3 Elementary Functions

3.1 The Exponential Function For z = x + iy we have ez = exeiy where Euler’s formula gives diy = cos y + i sin y The note: • When y = 0 we have ex the usual exponential. √ • When z = 1/n for n = 1, 2, 3... we have ez = e1/n = n e, the positive nth root of e.

• In polar coordinates, ez = exeiy = reiθ, from which

| ez |= ex = r

and arg(ez) = y + 2πn = θ + 2πn.

• Since ex 6= 0, this means ez 6= 0.

• Other properties as expected,

ez1 d ez1 ez2 = ez1+z2 , = ez1−z2 , ez = ez ez2 dz and ez is entire.

• Some properties are unexpected,

ez+2πi = eze2πi = ez

so the function is periodic with period 2πi. And ez can be negative, since

ei(2n+1)π = ei2πn+iπ = ei2nπeiπ = (1)(−1)

Examples: 1. Find numbers z = x + iy such that √ ez = 1 + 3i. √ Write 1 + 3i as ez = exeiy = 2eiπ/3, x π then e = 2 and y = 3 + 2nπ for n = 0, ±1 ± 2, .... Then, π ln(ex) = x =⇒ x = ln(2) =⇒ z = ln(2) + iπ + 2nπ n ∈ 3 Z

1 3.2 The Logarithmic Function The log function comes from solving ew = z for w, where z is a non-zero complex number. Note

z = reiΘ with modulus | z |= r, and the principle argument −π < Θ ≤ π, and

w = u + iv so

ew = z =⇒ eueiv = reiΘ =⇒ eu = r and v = Θ + 2nπ =⇒ u = ln r and v = Θ + 2nπ =⇒ w = ln r + i(Θ + 2nπ).

Deﬁnition: The logarithmic function is the multi-valued complex function given by

log(z) = ln r + i(Θ + 2nπ) n ∈ Z where Θ = Arg(z). The principle value of log(z) is

Log(z) = ln r + iΘ, and so log(z) = Log(z) + 2nπi n ∈ Z.

Some notes:

• log(z1z2) = log(z1) + log(z2) • One usual composition holds, for z = reiΘ,

elog(z) = eln r+i(Θ+2nπ) = eln reiΘe2nπi = reiΘ · 1 = z

• The other composition doesn’t necessarily hold, since for z = x + iy

log(ez) = log ex+iy = log exeiy = ln(ex) + i(y + 2nπi) = x + iy + 2nπi

takes multiple values.

2 Examples: √ 1. Let z = −1 − 3i. Then r = 2 and Θ = −2π/3. Hence

Log(z) = ln(2) − 2πi/3

and 1 log(z) = ln(2) − 2πi/3 + 2nπi = ln(2) + 2π n − i n ∈ . 3 Z 2. Since the complex number 1 has r = 1 and θ = 0 we have

log(1) = ln(1) + 2nπi = 2nπi

and Log(1) = 0 as we would want. 3. The complex number −1 has r = 1 and θ = π so

log(−1) = ln(1) + i(π + 2nπ) = (1 + 2n)πi

and so Log(−1) = πi. 4. Some properties carry over from calculus, for example √ 2 (1 + i)2 = 2eπi/4 = 2eπi/2

so √ Log(1 + i) = ln( 2) + πi/4 and h i Log (1 + i)2 = ln(2) + πi/2 imply h i Log (1 + i)2 = 2Log(1 + i). √ But this isn’t always true, for example −1 + i = 2e3πi/4 gives √ 2 (−1 + i)2 = 2e3πi/4 = 2e3πi/2

−π so r = 2 while Arg(−1 + i) = 2 . So we have √ 3πi Log(−1 + i) = ln( 2) + 4 and h i πi Log (−1 + i)2 = ln(2) − 2 and so h i Log (−1 + i)2 6= 2Log(−1 + i)

3 3.3 Branches and Derivatives of Logarithm Functions For z = reiθ a non-zero complex number, if

Arg(z) = Θ and arg(z) = θ = Θ + 2nπ for n ∈ Z, and hence log(z) = ln r + i(Θ + 2nπ) = ln r + iθ for n ∈ Z. Let α ∈ R and restrict θ so that

α < θ < α + 2π then we have a complex function

log(z) = ln r + iθ (r > 0, α < θ < α + 2π) with components u(r, θ) = ln r and v(r, θ) = θ deﬁned on the domain C \ rayα where

iα rayα = {re : 0 < r < ∞}.

On this domain, log(z) is single-valued, continuous, and analytic on its domain, since

ur = 1/r vr = 0

uθ = 0 vθ = 1 =⇒ rur = vθ and uθ = −rvr and from C-R it follows d log(z) = e−iθ(u + iv ) = 1/z (| z |> 0, α < arg(z) < α + 2π) dz r r and d Log(z) = e−iΘ(u + iv ) = 1/z (| z |> 0, −π < Arg(z) < π) dz r r

Away from this domain, i,e, on rayα the function is not continuous, since [IMAGE]

Deﬁnition: For a multiple valued function, f, a branch of f is any single-valued function F that is analytic in some domain at each point z for which F (z) is a value of f.

• The portion of line or curve being removed to deﬁne the branch is called the branch cut.

• The branch cut for the example above is the origin and rayα.

4 • Log(z) = ln r + iΘ(r > 0, −π < Θ < π) is the principle branch of log(z).

• The branch cut for the principle branch is the origin and rayπ = ray−π. Examples:

1. A non-prinicple branch for the logarithm function is

π 9π log(z) = ln r + iθ r > 0, < θ < 4 4

We will show that for this branch,

log(i2) = 2 log(i).

Write, 4πi log(i2) = log(−1) = ln1 + = πi 4 and notice 2πi 2 log(i) = 2 ln 1 + = πi. 4

2. Another non-principle branch for the logarithm function is

3π 11π log(z) = ln r + iθ r > 0, < θ < . 4 4

In this case, we write 5πi 5πi log(i2) = log(−1) = ln 1 + = 4 4 while 10πi 20πi 2 log(i) = 2 ln 1 + = 4 4

3. What does it mean when we say that

log(z1z2) = log(z1) + log(z2) (1)

holds? Let z1 = z2 = −1. Recall,

log(1) = ln 1 + i (0 + 2nπ) = 2nπi

and log(−1) = ln 1 + i (π + 2nπ) = (1 + 2n)πi

for n ∈ Z. Then, since z1z2 = 1, we have

log(z1z2) = 0 for n = 0

5 and log(z1) = πi for n = 0 and so log(z2) = −πi for n = −1 makes the statement true. It really means

log(z1) + log(z1) = (1 + 2n)πi + (1 + 2m)πi = 2(1 + n + m)πi = log(z1z2).

4. Now we will show that sometimes (but not always)

Log(z1z2) = Log(z1) + Log(z2)

holds. Let z1 and z2 be non-zero complex numbers with

Re z1 > 0 and Re z2 > 0.

Then, iΘ1 iΘ2 z1 = r1e and z2 = r2e π π where − 2 < Θi < 2 , and thus

−π < Θ1 + Θ2 < π.

Then i(Θ1+Θ2) z1z2 = r1r2e

and hence Arg (z1z2) = Θ1 + Θ2 is principal. Therefore,

Log(z1z2) = ln | z1z2 | +iArg (z1z2)

= ln r1r2 + i(Θ1 + Θ2)

= ln r1 + ln r2 + i(Θ1) + i(Θ2)

= Log(z1) + Log(z2)

5. Just as we would expect from calculus,

zn = en log(z)

To see this, just notice z = reiθ =⇒ zn = rneinθ and en log(z) = en(ln r+iθ) = en ln reinθ = rneinθ. And we also see that for nonzero z the following holds,

1/n 1 log(z) z = e n .

6 To see this, iΘ z = re =⇒ log(z) = ln r + i(Θ + 2kπ) for k ∈ Z 1 ln r i(Θ + 2kπ) =⇒ log(z) = + for k ∈ n n n Z 1 log(z) ln r + i(Θ+2kπ) =⇒ e n = e n n for k ∈ Z 1 log(z) ln r i(Θ+2kπ) =⇒ e n = e n e n for k ∈ Z 1 √ Θ 2kπ log(z) n i( + ) =⇒ e n = r · e n n for k ∈ Z 1 log(z) 1/n =⇒ e n = z

3.4 The Power Function Deﬁnition: For z 6= 0 and complex constant c, the power function is deﬁned by zc = ec log z. Some properties: • We already know that zn = en log z and z1/n = e1/n log(z).

1 e0 0−z −z • Recalling that ez = ez = e = e , we have 1 1 = = e−c log z = z−c. zc ec log z • For a ﬁxed branch, (r > 0, α < θ < α + 2π) the log function is analytic, so we deﬁne zc there. Then, d d c zc = ec log z = ec log z. dz dz z then recalling elog z = z, we have d c ec log z zc = ec log z = c = ce(c−1) log z = czc−1 dz elog z elog z on the branch (r > 0, α < θ < α + 2π). • The principal value of zc is zc = ecLogz, and the principle branch is this function on the domain (r > 0, −π < Arg z < π). Examples: 1. Consider the function ii = ei log i. Then, π π log(i) = ln | i | +i + 2nπ = i + 2nπ 2 2 and hence i i·i π +2nπ −π 1 +2n i = e ( 2 ) = e ( 2 ) where all values are in R, and with principle value i − π i = e 2 .

7 2. Consider log(−1) = ln 1 + i(π + 2nπ) = (1 + 2n)πi from which it’s easy to see

1/π 1 log(−1) 1 ((1+2n)πi) (1+2n)i (−1) = e π = e π = e .

3. The principle branch of z2/3 is √ 2/3 2 Logz 2 r+ 2 iΘ 3 i 2Θ z = e 3 = e 3 3 = r2e 3 .

with −π < Θ < π, and so the principle value of z2/3 is

√3 2Θ √3 2Θ z2/3 = r2 cos + i r2 sin . 3 3 this is analytic in domain r > 0 − π < Θ < π.

4. Consider the non-zero complex numbers,

z1 = 1 + i z2 = 1 − i z3 = −1 − i.

Taking principle values of powers,

i i iLog 2 i ln 2 (z1z2) = 2 = e = e

and √ i i iLog (1+i) i(ln 2+i π ) −π/4 i(ln 2)/2 z1 = (1 + i) = e = e 4 = e e

and √ i i iLog (1−i) i(ln 2−i π ) π/4 i(ln 2)/2 z2 = (1 − i) = e = e 4 = e e thus i i i (z1z2) = z1z2. On the other hand,

i iLog (z2z3) i(ln 2+iπ) −π i ln 2 (z2z3) = e = e = e e

but √ i i iLog (1−i) i(ln 2−i 3π ) 3π/4 i(ln 2)/2 z3 = (−1 − i) = e = e 4 = e e and so i i π/4 i(ln 2)/2 3π/4 i(ln 2)/2 π i ln 2 z2z3 = e e · e e = e e hence i i i −2π (z2z3) = z2z3e

8 3.5 Trig Functions From Euler’s formula, we know that ( ( eix = cos x + i sin x eix − e−ix = 2i sin x =⇒ e−ix = cos x − i sin x eix + e−ix = 2 cos(x) Therefore, it’s not a stretch to deﬁne eiz − e−iz sin z = 2i and eiz + e−iz cos z = . 2 and other trig functions are as expected sin z cos z tan z = cot z = cos z sin z 1 1 sec z = csc = cos z sin z Some properties: • sin z and cos z are linear combinations of entire functions, and are thus entire. • Since d d eiz = ieiz and e−iz = −ie−iz dz dz we get d d sin z = cos z and cos z = − sin z dz dz

Deﬁnition: A zero of a function is a number z0 such that f(z0) = 0. • Extending a real valued function to the complex plane can add zeroes, for example, 2 f(x) = x + 1 has no zeroes in R but zeroes ±i in C. • Extending the domain to C, trig functions gain no zeroes, that is zeroes of sin = {nπ : n ∈ Z} and n π o zeroes of cos = nπ + : n ∈ 2 Z • Other trig functions are analytic except at singularities, those are

{nπ : n ∈ Z} for tan z and csc z, and n π o nπ + : n ∈ 2 Z for cot z and sec z.

9 Deﬁnition: The hyperbolic sine and cosine functions are deﬁned by

ez − e−z sinh z = 2 and ez + e−z cosh z = 2