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5. The Exponential Function and its Friends
The exponential function. Define
exp z =
∞
∑
zn
.
n!
n=0
This has radius of convergence R where
1/R = lim sup(1/n!)1/n = 0,
n→∞
so R = ∞ and the series converges absolutely for all z ∈ C.
We may differentiate term by term (Theorem!) to get
∞
∞
∞
∑
∑
∑
d
z n−1
z n−1
zn
exp z =
n
=
=
= exp z.
dz
n!
(n
−
1)!
n!
n=1
n=1
n=0
Exponential of a sum. Set f (z) = exp z exp(c − z), c ∈ C. Then
d
d
f ′ (z) =
exp z exp(c−z)+exp z exp(c−z) = exp z exp(c−z)−exp z exp(c−z) = 0.
dz
dz
By Theorem 3.5, this tells us f (z) is constant. This constant must be f (0) = exp c,
so exp z exp(c − z) = exp c. Set c = z1 + z2 , z = z1 , then
exp(z1 + z2 ) = exp z1 exp z2 .
From this it is easy to deduce (induction) that for n ∈ Z+ ,
exp nz = (exp z)n .
The number e. Define e = exp 1(= 2.7182818 . . . ). The above shows
exp n = en ,
n ∈ Z+ .
Clearly
exp 0 = 1 = e0 .
Hence
exp n exp(−n) = exp(n − n) = exp 0 = 1
so
exp(−n) = (exp n)−1 = e−n .
So
exp(n) = en ,
n ∈ Z.
For a rational number m/n (n > 0),
(exp(m/n))n = exp(nm/n) = exp m = em ,
so
exp(m/n) = (em )1/n = em/n .
It is then consistent with this to use the notation
exp z = ez ,
for all z ∈ C.
The formula (*) becomes
ez1 +z2 = ez1 ez2 .
If z = x + iy, we have
ez = ex eiy .
Notice also that
exp z exp(−z) = exp 0 = 1,
so exp z ̸= 0 for all z ∈ C.
1
Trigonometric functions. Define
cos z =
∞
∑
(−1)n
n=0
∞
∑
z 2n
,
(2n)!
z 2n+1
sin z =
(−1)
.
(2n + 1)!
n=0
n
By the Ratio Test, these converge absolutely for all z ∈ C.
Substituting −z, we see cos is an even function and sin is an odd function, i.e.,
sin(−z) = − sin z.
cos(−z) = cos z,
Also
cos 0 = 1,
sin 0 = 0.
Differentiating term by term,
d
cos z = − sin z,
dz
For example,
d
sin z = cos z.
dz
∞
∑
d
z 2n
sin z =
(−1)n · (2n + 1)
dz
(2n + 1)!
n=0
=
∞
∑
(−1)n
n=0
z 2n
= cos z.
(2n)!
Theorem. (Euler’s Formula)
eiz = cos z + i sin z.
Proof.
z2
iz 3
z4
iz 5
eiz = 1 + iz −
−
+
+
+ ···
2!
3!
4!
5!
(
)
(
)
z2
z4
z3
z5
= 1−
+
+ ··· + i z −
+
+ ···
2!
4!
3!
5!
= cos z + i sin z.
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