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^nwhere 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)