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)