In consequence
ez = ex+iy = ex eiy = ex (cos y + i sin y),
√
|e | = e · | cos y + i sin y| = e
cos2 y + sin2 y = ex .
and
z
x
x
Corollary.
(1)
(2)
(3)
(4)
(5)
cos z = 12 (eiz + e−iz )
1
sin z = 2i
(eiz − e−iz )
cos2 z + sin2 z = 1
sin(z + w) = sin z cos w + cos z sin w
cos(z + w) = cos z cos w − sin z sin w
Proof. Exercises.
Hyperbolic functions. Define
cosh z =
1 z
(e + e−z ),
2
sinh z =
1 z
(e − e−z ).
2
Thus:
sin iz = i sinh z,
cos iz = cosh z,
sinh iz = i sin z,
cosh iz = cos z,
cosh2 z − sinh2 z = 1.
For example, sin iz =
1
i2 z
2i (e
1
− e−i z ) = − 2i
(ez − e−z ) = i sinh z.
2
Zeros of sin z and cos z. Let z = x + iy. Then:
sin z = sin(x + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y.
Therefore
| sin z|2 = sin2 x cosh2 y + cos2 x sinh2 y
= sin2 x(1 + sinh2 y) + (1 − sin2 x) sinh2 y
= sin2 x + sinh2 y.
Now
sin z = 0 ⇐⇒ sin2 x + sinh2 y ⇐⇒ sin x = 0 and sinh y = 0
⇐⇒ x = nπ, n ∈ Z, y = 0.
Similarly
cos z = 0 ⇐⇒ z = (n + 1/2)π, n ∈ Z.
Definition. For a function f : C → C, a nonzero number k ∈ C is called a period if
f (z + k) = f (z), for all z ∈ C.
1
Periods of sin z, cos z. Suppose sin(z + k) = sin z for all z.
Put z = 0 then k = nπ.
Now
sin(z + nπ) = sin(z + (n − 1)π + π)
= sin(z + (n − 1)π) cos π + cos(z + (n − 1)π) sin π
= − sin(z + (n − 1)π)
= (−1)n sin z.
So sin(z + nπ) = sin z if only only if n is even. Hence the periods of sin are 2πz,
n ∈ Z \ {0}.
Similarly for cos.
Periods of exp z. Suppose exp(z + k) = exp z for all z ∈ C.
Put z = 0 then exp k = 1.
Let k = α + iβ. Then
1 = exp k = exp α(cos β + i sin β).
So
exp α cos β = 1
(1)
exp α sin β = 0
(2)
(2) =⇒ β = nπ, n ∈ Z.
But (1) requires β = 2nπ, n ∈ Z, as cos β must be > 0.
Then cos β = 1 so exp α = 1, i.e. α = 0.
So k = 2nπi, n ∈ Z.
Conversely
exp(z + 2nπi) = exp z exp 2nπi = exp z(1 + 0i) = exp z.
So the periods of exp are k = 2nπi, n ∈ Z, n ̸= 0.
The logarithmic function.
Let z ∈ C. Consider
exp w = z.
(∗)
By above, if w1 is a solution of (*) then so is w1 + 2nπi. Each of these values is
called a logarithm of z and is written log z.
If x ∈ R, x > 0 then
exp w = x
has a unique real solution. We call this log x.
In (*), let w = u + iv:
z = exp w = exp u(cos v + i sin v),
so |z| = exp u and v is a value of arg z.
2