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Appendix E Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Absolute Value and Argument of a Complex Number

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Problem:

Find the absolute value \(|z|\) and the argument \(\arg(z)\) of \(z = 1 + i\). Express it in polar form \(z = re^{i\theta}\).

\(|z| = \sqrt{1^2 + 1^2} = \sqrt{2}\)

\(\arg(z) = \arctan(1/1) = \pi/4\) (first quadrant)

Polar form: \(z = \sqrt{2}\, e^{i\pi/4}\)

B-2. Euler's Formula \(e^{i\pi}+1=0\)

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Problem:

Using Euler's formula, verify that \(e^{i\pi} + 1 = 0\). List the five mathematical constants (\(e, i, \pi, 1, 0\)) contained in this expression.

\(e^{i\pi} = \cos\pi + i\sin\pi = -1 + i \cdot 0 = -1\)

\(e^{i\pi} + 1 = 0\) ✓

The five mathematical constants included: \(e\) (base of the natural logarithm), \(i\) (imaginary unit), \(\pi\) (pi), \(1\) (multiplicative identity), and \(0\) (additive identity).

B-3. Product in Polar Form

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Problem:

Given the complex numbers \(z_1 = 2e^{i\pi/3}\) and \(z_2 = 3e^{i\pi/6}\), find the product \(z_1 z_2\) in polar form, and then convert it to rectangular form \(a + bi\).

\(z_1 z_2 = 2e^{i\pi/3} \cdot 3e^{i\pi/6} = 6\, e^{i(\pi/3 + \pi/6)} = 6\, e^{i\pi/2}\)

\(= 6(\cos(\pi/2) + i\sin(\pi/2)) = 6i\)

B-4. Cauchy-Riemann: Verification with \(z^2\)

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Problem:

Verify that \(f(z) = z^2 = (x+iy)^2 = (x^2-y^2) + 2ixy\) satisfies the Cauchy-Riemann relations.

\(f(z) = (x+iy)^2 = (x^2 - y^2) + 2ixy\)

\(u = x^2 - y^2, \quad v = 2xy\)

Verification of the CR relations:

\(\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = 2x\) → \(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\) ✓

\(\frac{\partial u}{\partial y} = -2y, \quad -\frac{\partial v}{\partial x} = -2y\) → \(\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\) ✓

B-5. Cauchy-Riemann: \(|z|^2\) fails

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Problem:

Verify that \(f(z) = |z|^2 = x^2 + y^2\) does not satisfy the Cauchy-Riemann equations.

\(f = x^2 + y^2\) (a real-valued function, so \(v = 0\))

\(u = x^2 + y^2, \quad v = 0\)

\(\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial v}{\partial y} = 0\)

\(2x = 0\) holds only at \(x = 0\). In general, \(\frac{\partial u}{\partial x} \neq \frac{\partial v}{\partial y}\).

Since the CR relations are not satisfied, \(|z|^2\) is not holomorphic.

B-6. \(\partial_z(z^2) = 2z\)

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Problem:

Using the definition of the Wirtinger derivative

\[ \partial_z = \frac{1}{2}(\partial_x - i\partial_y) \]

show that \(\partial_z(z^2) = 2z\) for \(f(z) = z^2\) (where \(z = x + iy\)).

\(z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy\)

\(\partial_z = \frac{1}{2}(\partial_x - i\partial_y)\)

\(\partial_x(z^2) = 2x + 2iy = 2(x + iy) = 2z\)

\(\partial_y(z^2) = -2y + 2ix = 2i(x + iy) = 2iz\)

\(\partial_z(z^2) = \frac{1}{2}(2z - i \cdot 2iz) = \frac{1}{2}(2z + 2z) = 2z\) ✓

B-7. Residue of \(1/(z-1)\)

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Problem:

Find the residue of \(f(z) = 1/(z-1)\) at \(z = 1\).

\(z = 1\) is a simple pole (pole of order 1).

\(\text{Res}_{z=1} \frac{1}{z-1} = \lim_{z \to 1}(z-1) \cdot \frac{1}{z-1} = 1\)

B-8. Laurent Expansion and Residue of \(1/z^2\)

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Problem:

Write the Laurent expansion of \(f(z) = 1/z^2\) about \(z = 0\), and confirm that the residue is \(0\).

\(f(z) = z^{-2}\)

This is already in the form of a Laurent expansion. \(a_{-2} = 1\), \(a_{-1} = 0\), and all others are zero.

Residue \(= a_{-1} = 0\).

\(z = 0\) is a pole of order 2 (\(a_{-2} \neq 0\), \(a_n = 0\) for \(n < -2\)).

B-9. Laurent Expansion of \(e^{1/z}\)

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Problem:

Write the Laurent expansion of \(e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n! z^n}\) at \(z = 0\), and confirm that the residue (the coefficient of \(z^{-1}\)) is \(1\). What type of singularity is \(z = 0\)?

Substituting \(w = 1/z\) into \(e^w = \sum_{n=0}^{\infty} \frac{w^n}{n!}\):

\(e^{1/z} = \sum_{n=0}^{\infty} \frac{1}{n!\, z^n} = 1 + \frac{1}{z} + \frac{1}{2z^2} + \frac{1}{6z^3} + \cdots\)

Residue (coefficient of \(z^{-1}\)) \(= a_{-1} = 1\) ✓

Since the principal part (terms with \(n < 0\)) continues indefinitely, \(z = 0\) is an essential singularity.

Medium

M-1. Residue Theorem: Two Poles

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Problem:

Evaluate \(\oint_{|z|=2} \frac{dz}{z(z-1)}\) using the residue theorem. (There are two poles at \(z = 0\) and \(z = 1\) inside the circle \(|z| = 2\).)

Inside the circle \(|z| = 2\), there are two poles at \(z = 0\) and \(z = 1\).

Residue at \(z = 0\): \(\lim_{z \to 0} z \cdot \frac{1}{z(z-1)} = \frac{1}{0-1} = -1\)

Residue at \(z = 1\): \(\lim_{z \to 1} (z-1) \cdot \frac{1}{z(z-1)} = \frac{1}{1} = 1\)

Residue theorem:

\(\oint_{|z|=2} \frac{dz}{z(z-1)} = 2\pi i(-1 + 1) = 0\)

M-2. Residues of \(z/[(z-1)(z-2)]\)

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Problem:

Find the residues of \(f(z) = \frac{z}{(z-1)(z-2)}\) at \(z = 1\) and \(z = 2\), respectively, and compute \(\oint_{|z|=3} f(z) dz\).

Residue at \(z = 1\): \(\lim_{z \to 1}(z-1) \cdot \frac{z}{(z-1)(z-2)} = \frac{1}{1-2} = -1\)

Residue at \(z = 2\): \(\lim_{z \to 2}(z-2) \cdot \frac{z}{(z-1)(z-2)} = \frac{2}{2-1} = 2\)

Since both poles are contained inside the circle \(|z| = 3\):

\(\oint_{|z|=3} f(z)\,dz = 2\pi i(-1 + 2) = 2\pi i\)

M-3. Composition of Möbius Transformations

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Problem:

Given two Möbius transformations \(w_1(z) = (z+1)/(z-1)\) and \(w_2(w) = 2w + 3\), find the composed transformation \(w_2(w_1(z))\) and verify that the corresponding matrix product is consistent.

Solution:

First, \(w_1(z) = (z+1)/(z-1)\). The matrix is:

\[M_1 = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\]

Next, \(w_2(w) = 2w + 3\). Since \(w_2(w) = (2w+3)/(0\cdot w + 1)\):

\[M_2 = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}\]

Direct computation of the composition \(w_2(w_1(z))\):

\[w_2(w_1(z)) = 2 \cdot \frac{z+1}{z-1} + 3 = \frac{2(z+1) + 3(z-1)}{z-1} = \frac{5z - 1}{z-1}\]

Matrix product \(M_2 M_1\):

\[M_2 M_1 = \begin{pmatrix} 2 & 3 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 2+3 & 2-3 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 5 & -1 \\ 1 & -1 \end{pmatrix}\]

The corresponding Möbius transformation is \((5z - 1)/(z - 1)\), which agrees with the direct computation ✓

Advanced

A-1. Conformal Mapping \(w = 1/z\)

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Problem:

Determine where the conformal mapping \(w = 1/z\) maps the unit circle \(|z| = 1\) in the \(z\)-plane to in the \(w\)-plane. Also, determine where it maps the real axis in the \(z\)-plane to in the \(w\)-plane.

Image of the unit circle \(|z| = 1\):

When \(|z| = 1\), we have \(z = e^{i\theta}\). Then \(w = 1/z = e^{-i\theta}\).

Since \(|w| = 1\), the unit circle is mapped to the unit circle itself (though the direction of traversal is reversed).

Image of the real axis:

When \(z = x\) (\(x\) is real, \(x \neq 0\)), we have \(w = 1/x\) (real).

The real axis is mapped to the real axis itself. Specifically, \(x > 0\) maps to \(w > 0\), \(x < 0\) maps to \(w < 0\), as \(x \to 0^+\) we get \(w \to +\infty\), and as \(x \to \pm\infty\) we get \(w \to 0\).

A-2. Cross Terms of \(\partial X\) and \(\bar\partial X\)

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Problem:

In the main text E.8 "Green's Function for the 2D Free Field", the two-point function \(\langle X(z,\bar z)\, X(w,\bar w)\rangle = -\frac{\alpha'}{2}\ln\lvert z-w\rvert^2\) was derived. From this result, show the following:

(a) \(\langle \partial X(z)\, \bar\partial X(w)\rangle\) vanishes for \(z \neq w\). (b) \(\langle \bar\partial X(\bar z)\, \bar\partial X(\bar w)\rangle = -\frac{\alpha'}{2}\, \frac{1}{(\bar z - \bar w)^2}\) (the anti-holomorphic OPE).

Hint: Decompose \(\ln\lvert z-w\rvert^2 = \ln(z-w) + \ln(\bar z - \bar w)\), and use the fact that \(\partial_z\) acts only on the holomorphic part while \(\partial_{\bar z}\) acts only on the anti-holomorphic part.

Solution:

(a) Calculation of \(\langle \partial X(z)\, \bar\partial X(w)\rangle\):

\[\langle X(z,\bar z)\, X(w, \bar w)\rangle = -\frac{\alpha'}{2}\ln\lvert z-w\rvert^2 = -\frac{\alpha'}{2}[\ln(z-w) + \ln(\bar z - \bar w)]\]

Differentiating with respect to \(z\):

\[\partial_z \langle XX\rangle = -\frac{\alpha'}{2} \cdot \frac{1}{z-w}\]

(The \(\ln(\bar z - \bar w)\) term vanishes since it does not depend on \(z\).)

Then differentiating with respect to \(\bar w\):

\[\partial_{\bar w}\left[-\frac{\alpha'}{2}\cdot\frac{1}{z-w}\right] = 0\]

(\(1/(z-w)\) does not depend on \(\bar w\), so this is 0.)

Therefore:

\[\boxed{\langle \partial X(z)\, \bar\partial X(\bar w)\rangle = 0 \qquad (z \neq w)}\]

The holomorphic and antiholomorphic parts completely decouple. This is the basis for the "independence of left and right sectors" in conformal field theory.

(b) Calculation of \(\langle \bar\partial X(\bar z)\, \bar\partial X(\bar w)\rangle\):

Differentiating with respect to \(\bar z\) leaves only the antiholomorphic part:

\[\partial_{\bar z} \langle XX\rangle = -\frac{\alpha'}{2}\cdot\frac{1}{\bar z - \bar w}\]

Then differentiating with respect to \(\bar w\) (noting the sign when differentiating with respect to \(\bar w\)):

\[\partial_{\bar w}\frac{1}{\bar z - \bar w} = \frac{1}{(\bar z - \bar w)^2}\]

Therefore:

\[\boxed{\langle \bar\partial X(\bar z)\, \bar\partial X(\bar w)\rangle = -\frac{\alpha'}{2}\cdot\frac{1}{(\bar z - \bar w)^2}}\]

The antiholomorphic side has the same structure as the holomorphic side, in a form completely symmetric under \(z \leftrightarrow \bar z\).