Complex Analysis

Math 312 Spring 1998  MWF 10:30am - 11:25am
Buckmire  Fowler 112

Class 13 (Wednesday February 11)

SUMMARY Complex Logarithms and the debut of branch cuts CURRENT READING Brown & Curchill, pages 75-81 NEXT READING Brown & Curchill pages 81-85

The Complex Logarithm log z Let us de�ne w = log z as the inverse of z = e^{w} But we know that exp[ln |z| + i(θ + 2nπ)] = z, where n ∈ Z, from our knowledge of the exponential function. So we can de�ne

log z = ln |z| + i arg z = ln |z| + i Arg z + 2nπi = ln r + iθ

where r = |z| as usual, and θ is the argument of z If we only use the principal value of the argument, then we de�ne the principal value of log z as Log z, where

Log z = ln |z| + i Arg z = Log |z| + i Arg z

\underline{Examples} Compute Log (−2)and Log (2), log(−2), and log(2), Log (−4) and log(−4)

Logarithmic Identities z = e^{log z} but log e^{z} = z + 2kπi (Is this a surprise?) log(z_{1}z_{2}) = log z_{1} + log z_{2}

log (\frac{z_{1}}{z_{2}}) = log z_{1} − log z_{2}

However these do not neccessarily apply to the principal branch of the logarithm, written as Log z.

Log z: the Principal Branch of log z Log z is a single-valued function and is analytic in the domain D^{∗} consisting of all points of the complex plane except for those lying on the nonpositive real axis, where

\frac{d}{dz} Log z = \frac{1}{z}

Sketch the set D^{∗} and convince yourself that it is an open connected set. (Ask yourself: Is every point in the set an interior point?)

The set of points Re z ≤ 0 ∩ Im z = 0 is a line of discontinuities known as a branch cut. By putting in a branch cut we say that we �construct Log z from log z.� Why can we not evaluate log z along the entire positive x-axis?

Analyticity of Log z We can use a version of the Cauchy-Riemann Equations in polar coordinates to help us investigate the analyticity of Log z

Why don’t we investigate the analyticity of log z?

If x = r cos θ and y = r sin θ one can trewrite f(z) = u(x, y) + iv(x, y) into f = u(r, θ) + iv(r, θ) in that case, the CREs become:

u_{r} = \frac{1}{r}v_{θ}, v_{r} = −\frac{1}{r}u_{θ}

and the expression for the derivative f′(z) = u_{x} + iv_{x} can be re-written

f′(z) = e^{−iθ}(u_{r} + iv_{r})

\underline{Groupwork} Using this information, show that Log z is analytic and that \frac{d}{dz} Log z = \frac{1}{z}. (HINT: You will need to write down u(r, θ) and v(r, θ) for Log z)

Branching: Producing Single-valued functions from Multiple-valued ones A single-valued function F (z) is said to be a branch of a multiple-valued function f(z) in a domain D if F (z) is single-valued and analytic in D and has the property that for each z ∈ D, the value F (z) is one of the values of f(z)

Branch cuts do not have to be along the x-axis, they can be any line which when removed from the domain of de�nition of the multi-valued function, produce a single valued function.

\underline{Example} Determine the domain of analyticity for the function f(z) = Log (3z − i) and compute f′(z) What is f(i)? What about f′(i)?

Other branches of log z One can de�ne other analytic branches of log z by choosing different branch cuts. The usual way to do this is to make the branch cut along θ = α starting at the origin, so that

log z = ln |z| + iθ, where α < θ < α + 2π

These branches of log z can be denoted L_{α}, where θ = α is the branch cut.

A branch point of a function f is a point which is common to all branch cuts of f. So, 0 is a branch point of log z Roots of Complex Numbers (Reprise) From previous identities about complex logarithms and the complex exponential, we can show (for z ≠ 0) that

z^{n} = exp(n log z), as long as n ∈ Z

Similarly,

z^{1/n} = exp (\frac{1}{n} log z)

But

exp (\frac{1}{n} log z) = exp (\frac{1}{n}[ln |z| + i( Arg z + 2kπ)])  (k ∈ Z)
= |z|^{1/n} exp (i[ \frac{Arg z}{n} + \frac{2kπ}{n}])
= |z|^{1/n} exp (\frac{iθ + 2kπi}{n})  where θ = Arg z

But you should recognize the right hand side as the familiar formula for �nding the root of a complex number, where k is restricted to 0, 1, 2, …, n − 1. Why would we do that? [HINT: how many distinct values does exp(2kπi/n) have when k can be any integer and n is �xed?] What do you think happens if we try and raise a complex number to something besides an integer or rational number? How will we deal with complex exponents?

Complex Exponents If z ≠ 0 and c ∈ C, the function z^{c} is de�ned as

z^{c} = exp(c log z)

Since log z is a multi-valued function, z^{c} will have multiple values. How many values depends on the nature of c.

z^{c} = \{ \begin{matrix}z^{n/m} & if c is rational, i.e. n/m & �nite number of values (m) \\ z^{n} & if c = n, where n is an integer & single value \\ z^{c} & all other complex numbers & in�nite number of values \end{matrix}

\underline{Examples} Compute the following:

(a) (1 + i)^{1−i} =

(b) (0.5 − i)^{3} =

(c) (−1)^{2/3} =

Differentiating If you choose a branch of z^{c} which is analytic on an open set, then

\frac{d}{dz} (z^{c}) = cz^{c−1}

where the branch of the log used in evaluating z^{c} is the same branch used in evaluating z^{c−1} Similarly, we can de�ne the complex exponential function with base c

c^{z} = exp(z log c)

If a single value of c is chosen, then c^{z} is an entire function such that

\frac{d}{dz} (c^{z}) = \frac{d}{dz} exp(z log c) = c^{z} log c

\underline{Examples} If f(z) = (1 + i)^{z}, Find f′(1 − i) (Use the principal branch)

If g(z) = z^{(}1 − i), Find f′(1 + i) (Use the principal branch)