Complex Logarithm and Harmonic Functions

Lecture 6 Elementary functions Complex Logarithm

Note that: ez is an onto function from C to C∗ = C \{0}. Let w ∈ C∗ such that w = |w|eiθ for some θ ∈ (−π, π]. If we take z = log |w| + iθ then ez = elog |w|eiθ = w. But ez is not an injective function as ez+2πik = ez , k ∈ Z. If we restrict the domain then it (ez ) becomes injective. If take H = {z = x + iy : −π < y ≤ π} then z → ez is a bijective function from H to C \{0}. What is the inverse of this function? ∗ Definition: For z ∈ C , define log z = ln |z| + i arg z. Here ln |z| stands for the real logarithm of |z|.

Lecture 6 Elementary functions Complex Logarithm

Since arg z = Argz + 2kπ, k ∈ Z it follows that log z is not well defined as a function. (multivalued) ∗ For z ∈ C , the principal value of the logarithm is defined as Log z = ln |z| + i Argz. ∗ Log : C → {z = x + iy : −π < y ≤ π} is well defined (single valued). Log z + 2kπi = log z for some k ∈ Z. If z 6= 0 then eLog z = eln |z|+i Argz = z (What about Log (ez )?). Suppose x is a positive real number then Log x = ln x + i Argx = ln x.

The identity Log (z1z2) = Logz1 + Log z2 is not always valid. However, the above identity is true if and only if Arg z1 + Arg z2 ∈ (−π, π] (why?). Some examples: π iπ Log i = ln |i| + i 2 = 2 , Log (−1) = ln | − 1| + iπ = iπ, −π iπ Log (−i) = ln | − i| + i 2 = − 2 , Log (−e) = 1 + iπ (check!))

Lecture 6 Elementary functions Complex Logarithm

The function Log z is not continuous on the negative real axis − R = {z = x + iy : x ≤ 0, y = 0}. Proof: z → Logz is not continuous at z = −α for any α > 0. i(π− 1 ) i(−π+ 1 ) Let {an} = {αe n } and {bn} = {αe n }. Note that lim an = lim bn = z. n→∞ n→∞ 1 However, lim Log an = lim ln α + i(π − ) = ln α + iπ n→∞ n→∞ n 1 and lim Log bn = lim ln α + i(−π + ) = ln α − iπ. n→∞ n→∞ n ∗ − Log is an analytic on the set C \ R = C \ (−∞, 0]. Let z = reiθ 6= 0 and θ ∈ (−π, π). In polar coordinate

Log z = ln r + iθ = u(r, θ) + iv(r, θ).

1 1 Clearly it satisfy the C-R equations ur = r vθ = r and 1 vr = − r uθ and and partial derivatives are continuous.

Lecture 6 Elementary functions Complex Logarithm

Branch of a multiple valued function: Let F be a multiple valued function defined on a domain D. A function f is said to be a branch of the multiple valued function F if in a domainD0 ⊂ D if f (z) is single valued and analytic in D0. Branch Cut: The portion of a line or a curve introduced in order to define a branch of a multiple valued function is called branch cut. Branch Point: Any point that is common to all branch cuts is called a branch point. For the multiple valued log z Log z is a branch know as principal branch of the logarithm. The line (−∞, 0] is the branch cut. The point 0 (origin) is the branch point. α Now we can define the power z for any α ∈ C. For any z 6= 0, define z α = exp(α log z). Note that z → z α is a multiple valued function.

Lecture 6 Elementary functions Harmonic Functions

Theorem: Let f is analytic in a domain D. If f = u + iv Then

uxx (x, y) + uyy (x, y) = vxx (x, y) + vyy (x, y) = 0.

Proof:

By CR equations ux = vy and uy = −vx in D. It gives that uxx = vyx and uyy = −vxy . This implies uxx + uyy = vyx − vxy = 0. Similarly, one can show that vxx + vyy = 0. 2 A function u : D ⊂ R → R is said to be harmonic in a domain D if

uxx (x, y) + uyy (x, y) = 0 at each (x, y) ∈ Ω

If there exists a harmonic function v : D → R such that f (z) = u(x, y) + iv(x, y) is analytic in D, such harmonic function v is called the harmonic conjugate of u.

Lecture 6 Elementary functions Harmonic Conjugate

Given a harmonic function u. Suppose the harmonic conjugate of u exists. Is it unique? Ans:Yes, it is unique up to an additive constant.

Proof. Let v1 and v2 be two harmonic conjugates of u. That is there are two analytic functions f1 and f2 such that f1 = u + iv1 and f2 = u + iv2. Therefore f1 − f2 = i(v1 − v2) is an analytic function such that Re(f1 − f2) = 0. So (f1 − f2) = C (constant). This implies that v1 = C + v2. A function f (z) = u(x, y) + iv(x, y) is analytic if and only if v is the harmonic conjugate of u. The function v(x, y) = 2xy is a harmonic conjugate of u(x, y) = x 2 − y 2 2 2 2 in C. The function f (z) = z = (x − y ) + i (2xy) is analytic in C.

Lecture 6 Elementary functions Harmonic Conjugate

Construction of a harmonic conjugate Let u(x, y) = x 2 − y 2. We have to find the harmonic conjugate of u.

Step 1: Check that u is harmonic: clearly uxx + uyy = 2 − 2 = 0.

Step 2: Calculate ux and uy : ux (x, y) = 2x and uy (x, y) = −2y. Since the conjugate harmonic function v satisfied CR equations we have Z ux (x, y) = vy (x, y) = 2x =⇒ v(x, y) = ux (x, y) dy+φ(x) = 2xy+φ(x).

Consider

0 0 vx (x, y) = 2y + φ (x) = −uy (x, y) = 2y =⇒ φ (x) = 0.

So v(x, y) = 2xy + c, where c is a constant.

So f (x, y) = u(x, y) + iv(x, y) = x 2 − y 2 + 2ixy + ic = z 2 + ic is analytic.

Lecture 6 Elementary functions Harmonic conjugate

Does harmonic conjugate v always exist for a given harmonic function u in a domain D? Answer: ‘No’. 2 2 1 The function u(x, y) = log(x + y ) 2 is harmonic on G = C \{0} and it has no harmonic conjugate on G. Question: Under what condition harmonic conjugate v exists for a given harmonic function u in a domain D? Theorem: Let G be either the whole plane C or some open disk. If u : G → R is a harmonic function then u has a harmonic conjugate in G.

Lecture 6 Elementary functions