Complex Logarithm and Harmonic Functions Lecture 6 Elementary functions Complex Logarithm Note that: ez is an onto function from C to C∗ = C n f0g: Let w 2 C∗ such that w = jwjeiθ for some θ 2 (−π; π]. If we take z = log jwj + iθ then ez = elog jwjeiθ = w: But ez is not an injective function as ez+2πik = ez ; k 2 Z: If we restrict the domain then it (ez ) becomes injective. If take H = fz = x + iy : −π < y ≤ πg then z ! ez is a bijective function from H to C n f0g: What is the inverse of this function? ∗ Definition: For z 2 C , define log z = ln jzj + i arg z: Here ln jzj stands for the real logarithm of jzj: Lecture 6 Elementary functions Complex Logarithm Since arg z = Argz + 2kπ; k 2 Z it follows that log z is not well defined as a function. (multivalued) ∗ For z 2 C , the principal value of the logarithm is defined as Log z = ln jzj + i Argz: ∗ Log : C ! fz = x + iy : −π < y ≤ πg is well defined (single valued). Log z + 2kπi = log z for some k 2 Z: If z 6= 0 then eLog z = eln jzj+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 2 (−π; π] (why?). Some examples: π iπ Log i = ln jij + i 2 = 2 ; Log (−1) = ln j − 1j + iπ = iπ; −π iπ Log (−i) = ln j − ij + 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 = fz = x + iy : x ≤ 0; y = 0g. Proof: z ! Logz is not continuous at z = −α for any α > 0. i(π− 1 ) i(−π+ 1 ) Let fang = fαe n g and fbng = fαe n g. Note that lim an = lim bn = z. n!1 n!1 1 However, lim Log an = lim ln α + i(π − ) = ln α + iπ n!1 n!1 n 1 and lim Log bn = lim ln α + i(−π + ) = ln α − iπ: n!1 n!1 n ∗ − Log is an analytic on the set C n R = C n (−∞; 0]: Let z = reiθ 6= 0 and θ 2 (−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 2 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) 2 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 n f0g 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.