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Functions of Complex Variables

Date: 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 

     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 

Thank you, 

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) 

     ------- + cos(2x)

instead of:

     cosh(2y)+ cos(2x)

I would expect to see this last expression written as


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) = ----------------

(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) = -------------------------------------

               = cos(w)


     cosh(-iw) = cosh(i*(-w))

               = cos(-w)

               = cos(w)

Since i^4 = 1

     - = -i


     f(z) = 2*u(z/2,z/(2i)) - u(0,0)

          = 2 * -------------------------- - 0

          = 2 * ------------------
                cosh(-iz) + cos(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:

     sin(x) = --------------

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^(ix) + e^(i2x) + ... + e^(inx) = e^(ix)* -------------

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)


     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   
Associated Topics:
College Analysis
College Trigonometry

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