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A branch of a multivalued analytic expression is a single-valued holomorphic function on a specified domain that consistently selects one of the expression's possible values. A branch cut is a convenient way to choose such a domain, rather than an intrinsic locus of singularities. On $D=\mathbb{C}\setminus (-\infty ,0]$, the principal logarithm supplies two square-root branchesFor $z=i$, these give $B_{+}(i)=e^{i\pi /4}=(1+i)/\sqrt{2}$ and $B_{-}(i)=-(1+i)/\sqrt{2}$. Their derivatives follow by differentiating $B_{\pm }^2=z$: $B_{\pm }^{\prime }=1/(2B_{\pm })$, so $B_{+}^{\prime }(i)=(1-i)/(2\sqrt{2})$. There are exactly two branches on $D$: for any other holomorphic square root $B$, the ratio $B/B_{+}$ is continuous, takes values in $\{1,-1\}$, and hence is constant on the connected domain.

The cut can move. Choosing $0<\theta <2\pi$ defines a holomorphic logarithm $L_{+}(z)=\log |z|+i\theta$ on $D_{+}=\mathbb{C}\setminus [0,\infty )$. At $-1$ this gives $L_{+}(-1)=i\pi$ and $\exp (L_{+}(-1)/2)=i$, both regular values despite $-1$ lying on the previous cut. In the complex upper half-plane, $L_{+}$ agrees with $\mathrm{Log}$; in the lower half-plane it equals $\mathrm{Log}+2\pi i$, so the corresponding square roots have opposite signs there. The overlap $D\cap D_{+}$ has two components, exactly why the identity theorem cannot force agreement between them. ^[ref]

For example, the limits of $B_{+}$ at $-r$, $r>0$, are $i\sqrt{r}$ from above and $-i\sqrt{r}$ from below. This jump records the chosen branch cut; each point $-r\ne 0$ still has a neighbourhood supporting a holomorphic square root. Moving the cut changes the bookkeeping, while the branch point at $0$ and its monodromy persist. Specifying a branch of a multivalued function is therefore essential before differentiating or performing contour integration with holomorphic square roots, complex logarithms or complex exponentiation.

Using principal complex logarithms, the inner power is $i^i=e^{-\pi /2}$. Consequentlywhich are different. Thus parentheses in nested complex exponentiation cannot be dropped or moved. There is an additional issue when all complex logarithm branches are allowed: each nested operation must retain its own branch choices. In the first expression, the complete set of values is $-i{e}^{-2\pi m}$, $m\in \mathbb{Z}$; in the second, it is $\exp [i(\pi /2+2\pi m)e^{-(\pi /2+2\pi k)}]$, $k,m\in \mathbb{Z}$. The principal-value comparison already establishes nonassociativity; the multivalued sets make the role of branching explicit.

A holomorphic logarithm of a nowhere-zero holomorphic function $f$ on a domain $U$ is a holomorphic function $g$ such that $e^{g(z)}=f(z)$ throughout $U$. Differentiating with the chain rule gives $g^{\prime }=f^{\prime }/f$. For $f(z)=z$, a holomorphic logarithm is a branch of the complex logarithm, and its derivative is $1/z$.

The principal logarithm is holomorphic on the slit complex plane $D=\mathbb{C}\setminus (-\infty ,0]$, withHere $\mathrm{Arg}$ is the chosen complex argument, and $\mathrm{Im}\mathrm{Log}z\in (-\pi ,\pi )$. One may conventionally assign a principal value on the negative real axis, but that axis is excluded from this holomorphic domain. ^[ref] For example, $1+i=\sqrt{2}{e}^{i\pi /4}$ gives $\mathrm{Log}(1+i)=\frac{1}{2}\log 2+i\pi /4$ and $(\mathrm{Log}{)}^{\prime }(1+i)=(1-i)/2$; all its other complex logarithms differ by $2\pi ik$, $k\in \mathbb{Z}$.

There is no holomorphic logarithm of $z$ on the punctured complex plane $\mathbb{C}\setminus \{0\}$. Such a function would be a holomorphic primitive of $1/z$, so its contour integral around every closed path would vanish. On the counterclockwise unit circle, however,This is the period obstruction to a holomorphic antiderivative, and also exactly the logarithm change in the monodromy diagram.

Every nowhere-zero holomorphic function $f$ on a simply connected domain $U$ does admit a holomorphic logarithm. Choose $a\in U$ and $b\in \mathbb{C}$ with $e^b=f(a)$, and defineThe Cauchy integral theorem makes this contour integral independent of the path in $U$. Then $g^{\prime }=f^{\prime }/f$ and $(f{e}^{-g}{)}^{\prime }=e^{-g}(f^{\prime }-f{g}^{\prime })=0$; its value at $a$ is $1$, proving $e^g=f$. Any two choices on a connected domain differ by a constant $2\pi ik$, since their difference is continuous and takes values in the discrete set $2\pi i\mathbb{Z}$. ^[ref]

For a computed example, take $f(z)=1+z$ on the unit disc, with $a=b=0$. Expanding $f^{\prime }/f=1/(1+z)$ as a geometric series and integrating givesThe construction also produces holomorphic square roots: $h=e^{g/2}$ satisfies $h^2=f$, so here $h(1/2)=\sqrt{3/2}$. More generally, $e^{\alpha g}$ gives a consistent branch of a multivalued function for complex exponentiation by $\alpha$. A holomorphic logarithm converts multiplicative expressions into additive ones while making the domain restrictions explicit.

For $z\ne 0$, every logarithmic value is $\log |z|+i(\arg z+2\pi k)$ with integer $k$. Multiplying by $z$ and taking the real part proves the displayed formula for the modulus of each value of complex exponentiation. For nonreal $z$, these values form a geometric progression with nonunit positive ratio $e^{-2\pi \mathrm{Im}z}$. For real nonzero $z$, changing the logarithm branch changes only the phase and leaves the modulus fixed. A fixed initial argument changes only the indexing of the set of values.

All complex logarithms of $i$ are $i(\pi /2+2\pi k)$ for integer $k$. By complex exponentiation,Every value is positive and real. If the power is defined using the principal logarithm, the single value is $\boxed{e^{-\pi /2}}$. Keeping the logarithm branch explicit distinguishes this value from the full multivalued set.

Use the principal complex logarithm at each stage of the nested complex exponentiation. Since $\mathrm{Log}i=i\pi /2$,This inner value is positive real, so its principal complex logarithm is $-\pi /2$. ConsequentlyOne must retain the parentheses: complex exponentiation is not associative. If all complex logarithm branches are allowed, the inner values are $e^{-\pi /2-2\pi k}$. Taking the outer logarithm with an independent integer branch index $\ell$ yields $\exp [-i\pi /2-2\pi ik-2\pi \ell ]=-i{e}^{-2\pi \ell }$. Thus the complete multivalued set is $\{-i{e}^{-2\pi \ell }:\ell \in \mathbb{Z}\}$, and the boxed answer selects its principal evaluation.

The polar form is $z=e^{3\pi i/4}$. Interpreting the fractional complex exponentiation as all its values, a candidate $w$ satisfies $w^3=z^2=-i$. The three roots of a complex number areThey have arguments $\pi /2$, $7\pi /6$ and $11\pi /6$, respectively. Since all have modulus one, their squared distances from $z$ in the complex plane are $|w-z{|}^2=2-2\cos (\arg w-3\pi /4)$, givingThe last is strictly largest, so the required root of a complex number isRestricting complex exponentiation to its principal value would give only $i$; the request to choose among roots requires all three values.

The principal complex exponentiation gives $z^i=e^{-\theta }{e}^{i\log r}$. Equality to one requires its complex modulus to be one, hence $\theta =0$, and then its complex argument requires $\log r\in 2\pi \mathbb{Z}$. Thus all solutions are positive real, namelySubstitution verifies every value, and the modulus condition excludes every nonreal value.

For a nonzero complex number, write $z=r{e}^{i\theta }$ with $r=|z|>0$ and principal complex argument $-\pi <\theta \le \pi$. The principal complex logarithm and corresponding complex exponentiation areThe value at zero is not defined by this formula. The choice of complex argument matters: these are single-valued principal powers, rather than all values of a multivalued complex logarithm.

For the requested sketch, $(1+i)\mathrm{Log}z=(\log r-\theta )+i(\log r+\theta )$. Its complex exponential has complex modulus $e^{\log r-\theta }$, soThis is one turn of a logarithmic spiral, with increasing radius as the polar angle increases. It passes through $1$ at $\theta =0$, approaches the excluded endpoint $-e^{-\pi }$ from below the negative real axis, and ends at the included endpoint $-e^{\pi }$ from above that axis. The endpoint distinction follows from the chosen principal complex argument.

For $z\ne 0$, the principal value of complex exponentiation is $z^w=\exp (w\mathrm{Log}z)$, where the principal complex logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z\le \pi$. De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for every integer $n$.

Since $\sqrt{3}+i=2e^{i\pi /6}$, the six roots are

Use the principal complex logarithm, for which $\mathrm{Log}i=i\pi /2$, and setThen the definition of complex exponentiation givesand thereforeOther choices of branch of the complex logarithm produce further valid answers.

By complex exponentiation, a value of $z^i$ has the formFor this to equal one, its exponent must be $2\pi in$ for some $n\in \mathbb{Z}$. Hence a logarithm of $z$ must equal $2\pi n$, so $z$ is positive real andWith the principal branch this is the same list; in the multivalued convention these are precisely the $z$ for which one value of $z^i$ equals one.

The logarithms of $1+\omega =e^{i\pi /3}$ areThe definition of complex exponentiation therefore gives all values asSince $1+\omega =\frac{1}{2}+i\frac{\sqrt{3}}{2}$, these can be written

Complex exponentiation uses a branch of the complex logarithm. All logarithms of $i$ are $i(\pi /2+2\pi k)$ with $k\in \mathbb{Z}$, soSince the trigonometric factor is always $i$, all values in Cartesian form areThe principal logarithm selects $k=0$ and gives the principal value $i{e}^{\pi /2}$.

Using the principal complex logarithm, $\mathrm{Log}i=i\pi /2$ and $\mathrm{Log}(-e)=1+i\pi$. Thus the values and their real parts and imaginary parts areFor the final row, first calculate $i^i=e^{-\pi /2}$ and only then raise this positive real number to the power $i$. Treating nested powers as ordinary real powers would ignore the branching of complex exponentiation.

If all branches are requested, $\sqrt{i}$ also has the negative of its principal value, and the complex logarithm of $-e$ is $1+i(\pi +2\pi k)$. Allowing nonprincipal logarithms of the positive base $2$ gives $e^{-2\pi k}{e}^{i\ln 2}$. Finally, all values of $(i^i{)}^i$ are $-i{e}^{-2\pi m}$, for integers $k,m$ as appropriate. These alternatives do not alter the principal-value table.

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.

The inner power takes the values $r_k=e^{-(\pi /2+2\pi k)}$. For each such real exponent, all values of the outer complex exponentiation are obtained from all complex logarithm values of its base $i$:Writing ${\Theta }_{k,m}=(\pi /2+2\pi m)e^{-(\pi /2+2\pi k)}$, the answer isAll values lie on the unit circle. The principal choice is $k=m=0$, giving $\cos [(\pi /2)e^{-\pi /2}]+i\sin [(\pi /2)e^{-\pi /2}]$. In particular the two nestings in parts (i) and (ii) are different: complex exponentiation is not associative.

Use the full set of complex logarithm values of $i$:By the definition of complex exponentiation, the inner power can therefore take any value $i^i=\exp [-(\pi /2+2\pi k)]$, a positive real number. Its complex logarithm values are $-(\pi /2+2\pi k)+2\pi im$, where $m\in \mathbb{Z}$. ThusAll possible real parts and imaginary parts are consequentlyChoosing the principal complex logarithm at both stages gives the particular value $-i$. The independent logarithm choices are needed because the question asks for all possible values.

For $z=r{e}^{i\theta }\ne 0$, fix one argument $\theta$. Every value of the complex logarithm is $\log r+i(\theta +2\pi k)$, $k\in \mathbb{Z}$. By complex exponentiation, $z^z=\exp [z(\log r+i(\theta +2\pi k))]$. Taking the real part of its exponent gives all possible moduli:Equivalently these are $r^{\mathrm{Re}z}{e}^{-\mathrm{Im}z(\theta +2\pi k)}$. If $z$ is real and nonzero the modulus is the single value $r^{\mathrm{Re}z}$; otherwise the values form a geometric progression with ratio $e^{-2\pi \mathrm{Im}z}$. Changing the initially chosen argument just reindexes the same set. The logarithmic definition excludes $z=0$. This is the modulus of a complex self-power.

Put $s=1/n$, so $0<s<1$. For the complex exponentiation $z^s$, choose $-\pi /2<\arg z<3\pi /2$: the branch cut is the negative imaginary axis. Integrate $z^s/(a^2+z^2)$ around the upper half-plane semicircle, with a small semicircular indentation excluding the origin. The only enclosed simple pole is $z=ia$, with residueThe large arc is $O(R^{s-1})$ and the small arc is $O({ϵ}^{s+1})$, so both vanish. On the negative real segment, $z^s=e^{i\pi s}|z{|}^s$; accounting for the direction of traversal, the two real segments therefore sum to $(1+e^{i\pi s})I$, where $I$ is the desired positive-axis integral. The residue theorem givesSince $1+e^{i\pi s}=2e^{i\pi s/2}\cos (\pi s/2)$,The endpoint estimates also prove convergence. More generally the same calculation is the fractional-power integral from an upper-half-plane contour, valid for $-1<s<1$.