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A sufficient condition is that $U$ is a connected space. Suppose, towards a contradiction, that $|f|$ is the same constant $R$ throughout $U$. If $R=0$, then $f=0$, contradicting $f(a)\ne f(b)$.

If $R>0$, differentiate $u^2+v^2=R^2$ in $x$ and $y$, and use the Cauchy-Riemann equations from part (a). They giveThe coefficient matrix for $(u_x,v_x)$ has determinant $-(u^2+v^2)=-R^2\ne 0$. Hence $u_x=v_x=0$, so $f^{\prime }=0$ everywhere. A holomorphic function with zero complex derivative is locally constant: within a small disc, differentiate along a straight segment. Local constancy and connectedness imply constancy on $U$, again a contradiction.

Consequently,This is the constant-modulus holomorphic function principle. The extra hypothesis matters: on two disjoint open discs, define $f=1$ on one and $f=-1$ on the other. It is holomorphic and nonconstant, yet $|f|=1$ everywhere.

The function $h(z)=1/(z+1)$ is holomorphic on $U$ and hasIts imaginary part is exactly the prescribed $v$. Thus $f-h$ is a real-valued holomorphic function. The Cauchy-Riemann equations force both partial derivatives of its real part to vanish. Since $\mathbb{C}\setminus \{-1\}$ is a connected space, $f-h$ is a single real constant. All possibilities are thereforeConversely every such function has the stated imaginary part. The data determine $f$ only up to this real additive constant.

For a planar incompressible flow, a stream function satisfies $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ where the flow is irrotational; such a potential exists locally, and globally on a simply connected nonsingular region. The irrotational flow condition is $v_x-u_y=0$. Where both exist, the Cauchy-Riemann equations make the complex potential $w=\varphi +i\psi$ analytic with $w^{\prime }=u-iv$.

For the logarithmic potential, write $z-z_0=r{e}^{i\theta }$. ThenThe outward volume flux per unit span across a circle is $2\pi r{u}_r=m$. Hence it is a line source of strength $m$. The potential may use local logarithm branches; its velocity is single-valued away from the source.

By superposition, the full flow hasFor positive $m$ and $U$, the stagnation points areThey are real, one upstream of the source and one downstream of the sink. Away from the singularities, streamlines are level curves ofThe zero level through the two stagnation points contains an upper and a lower separating curve, symmetric about the real axis and about $x=1/2$. Together they bound a Rankine oval. Flow inside goes from source to sink; the external stream divides around this oval and tends to horizontal flow far away. This also specifies the sketch for every positive parameter ratio; the figure illustrates one ratio.

Write $a={\partial }_1{f}_1={\partial }_2{f}_2$ and $b={\partial }_2{f}_1=-{\partial }_1{f}_2$. The Cauchy-Riemann equations giveConsequentlyEquivalently, it is singular exactly when all four partial derivatives vanish. The sum of squares is over $\mathbb{R}$, so no cancellation is possible.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

The chain rule in polar coordinates gives $u_r=u_x\cos \theta +u_y\sin \theta$ and $u_{\theta }=r(-u_x\sin \theta +u_y\cos \theta )$, with analogous expressions for $v$. Substituting the Cauchy-Riemann equations givesThe restriction excludes the coordinate singularity at the origin.

Choose a continuous branch of a multivalued function of the complex logarithm on a simply connected region avoiding zero. ThenAlternatively $u_r=-e^{-\theta }\sin (\ln r)/r=v_{\theta }/r$ and $v_r=e^{-\theta }\cos (\ln r)/r=-u_{\theta }/r$, verifying the Cauchy-Riemann equations. The holomorphy is local to a chosen branch of a multivalued function: increasing $\theta$ by $2\pi$ multiplies $g$ by $e^{-2\pi }$, so this is not a single-valued holomorphic function on the whole punctured plane.

Conformality and analyticity. Use the orientation-preserving convention for a conformal map, with a continuously differentiable, nonzero differential. Its Jacobian matrix must be a positive scale times a rotation,Thus $u_x=v_y$, $u_y=-v_x$, the Cauchy-Riemann equations; continuous partial derivatives then imply that $u+iv$ is a holomorphic function. If angle preservation is allowed to reverse orientation, the statement needs this qualification: $z\mapsto \bar{z}$ preserves unsigned angles but is an anticonformal map and is not a holomorphic function.

The mapping. Put $a=\pi /L$. The displayed components combine to giveIts derivative does not vanish in the open half-strip $x>0$, $0<y<L$, so it is a conformal map there. Also $\mathrm{Im}w=\sinh (ax)\sin (ay)>0$. The lower edge maps to $[1,\infty )$, the upper edge to $(-\infty ,-1]$, and the vertical edge to $[-1,1]$. The open half-strip maps bijectively onto the upper half-plane. Indeed, $\cosh \zeta =\cosh \eta$ implies $\zeta =\pm \eta +2\pi im$; the half-strip allows only equality. Surjectivity follows by solving $t+t^{-1}=2w$ and choosing the root $t=e^{az}$ with $|t|>1$ and $0<\arg t<\pi$. The two corners are boundary critical points, not interior failures of conformality.

Transformed boundary values. A conformal map preserves Laplace's equation up to a positive scale. Thus the transformed Dirichlet problem is to find a real harmonic function $U(u,v)$ on $v>0$ withThe conflicting values at the original corners are interpreted as one-sided edge limits.

Logarithmic solution. Choose the complex logarithms of $w\pm 1$ with arguments between zero and $\pi$ in the upper half-plane. A linear combination $F(w)=\alpha \log (w+1)+\beta \log (w-1)+\gamma$ is a complex-valued harmonic function, since it is a holomorphic function; the real physical potential must be a real or imaginary part. In particular, setThenOn the three real boundary intervals, the argument difference is respectively $0,\pi ,0$, which gives the required values. This is the bounded harmonic solution, with $U\to V_0$ at infinity; uniqueness holds in this bounded class by the maximum principle. Without a boundedness or far-field condition, the edge data alone do not give uniqueness: one may add $C\sinh (\pi x/L)\sin (\pi y/L)$.

Back substitution. Sincethe required continuous argument difference is $2\arctan [\sin (ay)/\sinh (ax)]$ in the open half-strip. ConsequentlyAs $y\to 0,L$ with $x>0$, the arctangent tends to zero; as $x\to 0^{+}$ with $0<y<L$, it tends to $\pi /2$. These limits verify every open edge of the Dirichlet problem.

A complex derivative computed along the real and imaginary directions gives $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$. Equality yields the Cauchy-Riemann equationsThe gradients satisfy $\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=0$. Since a gradient is normal to a regular level set, the two families meet at right angles by orthogonality wherever $f^{\prime }\ne 0$. At a critical point the regular-curve argument fails: for $f=z^2$ both gradients vanish at zero, so the assertion requires this qualification.

Set $A=x^2-y^2$ and $B=2xy$. Direct differentiation givesTheir sum vanishes, proving that $u$ is a harmonic function. More generally the Cauchy-Riemann equations give $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$; the mixed partial derivatives commute because a holomorphic function is smooth. Since $e^{i{z}^2}=e^{-B}(\cos A+i\sin A)$, a harmonic conjugate and the resulting holomorphic function areThese expressions also verify the Cauchy-Riemann equations directly.

The chain rule in polar coordinates gives $u_r=u_x\cos \theta +u_y\sin \theta$ and $u_{\theta }=r(-u_x\sin \theta +u_y\cos \theta )$, with analogous expressions for $v$. Substituting the Cauchy-Riemann equations givesThe restriction excludes the coordinate singularity at the origin.

Choose a continuous branch of a multivalued function of the complex logarithm on a simply connected region avoiding zero. ThenAlternatively $u_r=-e^{-\theta }\sin (\ln r)/r=v_{\theta }/r$ and $v_r=e^{-\theta }\cos (\ln r)/r=-u_{\theta }/r$, verifying the Cauchy-Riemann equations. The holomorphy is local to a chosen branch of a multivalued function: increasing $\theta$ by $2\pi$ multiplies $g$ by $e^{-2\pi }$, so this is not a single-valued holomorphic function on the whole punctured plane.

Choose a continuous argument $\theta$ on a simply connected domain avoiding zero, and put $r=\sqrt{x^2+y^2}$. With $u=x\log r-y\theta$, direct differentiation givesHere ${\theta }_x=-y/r^2$, ${\theta }_y=x/r^2$; these identities specify the branch needed in the written arctangent. Integrating the Cauchy-Riemann equations givesThe literal principal $\arctan (y/x)$ works only separately on suitable half-planes. It cannot stand for a single continuous argument across the imaginary axis.

planar Brownian motion is recurrence of planar Brownian motion when, almost surely, it returns to every nonempty open neighborhood infinitely often, at arbitrarily large times. This does not mean that it hits each specified point; a prescribed point is polar for planar Brownian motion.

recurrence of planar Brownian motion and continuity imply that all the circle-hitting times $S_k,T_k$ in the question are finite. Starting on radius $ϵ$, it eventually visits the exterior of radius $2ϵ$ and must cross that circle; it later visits the interior of radius $ϵ$ and must cross back. Each duration ${\Delta }_k=T_k-S_k$ is strictly positive by path continuity, because the two radii are separated.

At $S_k$, the Strong Markov property restarts Brownian motion at its current point on the inner circle. rotational invariance of Brownian motion makes the law of the subsequent first exit time from the radius-$2ϵ$ disk independent of that starting angle. Its conditional distribution given the past is therefore the same fixed law at every cycle, proving that the ${\Delta }_k$ are independent and identically distributed.

To prove divergence of the conformal clock for any nonconstant entire $f$, choose a point $z_0$ with $f^{\prime }(z_0)\ne 0$. By continuity, there exist $\rho ,c>0$ such that $|f^{\prime }(z){|}^2\ge c$ throughout the closed disk $|z-z_0|\le 2\rho$. Use the same recurrence of planar Brownian motion circle cycles centered at $z_0$, with radii $\rho ,2\rho$. On each segment from the inner circle to the outer circle, the path remains in that disk, givingThe durations are positive iid variables. If their mean has not been computed, apply the strong law of large numbers to the bounded iid variables $\min ({\Delta }_k,1)$: their mean is strictly positive, so their sum, and hence $\sum {\Delta }_k$, diverges almost surely. ThereforeCentering the disks at a noncritical point avoids an unjustified positive lower bound on $|f^{\prime }|$ around the original origin.

Write $f=u+iv$. Harmonicity and Itô's formula make $u(Z_t),v(Z_t)$ continuous local martingales. The Cauchy-Riemann equations give equal brackets ${\int }_0^t|f^{\prime }(Z_s){|}^{2}ds$ and zero mutual bracket. Consequently Lévy's characterization of Brownian motion after time change of a continuous process makes $W$ a planar Brownian motion started at $f(Z_0)$, with no finite clock lifetime because the clock just proved is unbounded. The clock is strictly increasing: otherwise $f^{\prime }$ would vanish along a nonconstant path interval, contradicting its isolated zeros.

For the notation in the source, $|\nabla f{|}^2$ is interpreted as the conformal scale $|f^{\prime }{|}^2=|\nabla u{|}^2=|\nabla v{|}^2$. If it instead means the sum of the squared magnitudes of both complex coordinate derivatives, it is $2|f^{\prime }{|}^2$, and the resulting $W$ runs at half the usual Brownian speed. This convention does not change divergence of the integral. The mechanism is recurrent disk occupation makes an entire conformal clock infinite.

For $0<\alpha <2$, write $z=r{e}^{i\theta }$ with $0<\theta <\pi$. The chosen power map sends this to $r^{\alpha }{e}^{i\alpha \theta }$, soAt $\alpha =2$ this is the plane slit along the nonnegative real axis. For every $0<\alpha \le 2$, the map is conformal on the upper half-plane.

To check finiteness of the physical exit time, start a Brownian motion $B$ at $i$ and let $T$ be its real-axis exit. The time change of a continuous process in conformal invariance of planar Brownian motion isWe know $T<\infty$ almost surely. The Cauchy exit law from a Brownian half-plane assigns probability zero to $B_T=0$. On such a path, $B_s\ne 0$ throughout the compact interval $[0,T]$, because it is in the open upper half-plane before exit and has nonzero endpoint. Its modulus therefore has a positive minimum and finite maximum. The displayed integrand is bounded on that interval, proving $A_T<\infty$. The image Brownian motion exits $D$ at this finite changed time.

The endpoint $\alpha =0$ printed in the source is degenerate: $z^0=1$, so its image is the singleton $\{1\}$, not an open domain, and the map is not conformal. If exit from that set is defined for Brownian motion started at one, its exit infimum is zero. It cannot be included in a nondegenerate conformal-domain argument.

Here is the martingale argument for maximal comparisons, including the starting-value issue. For $P=f+ig$, the Cauchy-Riemann equations and Itô's formula show thatare orthogonal continuous local martingales with the same quadratic variation. Localize before large disks and before exit. The Burkholder-Davis-Gundy inequalities then givewith a constant depending only on $p$; Fatou's lemma removes the localization. These are centered martingales. For the uncentered processes this impliessince $|f(i^{\alpha })|\le \sup |f(Z_t)|$.

The source normalizes $P(0)=0$ at a boundary point, rather than the value of $g$ at the starting point. Consequently its displayed constant cannot be uniform in $\alpha$ with only $p$ dependence. To see this, take $P(z)=iz$ and $0<\alpha <\min (1/2,1/p)$. Then $f=-\mathrm{Im}z$ and $g=\mathrm{Re}z$. Inside this narrow wedge,Both $p$th moments are finite and the right one is positive; finiteness follows from the wedge radial-maximum tail of order $R^{-1/\alpha }$, justified below. The source inequality would force $C_p\ge {\cot }^p(\alpha \pi )$, unbounded as $\alpha \downarrow 0$. At $\alpha =0$, the immediate-exit interpretation gives the direct contradiction $1\le C_{p}0$ for the same polynomial.

For a fixed nondegenerate wedge, an uncentered inequality does hold with a constant allowed to depend on $\alpha$ as well as $p$. One rigorous outline for the polynomial class in this question is as follows. Separation of variables in a sector of radius $R$ gives, for exit through its circular arc, the expansionThus at the specified starting point the radial-maximum tail is comparable to $R^{-1/\alpha }$. Hitting any fixed interior subarc has that same power, with positive leading coefficient. If a nonzero polynomial has degree $d$, its real part has magnitude at least $c{R}^d$ on some fixed interior subarc for sufficiently large $R$. Therefore its maximal real-part $p$th moment is infinite whenever $dp\ge 1/\alpha$. In that case the desired extended-real inequality is automatic.

The remaining polynomials have $d<1/(\alpha p)$, so belong to a finite-dimensional vector space space. On that spaceis a genuine norm: if it is zero, Brownian motion's positive probability of reaching any interior neighborhood forces $\mathrm{Re}P=0$ throughout $D$, so analyticity makes $P$ a purely imaginary constant, and $P(0)=0$ makes it zero. The corresponding imaginary-part functional is bounded on this finite-dimensional vector space normed space; its finiteness follows from $dp<1/\alpha$ and the radial tail bound. equivalence of norms in finite dimensions givesThis establishes the fixed-domain interpretation and identifies exactly why the universal centered martingale inequality cannot silently be applied to the printed uncentered boundary normalization. It is the distinction in starting-value obstruction in wedge conjugate maximal inequalities.

The two-dimensional Itô formula gives the local martingale part of $u(X_t,Y_t)$ as ${\int }_0^t{u}_{x}dX+{\int }_0^t{u}_{y}dY$, with all derivatives evaluated along $(X_s,Y_s)$. Independent Brownian coordinates have brackets $[X{]}_t=[Y{]}_t=t$ and $[X,Y{]}_t=0$. Hence the quadratic variation isSimilarly, the quadratic covariation with $v(X,Y)$ isSufficient pointwise conditions are equal gradient lengths and perpendicular gradients:The Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$ imply both. These are the quadratic covariations of an analytic Brownian image; harmonicity is needed to remove the drift, but the two bracket identities themselves follow just from the displayed derivative relations.

Conformal invariance of planar Brownian motion means invariance after changing the clock, not preservation of the original speed. If $f:D\to D^{\prime }$ is a conformal bijection and $Z$ is planar Brownian motion stopped on exiting $D$, putThen $f(Z_t)$ has the law of planar Brownian motion started at $f(Z_0)$ and run at time $A_t$, up to the corresponding lifetime. Equivalently, composing with the inverse clock produces ordinary Brownian motion in $D^{\prime }$.

Indeed, the real and imaginary parts of a holomorphic function are harmonic, so Itô's formula gives zero drift. The Cauchy-Riemann equations give equal coordinate brackets $A_t$ and zero cross bracket. The Lévy characterization of multidimensional Brownian motion after time change identifies the image. In particular the harmonic measure of boundary arcs is carried to their conformal images.

Assume that a nonconstant polynomial $f$ has no zero. Then $g=1/f$ is entire. The leading polynomial term makes $|f(z)|\to \infty$ as $|z|\to \infty$, so $g(z)\to 0$ there. On any remaining compact disk, continuity and absence of zeros bound $1/|f|$. Consequently $g$ is bounded on the entire plane.

Part (c)(i) makes $u(Z_t)=\mathrm{Re}g(Z_t)$ a bounded local martingale, hence a true bounded martingale. The martingale convergence theorem says it converges almost surely as $t\to \infty$.

The harmonic function $u$ cannot be constant. Otherwise the Cauchy-Riemann equations would make both derivatives of $\mathrm{Im}g$ zero, so $g$ and then $f$ would be constant. Choose points $z_1,z_2$ with $u(z_1)<u(z_2)$. Continuity supplies two disjoint disks on which the ranges of $u$ are separated by a positive gap. By part (b), the Brownian path visits these disks alternately at times tending to infinity. The corresponding subsequences of $u(Z_t)$ then remain separated, contradicting its almost sure convergence.

The zero-free assumption is impossible. Hence
Every nonconstant complex polynomial has a complex root.
This proves the fundamental theorem of algebra using recurrence and martingale convergence, without invoking Liouville's theorem as a prior algebraic shortcut.

If $f:D\to D^{\prime }$ is a conformal map, the image of planar Brownian motion in $D$ is Brownian motion in $D^{\prime }$ after the increasing clock changeMore precisely, $f(Z_{A^{-1}(u)})$ is planar Brownian motion, starting at $f(Z_0)$ and run up to the relevant exit time. The Itô formula and the Cauchy-Riemann equations give image-coordinate local martingales with common bracket $A$ and zero cross bracket; the Lévy characterization of multidimensional Brownian motion identifies the time-changed process. In particular, conformal invariance of planar Brownian motion transports exit locations and harmonic measure, while changing traversal speed.