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UNSW — School of Mathematics and Statistics
MATH2521 Complex Analysis
5. COMPLEX LOGARITHMS AND POWERS
In real calculus, we may define the logarithm function ln to be the
inverse of the exponential function . That is, lnx is the unique real
solution y of the equation ey = x. Let’s attempt to do the same with
the complex exponential function: if z = reiθ is a complex number, then
we aim to define its logarithm w = log z to be the solution of ew = z.
Using the procedure from chapter 4, page 3, we have
ew = z eu = |z| and v = arg z
u = ln r and v = θ + 2kpi , k ∈ Z
w = ln r + i(θ + 2kpi) , k ∈ Z .
Note that if z = 0 this does not work and there is no solution; and
that for each z 6= 0 there are infinitely many solutions. Thus, solving
ew = z does not give w as a function of z. However, we can regard log z
as an expression taking multiple values, just as we did for the complex
argument; and we can then choose one particular value as the principal
value of the logarithm – just as we did for the principal value argument.
Definition. Let z = reiθ be a non–zero complex number. A (complex)
logarithm of z, denoted log z, is any complex number of the form
log z = ln r + i(θ + 2kpi) , k ∈ Z .
The principal logarithm of z is
Log z = ln |z|+ iArg z .
Notation. We shall use ln for the natural logarithm of a positive real
number. We shall write log (with a lowercase “l”) for the multi–valued
Or we may do it the other way around, as you did in first year calculus
lectures.
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logarithm of a complex number, and Log (with a capital “L”) for the
principal value logarithm.
Example. Find the logarithms, and the principal logarithms, of 3 and
1 + i and e5i.
Solution. We have 3 = 3eipi; therefore
log( 3) = ln 3 + (2k + 1)pii , k ∈ Z ,
Log( 3) = ln 3 + pii .
Likewise, 1 + i = √2 e3pii/4, so
log( 1 + i) = 1
2
ln 2 + (3
4
+ 2k)pii , k ∈ Z ,
Log( 1 + i) = 12 ln 2 + 34pii .
Moreover,
log(e5i) = (5 + 2kpi)i , k ∈ Z .
Note, however, that 5 is not the principal argument of e5i, because
pi < 5 < 3pi; we have Log(e5i) = (5 2pi)i . Notes. If w = Log z then, as we have seen above, ew = z: that is, exp(Log z) = z for all z 6= 0. However, it is not generally true that Log(exp z) = z. If z = x+ iy then Log(exp z) = Log ( ex+iy ) = ln(ex) + i(y + 2mpi) = x+ iy + 2mpii = z + 2mpii ; ( ) 2 here m is a specific integer with the property that y + 2mpi is the principal argument of ez , that is, pi < y + 2mpi ≤ pi. So we have Log(exp z) = z m = 0 pi < y ≤ pi pi < Im(z) ≤ pi . This means that the principal logarithm is not the inverse of the exponential function: in fact, exp is not one–to–one and hence has no inverse. The identity exp(Log z) = z, however, is sometimes expressed by saying that Log is the “right inverse” of exp, or exp is the “left inverse” of Log. We also have exp(log z) = z for all z 6= 0, and log(exp z) = z + 2kpii , k ∈ Z . The question of inverses does not arise in this case since log is not a (single–valued) function. Note carefully that k here is an arbitrary integer, whereas m in ( ) is a certain specific integer. If z = reiθ and w = seiφ then zw = (rs)ei(θ+φ) and we have log(zw) = ln(rs) + i(θ + φ+ 2kpi) log z + logw = ln r + i(θ + 2k1pi) + ln s+ i(φ+ 2k2pi) = ln(rs) + i(θ + φ+ 2(k1 + k2)pi) . Thus log(zw) = log z + logw , provided we interpret the equality as stating that the set of all possible values of the left hand side is the same as the set of all possible values of the right hand side: compare the similar situation for the argument function, chapter 1, pages 5–6. The equation Log(zw) = Log z + Logw is not generally true. 3 Exercise. Find the error: Log( 1) = Log ( 1 1 ) = Log( 1) Log( 1) = 0 . Continuity and differentiability. Consider the limiting behaviour of the principal logarithm function Log z as z approaches a negative real number, say z → 1. As we saw in chapter 2, for this limit to exist, Log z must approach the same value if z tends to 1 along any path whatsoever. Let z approach 1 along the unit circle, firstly in an anticlockwise direction and secondly in a clockwise direction. Writing z = eiθ, the two cases are θ → pi and θ → ( pi)+; we have Log z = iθ → ipi and Log z = iθ → ipi respectively. Since we have obtained two different results as z approaches 1 along two different paths, Log z has no limit as z → 1. A similar argument holds if 1 is replaced by any negative real number. On the other hand, let a be any non–zero complex number which is not a negative real number. It is clear from a diagram that if z → a, then |z| → |a| and Arg z → Arg a. Therefore we have lim z→a Log z = lim z→a ( ln |z|+ iArg z) = ln |a|+ iArg a = Log a , and so Log is continuous at a. Lemma. Continuity of the principal logarithm. The complex function Log : C { 0 } → C is continuous at all points of its domain except for negative real numbers. Next we investigate the differentiability of Log; we shall use the fact that exp(Log z) = z for all z 6= 0. Essentially, the chain rule for differentiable functions gives exp(Log z) = z exp(Log z) d dz (Log z) = 1 d dz (Log z) = 1 exp(Log z) = 1 z . 4 There is a problem with this, however: we don’t yet know that Log is differentiable at all. To make the above argument watertight, we need the following result. Theorem. Differentiability of a right inverse. Let A and B be open subsets of C. Suppose that f : A → C, that f(A) B, that g : B → C and that g(f(z)) = z for all z ∈ A. If f is continuous on A and a is an element of A such that g is differentiable at f(a) and g′(f(a)) 6= 0, then f is differentiable at a and f ′(a) = 1 g′(f(a)) . Proof. Let a ∈ A. Since g(f(z)) = z for all z ∈ A we can write f(z) f(a) z a = f(z) f(a) g(f(z)) g(f(a)) for all z ∈ A, other than z = a. If g is differentiable at b then we have by definition g′(b) = lim w→b g(w) g(b) w b . Now take b = f(a) and w = f(z). Since f is continuous, z → a implies w → b ; and we have assumed that g is differentiable at f(a); so g′(f(a)) = lim z→a g(f(z)) g(f(a)) f(z) f(a) . Since, again by assumption, g′(f(a)) 6= 0, the “limit of a quotient” rule, chapter 2, page 6, yields 1 g′(f(a)) = lim z→a f(z) f(a) g(f(z)) g(f(a)) = limz→a f(z) f(a) z a ; that is, f ′(a) = 1 g′(f(a)) , as claimed. 5 Corollary. Derivative of a complex logarithm. Let D be the set of all complex z other than zero and negative real numbers. Then Log is holomorphic on D, and d dz (Log z) = 1 z for all z ∈ D. Branches of the logarithm function. We have defined the principal logarithm of a non–zero complex number z by using the principal argu- ment, pi < θ ≤ pi. There is, however, nothing special about this choice: we could have taken 0 ≤ θ < 2pi, which you will find in some sources; we could have made our choice in many other ways. The only real essential is that we should choose θ in such a way that our logarithm function is continuous, except at points on a line (or curve) extending from the origin to infinity. In the case of Log, this line is the negative half of the real axis. Definition. Let D be a domain not containing the origin. A function f : D → C which is continuous on D, except at points on some line or curve extending from the origin to infinity, and which has the form f(z) = ln |z|+ iα(z) where α(z) is an argument of z, is called a branch of the complex logarithm function. The line or curve is known as a branch cut of f . Example. Define the functions Log1 and Log2 from C { 0 } to C thus: Log1(re iθ) = ln r + iθ where 0 ≤ θ < 2pi ; Log2(re iθ) = ln r + iθ where r ≤ θ < r + 2pi . These functions are branches of the logarithm function, and are contin- uous everywhere except on the branch cuts, shown in the diagrams on the following page (left for Log1, right for Log2). Taking, for example, z = 7e6i, we have pi < 6 < 2pi and hence Log z = ln 7 + (6 2pi)i Log1 z = ln 7 + 6i Log2 z = ln 7 + (6 + 2pi)i . 6 xy 0 θ = 0 x y 0 θ = r If f(z) is any branch of the logarithm, we can argue exactly as we did for the principal branch to show that f is holomorphic and f ′(z) = 1 z for every point z at which f is continuous. Exercise. Determine where the following functions are holomorphic, and find their derivatives: f(z) = Log(z i) z2 2i ; g(z) = Log(z 2 1) . Solution. The function f is differentiable wherever the numerator and denominator are differentiable and the denominator is not zero. The former condition excludes points z such that z i is a negative real number; the latter excludes the square roots of 2i, which (exercise!) are 1 + i and 1 i. So if we write S = {x+ i | x ∈ R, x ≤ 0 } ∪ { 1 + i, 1 i } , then f is differentiable on C S; since this is an open set, f is holo- morphic here too. The diagram shows where f is not holomorphic. By x y 0 i 1 + i 1 i standard differentiation procedures, f ′(z) = (z2 2i) 1 z i 2z Log(z i) (z2 2i)2 for z ∈ C S. 7 Similiarly, g is differentiable at all z except those for which z2 1 is a negative real number or zero, that is, z2 is a real number in the interval ( ∞, 1 ]. There are two cases. If z2 is real and 0 ≤ z2 ≤ 1, then z is real, z = x with 1 ≤ x ≤ 1. If z2 is real and z2 < 0, then z is purely imaginary, z = iy with y ∈ R. x y 0 1 1 Thus, g is holomorphic for all z except those in the set S = {x ∈ R | 1 ≤ x ≤ 1 } ∪ { iy | y ∈ R } , illustrated in the diagram. The derivative is g′(z) = 2z z2 1 for z ∈ C S. Comment. As is seen in these examples, a question about where a function is holomorphic is often most easily answered by considering where the function is not holomorphic. Complex powers. We know what is meant by a power zn where n is an integer; and we may interpret z1/n as an nth root of z, which is also a familiar concept. But what about something like zi We shall define such expressions by imitating the real identity xa = exp(a ln x) , noting that the introduction of logarithms of complex numbers will mean that, once again, we have to deal with the problems of multiple values and principal values. Definition. If z and a are complex, with z 6= 0, we define the multi– valued power za = exp(a log z) and its principal value pv(za) = exp(aLog z) . 8 Examples. Find all values, and the principal value, of i2i. Solution. We have log i = i (pi 2 + 2kpi ) , k ∈ Z and Log i = ipi 2 ; therefore i2i = exp(2i log i) = e (1+4k)pi , k ∈ Z and pv(i2i) = exp(2iLog i) = e pi . Amazingly, all of these values are real numbers!! We have (3 + 4i)5+6i = exp((5 + 6i) log(3 + 4i)) = exp((5 + 6i)(ln 5 + i(tan 1 43 + 2kpi))) = exp ( (5 ln 5 6 tan 1 43 12kpi) + i(6 ln 5 + 5 tan 1 4 3 + 10kpi) ) = 55 exp ( (6 tan 1 4 3 + 12kpi) + i(6 ln 5 + 5 tan 1 4 3 ) ) with k ∈ Z, and pv(3 + 4i)5+6i = 55 exp ( 6 tan 1 4 3 + i(6 ln 5 + 5 tan 1 4 3 ) ) . Since we have just given a general definition of powers, we should check that it is consistent with previous definitions, in which we have inter- preted zn as repeated multiplication, z1/n as an nth root and ez as a synonym for exp(z). This is the point of the following result. Lemma. Consistency of power definitions. Let z be a non–zero real number. We have z0 = 1. If n is a positive integer, then zn has one value only, and it is zn = n factors︷ ︸︸ ︷ z z · · · z . 9 If n is a negative integer, say n = m, then zn has one value only, and it is zn = z m = 1 zm . If n is a positive integer, then z1/n has exactly n different values, and each of these satisfies the equation wn = z. The principal value of ez (interpreted as a power) is ez (interpreted as exp(z)). Proof (sketch). The first result is easy. The second is proved by math- ematical induction using the relation zn+1 = exp((n+ 1) log z) = exp(log z) exp(n log z) = z exp(n log z) . For the third, properties of the exponential function give z m = exp( m log z) = 1 exp(m log z) = 1 zm . If z = reiθ then z1/n = exp ( 1 n (ln r + i(θ + 2kpi)) ) = r1/n exp ( iθ n ) exp (2kpii n ) with k ∈ Z; the value of the third factor here is the same for k + n as it is for k, and therefore taking k = 0, 1, 2, . . . , n 1 gives all possible values of the expression. The nth power of each is ( r1/n exp ( iθ n ) exp (2kpii n ))n = r exp(iθ + 2kpii) = z . For the last result we have pv(ez) = exp(z Log e) = exp(z(ln e+ 0i)) = exp(z) . Notation. We shall continue to use ez as a synonym for exp(z), unless it is clearly stated that we mean a multi–valued power. 10 Comments. Because the principal value power za is defined in terms of the principal value logarithm, it will be holomorphic wherever the latter is. Specifically, if z is in S = C {x ∈ R | x ≤ 0 } then Log is continuous at z; so aLog z is continuous at z; and the exponential function is continuous everywhere, so pv(za) = exp(aLog z) is continuous at z. Standard differentiation rules show that pv(za) is differentiable on S, with d dz ( pv(za) ) = d dz ( exp(aLog z) ) = exp(aLog z) a z = a exp(aLog z) exp( Log z) = a exp ( (a 1) Log z) = apv(za 1) – no surprise! In general, pv(za) is not continuous, and therefore not differentiable, on the negative real axis. For example, if z = reiθ with pi < θ ≤ pi, then pv(z1/2) = exp ( 1 2 (ln r + iθ) ) = r1/2eiθ/2 . If x is a positive real and we let z approach x, firstly anticlockwise around the circle |z| = x and then clockwise, we have pv(z1/2)→ x1/2eipi/2 = i√x and pv(z1/2)→ x1/2e ipi/2 = i√x respectively. So pv(z1/2) has no limit as z → x. On the other hand, if n is a non–negative integer, then we know already that zn has its usual meaning, and that zn is an entire 11 function; likewise, if n is a negative integer, then zn is holomorphic at all z except zero. You need to be aware that the familiar power laws for real numbers usually do not hold for complex powers. For example, if we take w = e2i and z = e3i and a = i, then pv(wz)a = pv ( (e5i)i ) = exp(i(5 2pi)i) = e2pi 5 and pv(wa) pv(za) = exp(i(2i)) exp(i(3i)) = e 5 , so that pv(wz)a 6= pv(wa) pv(za) . Similarly, taking z = e2pi and a = i and b = i gives pv(za) = exp(2pii) = 1 pv((za)b) = exp(0) = 1 pv(zab) = e 2pi , so that pv((za)b) 6= pv(zab) . Note that in finding pv((za)b) we have taken principal value powers (using principal value logarithms) twice. Problem. Find, if it exists, lim z→0 pv(za). Solution. Writing z = reiθ with pi < θ ≤ pi, and a = b+ ic, we have pv(za) = exp ( (b+ ic)(ln r + iθ) ) = exp(b ln r cθ) exp(i(bθ + c ln r)) , and we want to investigate this as r→ 0. Now b ln r |c|pi ≤ b ln r cθ < b ln r + |c|pi ; so if b > 0 then b ln r cθ → ∞ and∣∣pv(za)∣∣ = exp(b ln r cθ) → 0 ,
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while if b < 0 then ∣∣pv(za)∣∣ = exp(b ln r cθ) → ∞ . In the case b = c = 0 then pv(za) is the constant 1, and has limit 1 as z → 0. Finally, if b = 0 and c 6= 0 then ∣∣pv(za)∣∣ = exp( cθ) has different values for different θ, and hence has no limit. Therefore we have the following result. Proposition. Limit of a power at the origin. We have lim z→0 pv(za) = { 0 if Re(a) > 0
1 if a = 0;
for other a, the limit does not exist.
Definition. Let z = 0. Then we define pv(za) = 0 for Re(a) > 0 and
pv(za) = 1 for a = 0. For all other values of a the expression remains
undefined.
Notation. We shall, where convenient, use “radical” notation for roots
of complex numbers: this notation may also require consideration of
principal values. For example,
√z is an alternative notation for z1/2, the multiple valued square
root;
pv√z is the same as pv(z1/2).
Inverse trigonometric and hyperbolic functions can be treated in
a similar way. For instance, given z, we solve the equation sinw = z
thus:
sinw = z eiw e iw = 2iz
(eiw)2 2iz(eiw) 1 = 0
eiw = iz +
√
1 z2 ;
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as stated above, we use the symbol
√
for the multi–valued square root,
and so there is no need to write ± . Continuing, we note (exercise!) that
iz +
√
1 z2 can never be zero, and so for all z we have
sinw = z iw = log(iz +√1 z2)
w = i log(iz +√1 z2) .
Definition. The (multiple–valued) inverse sine function is
sin 1 : C→ C where sin 1(z) = i log(iz +√1 z2)
using the multiple–valued square root and logarithm functions. The
principal value inverse sine function is
pv sin 1 : C→ C where pv sin 1(z) = i Log(iz + pv√1 z2) .
By way of illustration we prove a couple of properties of the principal
value inverse sine.
Lemma. Properties of inverse sine.
1. The principal value inverse sine is holomorphic on C S, where
S = {x ∈ R : |x| ≥ 1 } .
2. The principal value inverse sine is an odd function.
3. If z = x is real with 1 ≤ x ≤ 1, then
pv sin 1 z = sin 1 x ,
where the sin 1 on the right hand side is the usual real inverse sine
function.
Proof. The inverse sine fails to be holomorphic only if
iz + pv
√
1 z2 = a , a ∈ R , a ≥ 0
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(giving a point where Log is not holomorphic), or
1 z2 = b , b ∈ R , b ≥ 0
(giving a point where the principal square root is not holomorphic).
Writing z = x+ iy, the first condition implies
1 z2 = ( a iz)2 1 = a2 + 2ai(x+ iy)
1 = a2 2ay , 0 = 2ax
a 6= 0 , x = 0 , z = iy ,
and now substituting back yields
a = iz + pv
√
1 z2 = y +
√
1 + y2 > 0 ,
which is impossible. The second condition gives z2 = 1 + b ≥ 1, so z
is a real number with absolute value at least 1, that is, z ∈ S. Hence
pv sin 1 is holomorphic in C S.
Secondly, we have
1
iz + pv
√
1 z2 =
iz + pv√1 z2
(1 z2) (iz)2 = iz + pv
√
1 z2 ;
since we have just shown that iz +pv
√
1 z2 cannot be a negative real
number,
Log
( iz + pv√1 z2) = Log(iz + pv√1 z2)
and hence pv sin 1( z) = pv sin 1(z).
Finally, if x is real and 1 < x < 1 then pv√1 x2 is a positive real number, so ix + pv √ 1 x2 is in the first or fourth quadrant and has polar form ix+ pv √ 1 x2 = eiθ with θ = tan 1 ( x√ 1 x2 ) . 15 Therefore pv sin 1 x = i ( i tan 1 x√ 1 x2 ) = tan 1 x√ 1 x2 = sin 1 x ; you can confirm the last step by drawing a triangle, or otherwise. It is easy to check that pv sin 1 x = sin 1 x also holds for x = ±1. Exercise. Following the same procedure as we used for the inverse sine, define other inverse trigonometric and hyperbolic functions (both as multi–valued expressions and as principal values). Give their domains, determine where they are holomorphic, find some properties, and show that they coincide with the corresponding real functions for suitable values of z. Suggestion. Inverse tangent is a good one to start with!