SOLUTION TO PROBLEM SET 1, FYS3140
disk of convergence
a)
∞
P
n=0
Sn+1 n(n + 1)(z − 2i) using test ratio:: limn→∞ Sn = R < 1
n
Sn+1 = n + 1(n + 1 + 1)(z − 2i)n+1 = n + 1(n + 2)(z − 2i)n+1
Sn = n(n + 1)(z − 2i)n
Sn+1
n + 1(n + 2)(z − 2i)n+1
=
Sn
n(n + 1)(z − 2i)n
(n + 2)(z − 2i)
=
n
(n + 2)(z − 2i) Sn+1 = lim lim n→∞ Sn n→∞ n
|z − 2i| < 1
Solution to the above complex equation in complex plane gives the region
with in which the series converges. In this case it is a disk of radius ”1” with
centrum at (0,2).
∞
P
b)
2n (z + i − 3)2n
n=0
Sn+1 = 2n+1 (z + i − 3)2n+2
Sn = 2n (z + i − 3)2n
Sn+1
= 2(z + i − 3)2
Sn
Sn+1 = lim 2(z + i − 3)2 lim n→∞
n→∞
Sn
= 2(z + i − 3)2 < 1
= |2(z + i − 3)|2 < 1
√
= |(z + i − 3)| < 1/ 2
neglecting -1 as absolute
√ value is positive real number, and the disk of convergence has radius 2/2 centred at (3,-1).
1
Euler’s formula, powers, and roots
√
√
5iπ
a) 2 exp 4 = 2(cos 5π
+ i sin( 5π
))
4
4
Ans. = −1 − i
√
b) using polar form: (1+i) implies r = 2, θ = π4 +2nπ; as it lies in the first quadrant.
√
( 3−i) implies r = 2, θ = 7π
+2nπ as it lies in the fourth quadrant of the complex plane.
6
Hence,
√ iπ
(1 + i)48
( 2e 4 +2nπ )48
√
=
−iπ
( 3 − i)25
(2e 6 +2nπ )25
2nπ terms√ do not yield unique value; thus can be neglected.
Ans. 43 + i 14
√
i5π
c) 8i 3 − 8 = 16e 6 +i2nπ ; It has four distinct roots.
i5π
i5π
i5π
iπ
i5π
i3π
Ans: 2e 6 ; 2e 6 + 2 ; 2e 6 +iπ ; 2e 6 + 2
d) the sum of the three cube roots of 8 is: (2 + 2ei2π/3 + 2ei4π/3 ) = 0.
nth root of any complex
reiθ+2kπ/n
k runs 1, ... n.
Pnwhere
Pn number:
i2kπ/n
iθ+i2kπ/n
iθ
= re
the sum of roots:
k=1 e
k=1 re
comparing it with geometric progression, its sum can be written as:
i2π ))
Ans. ei2π (1−(e
=0
i2π/n
1−e
Elementary functions
We use the follwing definitions for the remaining problems::
iz
−iz
iz
−iz
z
−z
z
−z
sin z = e −e
cos z = e +e
sinh z = e −e
cosh z = e +e
2i
2
2
2
i4x −e−i4x
a) use sin2 (4x) = ( e
2i
)2 =
ei8x +e−i8x −2
−4
similarly for (b) and (c).
d) For the inner argument
1−i
1+i
= −i
ln(−i) = Ln(1) + i(3π/2 ± 2nπ)
sin(i ln(−i)) = sin(3π/2 ± 2nπ) = −1
e) using ab = eb ln a ; (−e)iπ = eiπ ln(−e) = e(iπ(1+i(π±2nπ)) = eiπ−π
2
Ans. −e−π (1±2n)
2
2 (1±2n)
−w
w
−e
f) Let z = tanh w = eew +e−w
rewriting this equation for w gives
e2w − 1
1+z
z = 2w
→ e2w =
e +1
1−z
w = tanh−1 z =
1 1+z
ln
2 1−z
Extra problem(2.7.30)
z
e =
∞
X
zn
n=0
thus,
n!
e
√
X xn 2n einπ/4
=
n!
n
ex(1−i)
√
X xn 2n e−inπ/4
=
n!
n
x(1+i)
similarly,
using the definitions of cos x and sin x given above
ex cos x =
ex(1+i) + ex(1−i)
2
√
X ( 2x)n (einπ/4 + e−inπ/4 )
=
n!
2
n
=
√
X ( 2x)n
n
n!
cos(nπ/4); which has zero contribution for n = 2, 6, 10, ...
similarly,
ex sin x =
ex(1+i) − ex(1−i)
2i
√
X ( 2x)n (einπ/4 − e−inπ/4 )
=
n!
2i
n
√
X ( 2x)n
=
sin(nπ/4); which has zero contribution for n = 0, 4, 8, 12 ...
n!
n
3