NPTEL web course
on
Complex Analysis
A. Swaminathan
I.I.T. Roorkee, India
and
V.K. Katiyar
I.I.T. Roorkee, India
A.Swaminathan and V.K.Katiyar (NPTEL)
Complex Analysis
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Complex Analysis
Module: 3: Sequences and Series
Lecture: 4: Hyperbolic functions and Logarithmic functions
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Complex Analysis
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Elementary functions
Hyperbolic functions
Definition
The hyperbolic functions sinh z and cosh z are defined by
exp (z) − exp (−z)
,
2
exp (z) + exp (−z)
cosh z := hyp cos z =
.
2
sinh z := hyp sin z =
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Complex Analysis
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Hyperbolic functions
Note.
Clearly sinh(iz) = i sin z and sin(iz) = i sinh z.
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Complex Analysis
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Hyperbolic functions
Note.
Clearly sinh(iz) = i sin z and sin(iz) = i sinh z.
Similarly cosh(iz) = cos z and cos(iz) = cosh z.
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Complex Analysis
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Hyperbolic functions
Note.
Clearly sinh(iz) = i sin z and sin(iz) = i sinh z.
Similarly cosh(iz) = cos z and cos(iz) = cosh z.
Hence the properties of sin hz and cos hz can be extracted in the
similar way to that of cos z and sin z.
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Complex Analysis
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Hyperbolic functions
Note.
Clearly sinh(iz) = i sin z and sin(iz) = i sinh z.
Similarly cosh(iz) = cos z and cos(iz) = cosh z.
Hence the properties of sin hz and cos hz can be extracted in the
similar way to that of cos z and sin z.
Since exp(z) and exp(−z) are entire, sin hz and cos hz are also
entire functions.
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Complex Analysis
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Hyperbolic functions
Note.
Clearly sinh(iz) = i sin z and sin(iz) = i sinh z.
Similarly cosh(iz) = cos z and cos(iz) = cosh z.
Hence the properties of sin hz and cos hz can be extracted in the
similar way to that of cos z and sin z.
Since exp(z) and exp(−z) are entire, sin hz and cos hz are also
entire functions.
sinh(−z) = − sinh(z) and cosh(−z) = cosh(z).
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Complex Analysis
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Hyperbolic functions
Identities
cosh2 z − sinh2 z = 1
sinh(z1 + z2 ) = sinh z1 cosh z2 + sinh z2 cosh z1
cosh(z1 + z2 ) = cosh z1 cosh z2 + sinh z1 sinh z2
sinh z = sin hx cos y + i cosh x sin y
cosh z = cosh x cos y + i sinh x sin y
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Complex Analysis
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Hyperbolic functions
Identities
cosh2 z − sinh2 z = 1
sinh(z1 + z2 ) = sinh z1 cosh z2 + sinh z2 cosh z1
cosh(z1 + z2 ) = cosh z1 cosh z2 + sinh z1 sinh z2
sinh z = sin hx cos y + i cosh x sin y
cosh z = cosh x cos y + i sinh x sin y
Proof. . .
Hint: Use sin(iz) = i sinh z and cos(iz) = cosh z and use the
trigonometric identities.
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Complex Analysis
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Hyperbolic functions
Further deductions
| sinh z|2 = sinh2 x + sin2 y
| cosh z|2 = sinh2 x + cos2 y
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Complex Analysis
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Hyperbolic functions
Further deductions
| sinh z|2 = sinh2 x + sin2 y
| cosh z|2 = sinh2 x + cos2 y
Proof. . .
Use z = x + iy in the left hand side, expand the value inside the
modulus and then use the property |w|2 = ww.
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Complex Analysis
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Hyperbolic functions
Inequalities
| sinh x| ≤ | cosh z| ≤ cosh x
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Complex Analysis
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Hyperbolic functions
Inequalities
| sinh x| ≤ | cosh z| ≤ cosh x
Proof. . .
Consider | cosh z|2 = sinh2 x + cos2 y .
Since cos2 y ≥ 0 for all y , we get sinh2 x ≤ | cosh z|2 which upon
applying square root gives the first part of the inequality.
Since cos2 y ≤ 1 for all y , we have | cosh z|2 ≤ 1 + sinh2 x = cosh2 x
which leads to the second part of the inequality.
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Complex Analysis
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Hyperbolic functions
Periodicity
sinh(z + πi) = − sinh z
cosh(z + πi) = − cosh z
tanh(z + πi) = tanh z.
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Complex Analysis
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Hyperbolic functions
Example
1
2
Answer. cosh z = (ez + e−z )/2 = 1/2 =⇒ ez + e−z = 1. This by
multiplying by ez and taking m = ez gives
√
1±i 3
2
z
m − m + 1 = 0 =⇒m = e =
2
√ !
1
3
=⇒ z = log
+i
2
2
√
= Log1 ± i tan−1 ( 3) + 2nπi
1
= ±iπ/3 + 2nπi = (2n + )π.
3
Question: Find the roots of cosh z =
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Complex Analysis
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Elementary functions
Logarithmic function
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Complex Analysis
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Elementary functions
Logarithmic functions
Definition
Let D ∗ be the region defined as the entire complex plane (finite) except
non-positive real axis.
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Complex Analysis
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Logarithmic functions
Definition
Define f : D ∗ −→ D ∗ by,
f −1 (w) = exp(w).
This function f (z) is called logarithmic function and denoted by
log z, z ∈ D ∗ .
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Complex Analysis
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Logarithmic functions
1
2
d
1
f (z) = , z ∈ D ∗ .
dz
z
Clearly log z is not defined when Rez < 0, which is the domain D ∗ .
f (z) = log z is defined on D ∗ and
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Complex Analysis
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Logarithmic functions
Example
Let f (z) = u + iv ⇐⇒ log(x + iy ) = u + iv . To find u and v in terms of
x and y .
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Complex Analysis
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Logarithmic functions
Example
Let f (z) = u + iv ⇐⇒ log(x + iy ) = u + iv . To find u and v in terms of
x and y .
log(x + iy ) = u + iv ⇐⇒ x + iy = exp(u + iv )
= exp(u). exp(iv )
= exp(u)[cos v + i sin v ]
=⇒ x = exp(u) cos v ,
y = exp(u) sin v
1
=⇒ x 2 + y 2 = exp(2u) =⇒ u = log(x 2 + y 2 ) = log |r |
2
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Complex Analysis
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Logarithmic functions
. . . continued
Example
tan v =
A.Swaminathan and V.K.Katiyar (NPTEL)
y
y
=⇒ v = arc tan
x
x
y
= arc tan + 2k π,
x
= arg z + 2k π.
Complex Analysis
k ∈z
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Logarithmic functions
. . . continued
Example
In Polar form z = r exp(iθ). Then
log z = (u + iv ) = log(reiθ )
= log r + log(exp(iθ))
= log r + iθ
= log |z| + iArgz + 2kiπ,
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Complex Analysis
k ∈ Z.
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Logarithmic functions
Definition
log |z| + iArgz = Logz.
=⇒ log z = Logz + 2kiπ,
k ∈z
where Logz is defined in the fundamental region D ∗ with each k giving
various branches of log z. Hence for each k we have different domains
that have a slit (whole plane except a line). In particular, Logz is
analytic in the slit domain D ∗ .
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Complex Analysis
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Logarithmic functions
Branch cut
Definition
These set of points (for example non-positive real axis for logarithmic
function) where f (z) fails to be analytic are called Branch cut.
Note: For log z,
Re (z) ≤ 0 is the branch cut.
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Logarithmic functions
Branch cut
Definition
The intersection of all these slits have a point in common where f (z)
fails to be analytic.This point is called BRANCH POINT.
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Complex Analysis
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Logarithmic functions
Branch cut
Definition
The intersection of all these slits have a point in common where f (z)
fails to be analytic.This point is called BRANCH POINT.
Branch point of log z is z = 0.
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Complex Analysis
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Logarithmic functions
Example
√
Question. Find the region of analyticity of log( 3 − i).
Answer. For this function
−π
11π
∗
θ
D = z = re ∈ C, r ≥ 0,
<θ<
6
6
is the fundamental (principal) region of analyticity.
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Complex Analysis
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Logarithmic functions
Example
log(3z − i) is analytic in the region
∗
D = z = reiθ ∈ C, r > 0,
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Complex Analysis
π
9π
<θ<
4
4
.
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Logarithmic functions
Properties
log 1 = 2nπ
(n = 0, ±1, ±2, . . .).
log(−1) = (2n + 1)πi
(n = 0, ±1, ±2, . . .).
exp(log z) = z
log(z1 , z2 ) = log z1 + log z2
In general Log(z1 z2 ) = Logz1 + Logz2 + 2Nπi where N has one of
the values 0, ±1, but if Re z1 > 0 and Re z2 > 0, then
Log(z1 z2 ) = Logz1 + Logz2 .
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Complex Analysis
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Elementary functions
Complex exponents
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Complex Analysis
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Logarithmic functions
Complex Exponents
When z 6= 0 and the exponentc is any complex number, z c is defined
by means of the equation
z c = exp(c log z)
where log z denotes the multiple valued logarithmic function. The
principal value of z c occurs when log z is replaced by Logz, so that
z c = exp(cLogz)
it defines the principal branch of the function z c on the domain |z| > 0,
−π < Argz < π.
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Complex Analysis
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Complex Exponents
Example
The powers of z are, in general, multiple valued. As an example we
write
1
i −4i = exp(−4i log i) = exp[−4i(2n + )πi] = exp[(8n + 2)π],
2
where n = 0, ±1, ±2, . . ..
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Complex Analysis
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Complex Exponents
Properties
1
= z −c
zc
d c
z = cz c−1
dz
d z
c = c z log c.
dz
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(|z| > 0, α < argz < α + 2π).
Complex Analysis
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Complex Exponents
Example
(i) The principal value of (−i)2i is
π
exp[2iLog(−i)] = exp[2i(− )] = exp π.
2
(ii) The principle branch of z 2/3 can be written
√
4
4
4
i4Θ
3
exp( Logz) = exp( Logr + iΘ) = r 4 exp(
).
3
3
3
3
It is analytic in the domain r > 0, −π < Θ < π.
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Complex Analysis
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Powers of a complex number
Objective
Let z = x + iy . To find z α+β
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Powers of a complex number
Let
u + iv = w = (x + iy )α+iβ
log(u + iv ) = (α + iβ)log(x + iy )
or
u + iv = exp[(α + β)log(x + iy )].
So, the domain of ω, is given by the fundamental (principal) region of
x + iy .
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Complex Analysis
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Powers of a complex number
Further,if z = reiθ and ω = Re iφ . Then
log ω = (α + iβ) log z
=⇒ log R + iφ + 2kiπ = (α + iβ)(log r + iθ + 2kiπ)
R = α log r − βθ;
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Complex Analysis
φ = β log r + θ + α2k π.
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Multiple valued Function
Definition
A branch of a multiple valued function f is any single-valued function F
which is analytic in some domain at each point of which F (z) is one of
the values of f (z).
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Complex Analysis
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Multiple valued Function
Example
Consider w = z 1/2 .
Let z = r (cos θ + i sin θ), where r is fixed and θ lies between o and
2Π.
Then
√
w1 = | r |e(1/2)iθ
and
√ i(1/2)(θ+π)
√
w2 = | r |e
= −| r |e(1/2)iθ ,
are called the two branches of the two-valued function w.
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Complex Analysis
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Multiple valued Function
Observation
Consider c z =⇒ c z = exp(z log c).
ez is, in general, a multiple valued function.
If the principal value of the logarithm is taken, then ez is single
value.
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Complex Analysis
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