Navigation Panel: Previous | Up | Forward | Graphical Version

Assignment 9 Solutions

Page 143 #1
                     2      2            2n      2 n
           2        z     -z      z inf z     (-z )
   z cosh(z ) = z (e   + e   )  = - SUM --- + ------
                   ----------     2 n=0  n!     n!
                        2
   
                               n   2n
                z inf (1 + (-1) ) z                    n
              = - SUM ----------------   (note: 1 + (-1) = 2 if n even
                2 n=0       n!                               ie, n=2k
                                                           0 if n odd)
                         4k
                z inf 2 z
              = - SUM -----  (letting n = 2k, since odd terms are zero)
                2 k=0  (2k)!
   
   
                     4n+1
                inf z
              = SUM -----   (renaming k to n).
                n=0 (2n)!

Page 143 #3

   
          z      1         z inf    4   n 
   f(z) = - ------------ = - SUM (-z /9)   (provided |-z^4/9| < 1) 
          9 1 - (-z^4/9)   9 n=0
   
                     4n+1
          inf     n z
        = SUM (-1)  -----
          n=0       9^(n+1)
   
                     4n+1
          inf     n z
        = SUM (-1)  -----  .
          n=0       3^(2n+2)
   

This expansion is valid when |-z^4/9| < 1, i.e., when |z|^4 < 9, i.e., when |z| < sqrt(3).

Page 144 #4

Since

                3    5                   n
               w    w            inf  (-1)     2n+1
   sin w = w - -- + -- - ...  =  SUM  ------- w
               3!   5!           n=0  (2n+1)!
for all w, it follows that
                 6    10                  n
        2    2  z    z            inf  (-1)     4n+2
   sin z  = z - -- + -- - ...  =  SUM  ------- z
                3!   5!           n=0  (2n+1)!
for all z. Therefore, this series is the Maclaurin series for f(z) = sin(z^2).

However, we also know that the Maclaurin series is

        (k)
   inf f   (0)  k
   SUM ------- z
   k=0   k!
and therefore the coefficient f^(k)(0) must be zero when k is not of the form 4n+2. In other words, it is nonzero only when k is even but not a multiple of 4. It is therefore zero when k=2n+1 is odd, and it is zero when k=4n is a multiple of 4.

Page 144 #5

As indicated in the book, write

    1     1   1             z-i
   --- = --- ---  where w = ---.
   1-z   1-i 1-w            1-i

As long as |w| < 1 (which holds if |z-i| < |1-i| = sqrt(2)), this equals

    1  inf  n      1  inf     1         n     inf  (z-i)^n
   --- SUM w   =  --- SUM  ------  (z-i)   =  SUM  -------
   1-i n=0        1-i n=0 (1-i)^n             n=0  (1-i)^(n+1)

Page 144 #11

      1       1   1       1      1      
   -------- = - -----  =  --  -------- 
   4z - z^2   z 4 - z     4z  1 - (z/4)  
   
                         1  inf
                       = -- SUM (z/4)^n (provided |z/4| < 1)
                         4z n=0
   
                    n-1              k
              inf  z           inf  z
            = SUM  ------   =  SUM  ----      (letting k = n-1)
              n=0  4^(n+1)     k=-1 4^(k+2)
   
               -1         k
              z      inf z
            = ---  + SUM ------
              4^1    k=0 4^(k+2)
   
                          n
              1     inf  z
            = --  + SUM --------     (renaming k to n).
              4z    n=0  4^(n+2)

Page 151 #3

    1     1            1      1
   --- = ---------  =  -  ---------
   1+z   z(1/z + 1)    z  1 - (-1/z)
   
   
         1  inf       n
       = -  SUM (-1/z)     provided |-1/z| < 1, i.e. |z| > 1.
         z  n=0
   
         inf (-1)^n
       = SUM ------ .
         n=0 z^(n+1)

Page 151 #4

In addition to the point z=0 around which our Laurent series is to be based, f(z) is singular at z=1, but analytic everywhere else. Therefore, there will be two Laurent series for f(z) in powers of z: one on the region 0 < |z| < 1, the other on the region 1 < |z| < infinity .

On the region 0 < |z| < 1, we have |z| < 1, so

           1    1       1   inf  n    inf  n-2    -2    -1   inf  n
   f(z) = ---  ---  =  ---  SUM z   = SUM z    = z   + z   + SUM z.
          z^2  1-z     z^2  n=0       n=0                    n=0

On the region 1 < |z| < infinity . we have |1/z| < 1, so

           1    1       1   1    1      -1    1
   f(z) = ---  ---  =  ---  - ------- = --- --------
          z^2  1-z     z^2  z (1/z)-1   z^3 1 - (1/z)
   
          -1  inf      n      inf   1         inf 1
        = --- SUM (1/z)   = - SUM ------  = - SUM ---
          z^3 n=0             n=0 z^(n+3)     n=3 z^n
where in the last sum we have relabelled the index with the new n being the old n plus 3. (If this disturbs you you can do it in two stages: first switch to index k=n+3, then rename k to n).

Page 151 #5(b)

Note that we are looking for a Laurent series in powers of z. (In general, for a domain of the form a < |z-z_0| < b, the Laurent series is in powers of z-z_0).

On the domain 1 < |z| < infinity , |1/z| < 1, so we have

           z + 1       1       1   1       inf      n   1  inf      n
   f(z) = -------  = -----  +  - -----  =  SUM (1/z)  + -  SUM (1/z)
          z(1-1/z)   1-1/z     z 1-1/z     n=0          z  n=0
   
          inf  1     inf   1         inf   1    inf  1
        = SUM ---  + SUM ------   =  SUM  --- + SUM --- 
          n=0 z^n    n=0 z^(n+1)     n=0  z^n   n=1 z^n
   
           1    inf  1     inf  1             inf   1 
        = --- + SUM ---  + SUM ---   = 1 + 2  SUM  --- 
          z^0   n=1 z^n    n=1 z^n            n=1  z^n 

Page 151 #7

In addition to the point z=0 about which we are finding the Laurent series, the function f(z) is also singular at z= +/- i. These both lie on the circle |z|=1. Therefore, there will be two Laurent series for f(z) in powers of z: one in the region 0 < |z| < 1 and the other in the region 1 < |z| < infinity .

On the region 0 < |z| < 1 we have |-z^2| < 1, so

          1   1         1 inf    2 n    inf    n  2n-1
   f(z) = - --------  = - SUM (-z )   = SUM (-1) z    .
          z 1-(-z^2)    z n=0           n=0

On the region 1 < |z| < infinity we have |-1/z^2| < 1, so

          1       1           1      1             1  inf      2 n  
   f(z) = - -------------- = --- ------------  =  --- SUM (-1/z )
          z z^2[(1/z^2)+1]   z^3 1 - (-1/z^2)     z^3 n=0
   
               
          inf      n    1
        = SUM  (-1)  -------- .
          n=0        z^(2n+3)


Navigation Panel: 

  Go backward to Assignment 9 (due Thursday November 27th)
  Go up to Contents
  Go forward to Assignment 10 (due Thursday December 4th)
  Switch to graphical version (better pictures & formulas)