Functions of Complex VariablesDate: 02/20/2001 at 14:48:35 From: Catherine Subject: Advanced Complex Variables Dear Dr. Math, I have two problems in advanced complex variables that I am stuck on. The first is about a method of finding f(z) when u(x,y) is known. f(z) is analytic in a domain D. Since z = x+iy, and u(x,y) = 1/2[f(z)+(f(z))] bar But f(z)-bar is f-bar z-bar, thus making the above expression: f(z) = 1/2[ f(x+iy) + f-bar(x-iy)] Next Ahlfors uses the substitution x = z/2 and y = z/2i, which implies that: f(z) = 2U(z/2,z/2i) - f-bar(0) where f-bar(0) = Real[f-bar(0)] + i*Im[f bar(0)] Let u(x,y) = sin(2x) / cosh(2y)+cos(2x), find f(z). Here's where I'm stuck. There wasn't a worked-out example using this method; in fact there isn't much of a formal explanation of this method at all, but it works. Could you please show me how to work this one out? When I work it out by finding the harmonic conjugate of u, v; the answer is tan z. Is that right? The second problem I have is showing that sin(x) + sin(2x) + sin(3x) + ... + sin(nx) = 1/2 cot(x/2)cos[(n+1/2)x]/2sin(x/2) Here's what I tried. I know that sin(nx) = n cos^n - 1sin x - (nC3) cos^(n-3)sin^3x Once I apply this, I can factor it and get an expression of the type 1 + p + p^2 + p^3 + ... + p^n which is a convergent geometric series for abs(p) < 1. This applies since both sin and cos take on values between -1 and +1. In fact, the symbol used is theta, not x. Could you shed a little light on this problem? Thank you, Catherine Date: 02/26/2001 at 22:05:08 From: Doctor Fenton Subject: Re: Advanced Complex Variables Hi Catherine, Thanks for writing to Dr. Math. Your formula sent me on a wild goose chase for a while, because I interpreted sin(2x) / cosh(2y)+cos(2x) as: sin(2x) ------- + cos(2x) cosh(2y) instead of: sin(2x) ----------------- cosh(2y)+ cos(2x) I would expect to see this last expression written as sin(2x)/(cosh(2y)+cos(2x)) Ahlfors' argument, as he admits, is heuristic, and is more of a mnemonic, but he assures the reader that it can be proved. In this case, u(0,0) = 0, and the key is that for an imaginary argument iw (w real), e^(iw) + e^(-iw) cosh(iw) = ---------------- 2 (you get this from the formula for cosh(x) when x is real, by just replacing x by a complex value, in this case, a pure imaginary one). Using Euler's formula: e^(it) = cos(t) + i*sin(t) cos(w) + i*sin(w) + cos(w) - i*sin(t) cosh(iw) = ------------------------------------- 2 = cos(w) Also, cosh(-iw) = cosh(i*(-w)) = cos(-w) = cos(w) Since i^4 = 1 1 - = -i i so f(z) = 2*u(z/2,z/(2i)) - u(0,0) sin(2*(z/2)) = 2 * -------------------------- - 0 cosh(2*(z/2i))+cos(2*(z/2) sin(z) = 2 * ------------------ cosh(-iz) + cos(z) sin(z) = 2 * --------------- cos(z) + cos(z) = tan(z) In your second problem, I usually use the fact that sin(x) = Im[e^(ix)] the imaginary part of e^(ix). You can also write: e^(ix)-e^(-ix) sin(x) = -------------- 2i In either case, you obtain a series of complex exponentials. sin(x) + sin(2x) + sin(3x) + ... + sin(nx) is the imaginary part of e^(ix) + e^(i2x) + e^(i3x) + ... + e^(inx) which is a geometric series, with first term a = e^(ix) and ratio r = e^(ix), so e^(i(n+1)x)-1 e^(ix) + e^(i2x) + ... + e^(inx) = e^(ix)* ------------- e^(ix)-1 You can rewrite e^(ix)-1 = e^(i*(x/2)) * [e^(i(x/2)-e^(-i(x/2))] = e^(i*(x/2)) * 2i * sin(x/2) and e^(i(n+1)x)-1 = e^(i(n+1)x/2)*[ e^(i(n+1)x/2) - e^(i(n+1)x/2)] = e^(i(n+1)x/2)*[ 2i * sin((n+1)x/2) ] The factor e^(i*(x/2)) in the denominator will cancel half the factor e^(ix) in the numerator, but I'm not sure how to interpret the expression you're trying to obtain. Remember that you are looking for only the imaginary part of this expression, but there could be some trig identities needed to get it into exactly the form you want. I hope this is of some help to you. If you have further questions, please write us again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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