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Rationalizing the denominator givesHence its real part, imaginary part, complex modulus, and principal argument are respectivelyThe quadrant matters: the real part is positive and the imaginary part negative. For the exponential expression, Euler's formula gives $e^{2+i\pi /4}=e^2(1+i)/\sqrt{2}$, so the same four quantities areGeneral complex arguments are obtained from each stated principal argument by adding $2\pi k$, $k\in \mathbb{Z}$.

Put $z=x+iy$. The expression whose complex argument is constrained becomesIts principal argument is $\pi /2$ exactly when its real part is zero and its imaginary part is strictly positive. HenceThe locus consists of the upper-right half of the upper hyperbola branch, $y=\sqrt{x^2+1}$ with $x>0$, and the lower-left half of the lower branch, $y=-\sqrt{x^2+1}$ with $x<0$. The vertices $z=\pm i$ are excluded because $z^2+1=0$ there and zero has no complex argument. The remaining halves have negative imaginary part and give argument $-\pi /2$, not $\pi /2$. The right panel of the original illustration marks the excluded vertices with open circles.

The complex number $\sqrt{3}+i$ has modulus two and principal argument $\pi /6$. Therefore its principal complex logarithm isIf the source's logarithm means the full multivalued complex logarithm, rather than its principal value, the complete answer is $\ln 2+i(\pi /6+2\pi n)$ with $n\in \mathbb{Z}$. A branch convention is necessary to obtain a unique imaginary part.

A complex power requires a choice of complex logarithm. Taking the principal argument of $-e$ to be $+\pi$ gives $\mathrm{Log}(-e)=1+i\pi$, henceThus $a=0$ and $b=e^{-{\pi }^2/2}$ for this principal value. If the power is interpreted as multivalued, the logarithms are $1+i(2k+1)\pi$ and the possible values are $i\exp [-(2k+1){\pi }^2/2]$, $k\in \mathbb{Z}$. Applying an identity for real powers without specifying this branch would lose the ambiguity.

The two factors have moduli $\sqrt{2}$ and $2$, and complex arguments $\pi /4$ and $\pi /6$. Multiplication multiplies the moduli and adds the complex arguments, givingThe displayed angle is the principal argument in radians; equivalently the argument is $75^{\circ }$ modulo $360^{\circ }$.

For $z\ne 0$, the complex logarithm is $\ln z=\ln r+i(\theta +2\pi k)$, where $k\in \mathbb{Z}$. ConsequentlyA specified branch chooses one continuous determination of the complex argument on an appropriate cut domain. For the usual principal-value convention use the principal argument $\mathrm{Arg}(x+iy)\in (-\pi ,\pi ]$. A formula using only $\arctan (y/x)$ would lose the quadrant information; it is a valid local angle only with the appropriate additive multiple of $\pi$.

The factor $1+i$ has modulus $\sqrt{2}$ and principal argument $\pi /4$. Multiplication by $e^{i\pi /6}$ adds $\pi /6$ to its complex argument, while the real factor $1/2$ halves its modulus. HenceTaking the principal logarithm for $w$ givesThis complex number lies in the second quadrant. Its modulus follows from the sum of squared components, and its principal argument follows by the second-quadrant correction:All arguments of this fixed $w$ add integer multiples of $2\pi$ to the displayed one.

If $\ln z$ is interpreted as the multivalued complex logarithm, there is instead a family $w_k=-a+i{b}_k$, where $a=\frac{1}{2}\ln 2>0$ and $b_k=5\pi /12+2\pi k$. Then $|w_k|=\sqrt{a^2+b_k^2}$, and the principal argument is $\pi -\arctan (b_k/a)$ for $b_k>0$, or $-\pi -\arctan (b_k/a)$ for $b_k<0$. This records the logarithm-branch choice that the paper leaves implicit.

Take $r>0$ and select a complex argument $\theta$ for $z$. To specify the complex exponentiation, use $z^{1+i}=\exp ((1+i)(\ln r+i\theta ))$. Expanding its exponent givesIts complex logarithm therefore has valuesThe continuous logarithm obtained by retaining the displayed exponent uses $k=0$. If the outer logarithm is the principal logarithm, choose $k$ to put the imaginary part in $(-\pi ,\pi ]$ instead; the real part is unchanged. In particular one cannot generally assert $\mathrm{Log}(e^w)=w$ without checking its imaginary part.

The condition on the real part gives $\ln r=\theta$, so the required logarithmic spiral has polar coordinatesAs $\theta$ increases, it runs from the negative real axis near radius $e^{-\pi }$ through $-i{e}^{-\pi /2}$, $1$ and $i{e}^{\pi /2}$ to the negative real axis at radius $e^{\pi }$. The angle increases by one complete turn. Using the conventional principal argument $-\pi <\theta \le \pi$, the inner negative-axis endpoint is excluded and the outer endpoint is included. The convention $-\pi \le \theta <\pi$ reverses those endpoint choices. If both limiting arguments are admitted by the paper's wording, both limiting endpoints are drawn; the interior curve is the same. The origin is never included because the complex logarithm is undefined there.

For $z\in \mathbb{C}\setminus [1,\infty )$, the number $w=1-z$ avoids the nonpositive real axis. Define the branch of the complex logarithm byHere the principal argument is selected as a continuous map on that slit domain, so this is a holomorphic function, with derivative $-1/(1-z)$ and $L(1)=0$. For example, locally writing $w=\rho e^{i\varphi }$ gives $d(\log \rho +i\varphi )=dw/w$, verifying the complex derivative. The normalization also makes this branch of the complex logarithm unique: any two continuous choices differ by a constant integer multiple of $2\pi i$ on the connected slit domain, and their values at $w=1$ agree.

To evaluate the corresponding analytic branch of a square root on the unit circle, factorFor $0<\theta <2\pi$, we have $\sin (\theta /2)>0$ and $-\pi /2<\theta /2-\pi /2<\pi /2$. Thus this factorization already uses the selected principal argument, with no hidden change by $2\pi$. It follows thatHalving this expression and applying the exponential function givesSince $\sqrt{2}{e}^{-i\pi /4}=1-i$, the desired branch value isFor instance, $\theta =\pi$ gives $\sqrt{2}>0$, a quick check that the sign agrees with the normalization.

Choose each factor with its argument between zero and $\pi$ in the upper half-plane. On an interval between consecutive real prevertices, each factor has a constant boundary argument, so $f^{\prime }$ has constant argument and the image is a straight side. Crossing $X_j$ from left to right decreases $\arg (Z-X_j)$ by $\pi$, hence increases $\arg f^{\prime }$ by ${\beta }_j\pi$, giving the prescribed exterior turn. Locally,so an upper half-disc maps to angle $(1-{\beta }_j)\pi$, the required interior angle. The overall constant rotates and scales the polygon; integration adds translation. For a general prescribed polygon, the prevertex positions must also satisfy the side-length conditions: the derivative formula fixes angles but arbitrary choices do not fix arbitrary side lengths.

For the slit domain, the slit tip has interior angle $2\pi$, so its exterior parameter is $-1$. The two vertices at infinity give the factors at $\pm 1$, leading to $f^{\prime }(Z)$ proportional to $Z(1-Z^2{)}^{-3/2}$. Integrating and imposing the tip value givesThe square root is the principal one: for $\Im Z>0$, $1-Z^2$ never lies on the nonpositive real axis. Thus $\Re f>0$. Boundary points $-1<X<1$ map to the two banks of the slit, with $X=0$ at the tip; the outside intervals map to the imaginary axis. The derivative is nonzero in the open half-plane. Conversely each point of the slit domain determines exactly one root $Z$ of $Z^2=1-z^{-2}$ with positive imaginary part, proving bijectivity.

The Möbius map $(Z-1)/(Z+1)$ takes the upper half-plane to itself, so its principal argument lies in $(0,\pi )$. The imaginary part of its analytic logarithm is harmonic, equals $\pi$ on $(-1,1)$ and zero on the other real intervals. ThusIt lies between zero and one and has the required boundary limits away from the two discontinuities. Pulling this harmonic function back by the inverse conformal map yieldsThis specifies all branches and remains bounded even as $z$ tends to either end at infinity.

To exhibit nonuniqueness without the growth restriction, putThis is harmonic in the domain. On its boundary the corresponding $Z$ is real and the rational expression is real, except at $\pm 1$, which represent infinity. At $Z=\infty$, representing $z=0$, the expression tends to zero. Thus $H$ has zero finite boundary data. It is nonzero, since for $Z=iY$ its value is $Y/(1+Y^2)>0$, and it is unbounded as $Z$ approaches either pole $\pm 1$ non-tangentially. Therefore $u+cH$ gives distinct solutions with the same finite boundary data when boundedness at infinity is removed.

The principal argument selects a single complex argument of a nonzero complex number, conventionally in $(-\pi ,\pi ]$. It is the imaginary part of the principal logarithm. The alternative interval $[-\pi ,\pi )$ changes the convention on the negative real axis. Neither convention makes the selected complex argument continuous across that axis.