-ln(2-sqrt(5))...
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    MHB Solving Complex Analysis: Find z for z^(1+4i) = i

    Basically I might have done something wrong! I actually managed to solve the exercise by finding Logz and then had to say that my solution requires Im(Logz)=Argz to be in (-pi,pi)! So this will restrict the values of k! Then z=exp(...) in the cut plane Do, is valid for k in [-8,8]! And the...
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    MHB A second complex analysis question

    Terrific! There are so many ways to solve this exercise, the thing is that we have learned only a particular material in order to use! For example we did not do any other complex functions except sin, cos, exp and Log, we are not even supposed to use the multivalued logarithm! So I suppose the...
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    MHB A second complex analysis question

    That is a brilliant method avoiding any logarithms and any ambiguities! Thank you very much! The only thing I can add is that x could be (pi/2) + 2*k*pi where k is an integer for a complete solution, would you agree with that?
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    MHB A second complex analysis question

    Ok let's try to do it with the log instead of Log. I have to solve e^(iz) = i(2+sqrt(5)) and e^(iz) = i(2-sqrt(5)) Take the first equation. Taking logarithms yields i*z + 2*pi*i*k = ln(2+sqrt(5)) + i*((pi/2) + 2*pi*k), k is integer pi/2 is the principal argument plus 2*pi*k to obtain arg z...
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    MHB A second complex analysis question

    Thank you very much for your reply :) But does log(e^(iz)) always equal iz? For example Log(e^(4i)) = 4i-2*i*pi Basically by doing an exercise I have realized that Log(e^(iz)) = iz - 2*k*pi*i, where k is an integer. So does log(e^(iz)) always equal to iz but Log(e^(iz)) = iz - 2*k*pi*i...
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    MHB A second complex analysis question

    So this time I have to solve cos(z)=2i My approach: cos(z)= [ e^(iz) + e^(-iz) ] / 2 = 2i Rearranging and setting e^(iz) = w we get a quadratic w^2 - 4iw + 1 = 0 The quadratic yields two solutions: w=e^(iz) = i(2 + sqrt(5)) or e^(iz) = i(2-sqrt(5))And now my problem is here. In the...
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    MHB Solving Complex Analysis: Find z for z^(1+4i) = i

    Guys thanks for your replies but this was not what I was asking exactly!
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    MHB Solving Complex Analysis: Find z for z^(1+4i) = i

    Hello guys! I have to find for z in D= C \ {x in R: x<=0} with z^(1+4i) = i a) all possible values of Log(z) b)all possible values of z.Now my approach is: Write z^(1+4i) = exp((1+4i) * Logz) = i = exp(i*pi/2) which holds iff (1+4i) * Logz = i*pi/2 + i*k*2*pi where k in Z.After a lot of...
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    MHB How to Sketch and Interpret Streamlines in a Streamfunction?

    I have managed to plot the streamlines, any clue on the physical interpretation of the two terms?
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    MHB How to Sketch and Interpret Streamlines in a Streamfunction?

    Hello guys! I have the streamfunction \(\psi(r,\,\theta) = U\left(r-\frac{a^2}{r}\right)\sin{\theta}\), \(a\) is a positive constant and I have to sketch the streamlines and also interpret the two terms in the streamfunction! Any clue on how to do that? Thank you very much Kindest regards
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