Definition. For z ∈ C we define
cos z =
1 iz
(e + e−iz );
2
sin z =
1 iz
(e − e−iz );
2i
cosh z =
1 z
(e + e−z );
2
sinh z =
1 z
(e − e−z ).
2
Remark In all of what follows we will make use of the fundamental identity
ez+w = ez ew ,
(∗)
z, w ∈ C.
Here is a proof of (*) in which we leave out the nontrivial justification of some of the steps. We have
∞
X
(z + w)n
n!
n=0
!
Ã
∞
n
X
X
z m wn−m
=
m! (n − m)!
n=0 m=0
!
à ∞
∞
X X z m wn−m
=
m! (n − m)!
m=0 n=m
Ã
!
∞
∞
X
z m X wn−m
=
m! n=m (n − m)!
m=0
Ã
!
∞
∞
X zm X
wn
=
m! n=0 n!
m=0
ez+w =
=
∞
X
zm w
e
m!
m=0
= ez ew .
From the definitions we immediately deduce that
eiz = cos z + i sin z
(1)
and that
cos2 z + sin2 z = 1,
(2)
cosh2 z − sinh2 z = 1.
From (*) we obtain the addition formulae
(3)
sin(z + w) = sin z cos w + cos z sin w,
cos(z + w) = cos z cos w − sin z sin w.
Differentiating term by term the power series that define the exponential function we find that
d z
e = ez .
dz
(4)
d
f (az) = af 0 (az) provided f is differenOne could also use (*) to deduce (4). For any constant a we have dz
tiable in the complex; this fact, (1) and the definitions immediately imply that
(5)
d
cos z = − sin z;
dz
d
sin z = cos z;
dz
d
cosh z = sinh z;
dz
1
d
sinh z = cosh z.
dz
We have
eiz =
(6)
∞
∞
∞
∞
∞
X
X
X
X
X
(iz)n
(iz)2m
(iz)2m+1
(−1)m z 2m
(−1)m z 2m+1
=
+
=
+i
.
n!
(2m)!
(2m + 1)! m=0 (2m)!
(2m + 1)!
n=0
m=0
m=0
m=0
This gives
e−iz =
(7)
∞
∞
X
X
(−1)m z 2m
(−1)m z 2m+1
−i
.
(2m)!
(2m + 1)!
m=0
m=0
Combining (6) and (7) we get
(8)
cos z =
∞
X
(−1)m z 2m
,
(2m!)
m=0
sin z =
∞
X
(−1)m z 2m+1
.
(2m + 1)!
m=0
We have
ez =
(10)
∞
∞
∞
X
X
X
zn
z 2m
z 2m+1
=
+
.
n!
(2m)! m=0 (2m + 1)!
n=0
m=0
This gives
e−z =
(11)
∞
∞
X
X
z 2m
z 2m+1
−
.
(2m)! m=0 (2m + 1)!
m=0
Combining (10) and (11) we get
(12)
cosh z =
∞
X
z 2m
,
(2m)!
m=0
sinh z =
∞
X
z 2m+1
.
(2m + 1)!
m=0
If z = x + iy where x, y are real we have
´
1 ³ i(x+iy)
cos z =
e
+ e−i(x+iy)
2
¢
1 ¡ −y+ix
=
e
+ ey−ix
2
¢
1 ¡ −y
(13)
e (cos x + i sin x) + ey (cos x − i sin x)
=
2
¢
1¡
=
cos x(ey + e−y ) − i sin x(ey − e−y )
2
= cos x cosh y − i sin x sinh y.
Changing a sign and dividing by i in the third line of the preceding calculation one obtains
¢
1 ¡ −y
e (cos x + i sin x) − ey (cos x − i sin x)
sin z =
2i
¢
1 ¡
(14)
=
i sin x(ey + e−y ) − sin x(ey − e−y )
2i
= sin x cosh y + i cos x sinh y.
Finally,
(15)
cosh z = cos iz = cos(−y + ix) = cos(−y) cosh x + sin(−y) cosh x = cosh x cos y − cosh x sin y
and
(16)
sinh z = −i sin(iz) = −i sin(−y + ix) = −i(sin(−y) cosh x + i cos(−y) sinh x) = sinh x cos y + i cosh x sin y.
Whew!
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