Elementary Functions II

Lecture 9

As logzis a multi-valued function, we must interpret (9.1) and (9.2) to mean that if particular values are assigned to any two of their terms, then one can find a value of the third term so that the equation is satisfied.

Example 9.2.

Let z₁ = z₂ = −1. Then, z₁z₂ = 1. Thus, logz₁ = (2k₁+ 1)πi, logz₂ = (2k₂+ 1)πi, and log 1 = 2k₃πi where k₁, k₂, and k₃ are integers. If we select πi to be the value of logz₁ and logz₂, then equation (9.1) will be satisfied if we use 2πifor log 1.If we select 0 andπi to be the values of log 1 and logz₁,respectively, then equation (9.1) will be satisfied if we use−πifor logz₂.

Abranchof a multi-valued function is a single-valued function analytic in some domain. The principal value or branch of the logarithm, denoted by Logz,is the value inherited from the principal value of the argument:

Logz = Log|z|+iArgz.

The value of Argz jumps by 2πiaszcrosses the negative real axis. There- fore, Logz is not continuous at any point on the nonpositive real axis.

However, at all points off the nonpositive real axis, Logz is continuous.

Theorem 9.1.

The function Logzis analytic in the domainD^∗consist- ing of all points of the complex plane except those lying on the nonpositive real axis; i.e.,D^∗=C−(−∞,0).Furthermore,

d

dzLogz = 1

z for z in D^∗.

Figure 9.1

x y

0

D^∗

Proof.

Setw= Logz.Letz0∈D^∗ andw0= Logz0. We have to show that the

zlim→z₀

Logz−Logz₀ z−z0

exists and is equal to 1/z₀. Since Logz is continuous,w = Logz →w₀ =

Elementary Functions II 59 Logz₀as z→z₀.Thus,

zlim→z₀

Logz−Logz0

z−z0 = lim

w→w₀

w−w0

e^w−e^w0 = lim

w→w₀

1 e^w−e^w0

w−w₀

= 1

e^w0 = 1

eLogz₀ = 1 z0.

(9.3)

Note that (9.3) is meaningful since, for z =z0, w will not coincide with w0. This follows from the fact that z =eLogz =e^w. Thus, w = Logz is differentiable at every point inD^∗,and hence is analytic there.

A line used to create a domain of analyticity is called abranch line or branch cut. Any point that must lie on a branch cut-no matter what branch is used-is called abranch pointof a multi-valued function. For example, the nonpositive real axis shown inFigure 9.1is abranch cutfor Logz,and the pointz= 0 is abranch point.

Ifαis a complex constant andz= 0,then we definez^αbyz^α=e^αlogz. Powers ofzare, in general, multi-valued.

Example 9.3.

Find all values of 1ⁱ.Since log 1 = Log 1 + 2kπi= 2kπi, we have 1ⁱ=e^{ilog 1}=e^−2kπ,wherek= 0,±1,±2,· · ·.

Example 9.4.

Find all values of (−2)ⁱ. Since log(−2) = Log 2 + (π+ 2kπ)i, we have (−2)ⁱ = e^ilog(−2) = eⁱLog2e^−(π+2kπ), where k = 0,±1,±2,· · ·.Thus, (−2)ⁱ has infinitely many values.

Example 9.5.

Find all values ofi⁻²ⁱ.Since logi= Log 1+

2k+¹₂ πi= 2k+¹₂

πi, we have i⁻²ⁱ = e^−2ilogi = e⁻2i(2k+1/2)πi = e^(4k+1)π, where k= 0,±1,±2,· · ·.

Since logz= Log|z|+iArgz+ 2kπi,we can write

z^α = e^α(Log_|z|+iArgz+2kπi) = e^α(Log_|z|+iArgz)e^α2kπi, (9.4) where k = 0,±1,±2,· · ·. The value of z^α obtained by taking k =k₁ and k = k₂(=k₁) in equation (9.4) will be the same when e^α2k1πi = e^α2k2πi. This occurs if and only ifα2k₁πi=α2k₂πi+ 2mπi(m is an integer); i.e., α=m/(k₁−k₂). Hence, formula (9.4) yields some identical values of z^α only when α is a real rational number. Consequently, if α is not a real rational number, we obtain infinitely many different values of z^α, one for each choice of the integerkin (9.4).

On the other hand, ifα=m/n,wheremandn >0 are integers having no common factor, then there are exactly n distinct values (branches) of z^m/n, namely

z^m/n = e^(m/n)Log_|z|e^(m(Argz+2kπ)i)/n, k= 0,1,· · ·, n−1.

In summary, we find:

1.z^αis single-valued whenαis a real integer.

2.z^αtakes finitely many values whenαis a real rational number.

3.z^αtakes infinitely many values in all other cases.

If we use the principal value of logz,we obtain theprincipal branchof z^α,namelye^αLogz.

Example 9.6.

The principal value of (−i)ⁱ is eⁱLog(−i)=e^i(−πi/2) = e^π/2.

Sincee^z is entire and Logz is analytic in the domainD^∗=C\(−∞,0], the chain rule implies that the principal branch ofz^αis also analytic inD^∗. Furthermore, forzin D^∗,we have

d dz

e^αLogz

= e^αLogz d

dz(αLogz) = e^αLogzα z

.

Now we shall use logarithms to describe the inverse of the trigonomet- ric and hyperbolic functions. For this, we recall that if f is a one-to-one complex function with domain S and rangeS, then theinverse function of f,denoted as f⁻¹,is the function with domain S and rangeS defined by f⁻¹(w) =z if f(z) = w. It is clear that if the function f is bijective, thenf⁻¹mapsS ontoS.Furthermore, both the compositionsf◦f⁻¹and f⁻¹◦f are the identity function. For example, the inverse of the function f(z) =ax+b, a= 0,isf⁻¹(z) = (z−b)/a.

The inverse sine function w = sin⁻¹z is defined by the equation z = sinw.We shall show that sin⁻¹z is a multi-valued function given by

sin⁻¹z = −ilog[iz+ (1−z²)^1/2]. (9.5) From the equation

z = sinw = e^iw−e^−iw 2i ,

we have 2iz=e^iw−e^−iw.Multiplying both sides bye^−iw,we deduce that e^2iw−2ize^iw−1 = 0,

which is quadratic ine^iw.Solving fore^iw,we find

e^iw = iz+ (1−z²)^1/2, (9.6) where (1−z²)^1/2is a double-valued function ofz.Taking logarithms of each side of (9.6) and recalling thatw= sin⁻¹z,we arrive at the representation (9.5).

Elementary Functions II 61

Example 9.7.

From (9.5), we have

sin⁻¹(−i) = −ilog(1±√ 2).

However, since log(1 +√

2) = Log (1 +√

2) + 2kπi, k= 0,±1,±2,· · ·, (9.7) log(1−√

2) = Log (√

2−1) + (2k+ 1)πi, k= 0,±1,±2,· · ·, (9.8) and

Log(√

2−1) = Log 1 1 +√

2 = −Log (1 +√ 2), the lists (9.7) and (9.8) can be combined in one list as

(−1)^kLog (1 +√

2) +kπi, k= 0,±1,±2,· · ·. Thus, it follows that

sin⁻¹(−i) = kπ+i(−1)^k+1Log(1 +√

2), k= 0,±1,±2,· · ·. Similar to the expression for sin⁻¹zin (9.5), it is easy to show that

cos⁻¹z = −ilog[z+i(1−z²)^1/2] and

tan⁻¹z = i

2logi+z

i−z. (9.9)

Clearly, the functions cos⁻¹zand tan⁻¹z are also multi-valued.

The derivatives of these three functions are readily obtained from the representations above and appear as

d

dzsin⁻¹z= 1

(1−z²)^1/2, d

dzcos⁻¹z=− 1

(1−z²)^1/2, d

dztan⁻¹z= 1 1+z². Finally, we note that the inverse hyperbolic functions can be treated in a corresponding manner. It turns out that

sinh⁻¹z = log[z+ (z²+ 1)^1/2], cosh⁻¹z = log[z+ (z²−1)^1/2], tanh⁻¹z = 1

2log1 +z 1−z,

and d

dzsinh⁻¹z = 1 (z²+ 1)^1/2, d

dzcosh⁻¹z = 1 (z²−1)^1/2, d

dztanh⁻¹z = 1 1−z².

Problems

9.1. Find all values ofz such that (a).e^z=−2, (b).e^z= 1 +√

3i, (c). exp(2z−1) = 1, (d). sinz= 2.

9.2. If z₁, z₂ = (π/2) +kπ, show that tanz₁ = tanz₂ if and only if z₁=z₂+kπ,wherek is an integer.

9.3. Use (7.7) and (7.8) to prove Theorem 9.1.

9.4. Evaluate the following:

(a). Log(−ei), (b). Log(1−i), (c). log(−1 +√ 3i).

9.5. Show that

(a). Log(1 +i)²= 2Log(1 +i) but (b). Log(−1 +i)²= 2Log(−1 +i).

9.6. Find the limit lim_y→0+[Log(a+iy)−Log(a−iy)] when a > 0, and whena <0.

9.7. Evaluate the following and find their principal values:

(a). (1 +i)ⁱ, (b). (−1)^π, (c). (1−i)⁴ⁱ, (d). (−1 +i√ 3)^3/2. 9.8. Establish (9.9).

9.9. Evaluate the following:

(a). sin⁻¹√

5, (b). sinh⁻¹i.

9.10.Prove the following inequalities:

e^−2y

1 +e^−2y < |tan(x+iy)−i| < e^−2y

1−e^−2y, y >0 e^−2y

1 +e^−2y < |cot(x+iy) +i| < e^−2y

1−e^−2y, y >0 e^2y

1 +e^2y < |tan(x+iy) +i| < e^2y

1−e^2y, y <0 e^2y

1 +e^2y < |cot(x+iy)−i| < e^2y

1−e^2y, y >0

Answers or Hints

9.1. (a). e^z =−2 iffe^z=e^ln(2)e^iπ iff z= ln 2 +iπ+i(2kπ), k∈Z, (b).

e^z= 1 +√

3i= 2e^iπ/3 iffz= ln 2 +i^π₃ +i(2kπ), k∈Z,(c). e^2z−1=eⁱ⁰iff

Elementary Functions II 63 2z−1 =i(2kπ) iffz=¹₂ +i(kπ), k∈Z, (d). sinz= 2 iffe^iz−e^−iz= 4i iff e^2iz −4ie^iz−1 = 0 iff e^iz = (4i±√

−16 + 4)/2 = (4i±√

12i)/2 = (2±√

3)i= (2 +√

3)e^iπ/2, (2−√

3)e^iπ/2so z= ln(2 +√

3) +i_π

2 + 2kπ , or ln(2−√

3) +i_π

2 + 2kπ

, k∈Z.

9.2. tanz₁−tanz₂= 0 if and only if sin(z₁−z₂) = 0.

9.3. If Logz= Log|z|+iArgz= Logr+iΘ,thenu= Logr, v= Θ.Thus, ur= 1/r, uΘ= 0, vr= 0, vΘ= 1.

9.4. (a). Log(−ei) = lne+i(−π/2) = 1−iπ/2, (b). Log(1−i) = ln√

2−iπ/4,(c). log(−1 +√

3i) = ln 2 +i^2π₃ + 2kπi, k∈Z.

9.5. (a). Log(1 +i)²= Log(2i) = ln 2 +i^π₂ = 2 ln√

2 +i^π₄

= 2Log(1 + i), (b). Log(−1 +i)² = Log(−2i) = ln 2−iπ/2 and 2Log(−1 +i) = 2

ln√

2 +i^3π₄

= ln 2 +i^3π₂.

9.6. 0 whena >0,and 2πiwhena <0.

9.7. (a). (1 +i)ⁱ =e^ilog(1+i) =e^i(ln√2+i(π/4+2kπ)) =e^{−(π/4+2kπ)}e^iln√2, k∈Z,(b). (−1)^π=e^πlog(−1)=eπ(ln 1+i(π+2kπ))=e^iπ2(1+2k), k ∈Z, (c). e^πe^{i(2 ln 2)},(d). e3/2(ln 2+i2π/3)= 2√

2e^iπ =−2√ 2.

9.8. 1 +itanw= 2e^iw/(e^iw+e^−iw), 1−itanw= 2e^−iw/(e^iw+e^−iw).

9.9. (a). (4k+ 1)π/2±iLog(√

5 + 2),(b). (4k+ 1)πi/2.

9.10.Follow the same method as in Example 8.2.

Lecture 10