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Math 132, section 1a (Prof. Balmer)
Homework #9 — Solutions
V.7.1. Find the zeros and orders of zeros of the following functions.
(a) f (z) = z^2 + 1 z^2 − 1 Since we have f (z) = (^) ((zz−−1)(i)(zz++1)i) , wee see that i and −i are zeros of f. We can write f (z) = (z − i)g(z), where g(z) = (^) (z−1)(z+zi+1) is analytic at i and g(i) 6 = 0, so we conclude that z = i is a simple zero (i.e. a zero of the first order) of f. Similarly we see that z = −i is also a simple zero of f. (b) f (z) =
z
z^5 From f (z) = z (^4) + z^5 we see that^ ±^ √^1 2 ±^ √i 2 are zeros of^ f^ , and similarly as in problem (a) we conclude that all these zeros are simple. (c) f (z) = z^2 sin z Since f (z) = 0 if and only if z = 0 or sin z = 0, we obtain that all zeros of f are solutions of eiz^ − 2 ei− iz = 0. This can be simplified to e^2 iz^ = 1 = e^0 , i.e. 2 iz = 2πik for some k ∈ Z, so all zeros are of the form z = kπ for k ∈ Z, What is the order of 0 in the function g(z) = sin z? Since g(0) = 0, g′(0) = cos 0 = 1 6 = 0, we see that z = 0 is a first order zero of g. On the other hand, z = 0 is a second order zero in z^2 , so we conclude that z = 0 is a zero of the third order in the function f (z) = z^2 g(z). Alternatively we can see this from the power series expansion:
g(z) = sin z = z − z^3 3!
z^5 5! −... = z
z^2 3!
z^4 5!
= z h(z),
i.e. f (z) = z^2 g(z) = z^3 h(z), where we have denoted h(z) = 1− z 3!^2 + z 5!^4 −.. .. Since h(0) = 1 6 = 0, we conclude that z = 0 is a third order zero of f. What is the order of other zeros kπ, k ∈ Z, k 6 = 0 in f? We can use 2 πi-periodicity of sin z to conclude that all these are simple zeros of sin z. Since they are not zeros of z^2 , they also have to be simple zeros of f. (d) f (z) = cos z − 1 Similarly as in problem (c) we solve cos z = 1 to obtain all zeros of f and these are 2kπ, k ∈ Z. Since for any k ∈ Z we have f (2kπ) = 0, f ′(2kπ) = − sin(2kπ) = 0, f ′′(2kπ) = − cos(2kπ) = − 1 6 = 0, we conclude that these are zeros of the second order.
Another way to see that z = 0 is a second order zero is from the power expansion
f (z) = cos z − 1 = − z^2 2!
z^4 4! −... = z^2
z^2 4!
Then we can use 2π-periodicity to conclude that all 2kπ, k ∈ Z are also second order zeros. (e) f (z) =
cos z − 1 z Since (^1) z is analytic and nonzero at z = 2kπ for k ∈ Z, k 6 = 0, we conclude from problem (d) that these are still second order zeros of f. The power expansion at 0 is
f (z) = z
z^2 4!
so z = 0 is now a simple zero of f. (f) f (z) = cos z − 1 z^2 Since (^) z^12 is analytic and nonzero at z = 2kπ for k ∈ Z, k 6 = 0, we conclude from problem (d) that these are still second order zeros of f. The power expansion at 0 is
f (z) = −
z^2 4!
so f (0) = −^12 and z = 0 is not even a zero of f. (g) f (z) = ez^ − 1 This is very similar to problem (d). Here z = 2kπi, k ∈ Z are simple zeros of f.
(h) f (z) = sinh^2 z + cosh^2 z We first simplify: f (z) = e 2 z (^) +e− 2 z (2k+1)πi^2 = cosh(2z), and then deduce that^ z^ = 4 ,^ k^ ∈^ Z^ are simple zeros of^ f^. (i) f (z) = Log z z To solve Log z = 0, write z = |z|eiϕ, −π ≤ ϕ < π, and from log |z|+iϕ = 0 conclude |z| = 1, ϕ = 0, i.e. z = 1. Therefore z = 1 is the only zero of f. Since f (1) = 0, f ′(z) = 1 −Logz 2 z, f ′(1) = 1 6 = 0, we conclude that z = 1 is a simple zero of f.
In particular we know how to expand f (z) = (^) z−^11 − (^1) z. The function f is analytic on C \ { 0 , 1 }, so we are expanding it on domains 0 < |z| < 1 and |z| > 1. On the “punctured disk” 0 < |z| < 1 we have:
f (z) = −
1 − z
z
∑^ ∞
n=
zn^ −
z
∑^ ∞
n=− 1
zn,
while on the “outer disk” |z| > 1 we have:
f (z) =
z
1 − (^1) z
z
z
∑^ ∞
n=
z
)n −
z
∑^ −^1
n=−∞
zn^ − z−^1 =
∑^ −^2
n=−∞
zn.
(b) f (z) = z − 1 z + 1 The function f is analytic on C \ {− 1 }, so we are expanding it on |z| < 1 and |z| > 1. We first write f (z) = (z+1) z+1− 2 = 1 − (^) z+1^2 , and then simply apply the recipe from problem (a). On the disk |z| < 1 we have:
f (z) = 1 − 2 ·
1 − (−z)
∑^ ∞
n=
(−z)n^ = 1 − 2
∑^ ∞
n=
(−1)nzn,
while on the “outer disk” |z| > 1 we have:
f (z) = 1 −
z
1 − (−^1 z )
z
∑^ ∞
n=
z
)n
∑^ ∞
n=
(−1)n zn+^
∑^ −^1
n=−∞
(−1)nzn.
(c) f (z) =
(z^2 − 1)(z^2 − 4) Since f (z) = (^) (z+1)(z−1)(^1 z+2)(z−2) , we see that f is analytic on C{− 2 , − 1 , 1 , 2 }, so we are expanding it on |z| < 1, on 1 < |z| < 2 and on |z| > 2. Let us rewrite f (z) as
f (z) =
(z^2 − 1)(z^2 − 4)
(z^2 − 1) − (z^2 − 4) (z^2 − 1)(z^2 − 4)
z^2 − 4
z^2 − 1
(We could also decompose f completely to partial fractions, but the above decomposition is already enough and is much shorter.)
On |z| < 1 we have:
f (z) =
1 − z 42
1 − z^2
∑^ ∞
n=
z^2 n 4 n^
∑^ ∞
n=
z^2 n
∑^ ∞
n=
(1 − 4 −n−^1 )z^2 n.
On 1 < |z| < 2 we have:
f (z) =
1 − z 42
z^2
1 − (^) z^12
∑^ ∞
n=
z^2 n 4 n^
z^2
∑^ ∞
n=
z^2 n
∑^ −^1
n=−∞
z^2 n^ −
∑^ ∞
n=
z^2 n 4 n+^
Finally, on |z| > 2 we have:
f (z) =
z^2
1 − (^) z^42
z^2
1 − (^) z^12
z^2
∑^ ∞
n=
4 n z^2 n^
z^2
∑^ ∞
n=
z^2 n
∑^ ∞
n=
(4n^ − 1)
z^2 n+^
∑^ −^1
n=−∞
(4−n−^1 − 1) z^2 n.
VI.1.2. For each of the functions in the preceding exercise, find the Laurent expansion centered at z 0 = − 1 that converges at z = 12. Determine the largest open set on which each series converges. In these problems we substitute w = z − z 0 = z + 1, i.e. z = w − 1 and then expand the function g(w) = f (w −1) into the Laurent series centered at w 0 = 0, on the region that contains w = 12 + 1 = 32.
The expansion on 1 < |w| < 2 is:
g(w) = −
1 − w 3
w
1 − (− (^) w^1 )
1 − w 2
6 w
∑^ ∞
n=
wn 3 n+^
∑^ ∞
n=
(−1)n wn+^
∑^ ∞
n=
wn 2 n+^
6 w
∑^ −^1
n=−∞
(−1)nwn^ +
w−^1 +
∑^ ∞
n=
2 n^
3 n+
wn^ =
∑^ −^2
n=−∞
(−1)nwn^ +
w−^1 +
∑^ ∞
n=
(2−n^ − 3 −n−^1 ) wn
The desired Laurent expansion of f is
f (z) =
n=−∞
(−1)n(z + 1)n^ + (z + 1)−^1 +
∑^ ∞
n=
(2−n^ − 3 −n−^1 ) (z + 1)n
and the largest open set on which it converges is the annulus 1 < |z+1| < 2.
Vjekoslav Kovaˇc http://www.math.ucla.edu/∼vjekovac/
