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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 Complex Numbers

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

\(\arg(z) = \arctan(1/1) = \pi/4\) (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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\(e^{i\pi} = \cos\pi + i\sin\pi = -1 + i \cdot 0 = -1\)

\(e^{i\pi} + 1 = 0\) ✓

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

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

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

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

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

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

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

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

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

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

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

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\(z = 1\) 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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\(f(z) = z^{-2}\)

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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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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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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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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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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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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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 \(z \leftrightarrow \bar z\).