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For a holomorphic function, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. ConsequentlyA gradient is normal to its level set, so these identities give orthogonal level curves wherever the gradients are nonzero. At a critical point where $f^{\prime }=0$, both gradients vanish and a regular tangent to a level set need not exist; the geometric orthogonality statement is restricted to regular intersections.

For real-direction increments, complex differentiability gives $f^{\prime }=u_x+i{v}_x$. For imaginary-direction increments it gives $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equating these expressions gives the Cauchy-Riemann equationsA holomorphic function is smooth, so mixed partial derivatives commute. Differentiating these equations givesThus both real and imaginary parts are harmonic functions:

For a holomorphic function with $z=r{e}^{i\theta }$, the chain rule givesWriting $f=u+iv$ and equating real parts and imaginary parts,For $r>0$, division by $r$ gives the Cauchy-Riemann equations in polar coordinates:The restriction $r>0$ reflects the coordinate singularity at the origin; the Cartesian equations have no such restriction.

For a holomorphic function, the derivative supplied by complex differentiability must have the same value along real and imaginary increments. Those two computations giveEquating real and imaginary parts proves the Cauchy-Riemann equations:Conversely, if the first partial derivatives are continuous near a point and obey these equations there, real differentiability giveswhich proves complex differentiability under this regularity assumption.

The gradients of the real and imaginary parts have dot productand $|\nabla u{|}^2=|\nabla v{|}^2=|f^{\prime }(z){|}^2$. At every point with $f^{\prime }(z)\ne 0$, the two level sets are regular curves, their gradients are nonzero normal vectors, and their tangent directions are therefore perpendicular.

The printed phrase "at every point" needs this regular-point qualification. At a critical point with $f^{\prime }(z)=0$, the level sets may cross or fail to have a unique tangent. For example, $f(z)=z^2$ gives $u=x^2-y^2$ and $v=2xy$: at the origin the zero level sets are respectively the diagonals and the coordinate axes. Thus the valid conclusion is $\boxed{\text{orthogonality at every regular intersection, where }{f}^{\prime }(z)\ne 0}$.

Fix any bounded disk avoiding the integers, and choose $R$ with $|z|\le R$ on it. For sufficiently large $|n|$, say $|n|\ge 2R$ and $|n|\ge 1$,The convergent numerical series ${\sum }_{n\ne 0}{n}^{-2}$ and the Weierstrass M-test give absolute and uniform convergence of the tails. The finitely many remaining summands are holomorphic away from their integer poles. Thus the partial sums converge locally uniformly on $\mathbb{C}\setminus \mathbb{Z}$.

By part (a)(iii), the sum is a holomorphic function on $\mathbb{C}\setminus \mathbb{Z}$. The term $n=0$ is $-1/z^2$ and is harmless on this domain; it must not be omitted.

Use the generalized binomial theorem with exponent $-1/2$, choosing the analytic branch of a square root that equals $1$ at $z=0$. For $|z|<1$,The signs cancel in each binomial coefficient. For $n\ge 1$,The same expression equals $1$ for $n=0$, using the empty-product convention. Consequently the power series isIts coefficients are the normalized central binomial coefficients. As a convergence check, the ratio of successive absolute terms tends to $|z|$, so the ratio test gives absolute convergence inside the unit disc. Applying the Cauchy coefficient formula to the holomorphic function $(1-z{)}^{-1/2}$ now immediately yieldsThis value is independent of the chosen radius because the contour integral extracts the same coefficient on every such circle.

Choose $0<\rho <r$ and traverse the circle $|z|=\rho$ counterclockwise, so it has positive orientation of a contour. The Cauchy coefficient formula gives, for every nonnegative integer $n$,The smaller radius matters: the holomorphic function is only known on $|z|<r$, so its values on $|z|=r$ need not exist. To see directly why this contour integral selects $c_n$, the power series converges uniformly on $|z|=\rho$, allowing uniform convergence and contour integration to justify integrating term by term. The contour integral of $z^{k-n-1}$ is zero for $k\ne n$ and $2\pi i$ for $k=n$. Thus all other terms disappear.

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.

One standard form of Cauchy's residue theorem is the following. Let $\Gamma$ be a positively oriented contour that is a simple closed curve, and let $f$ be a holomorphic function on an open set containing $\Gamma$ and its interior except for finitely many interior isolated singularities $z_1,\dots ,z_m$. There are no singularities on $\Gamma$. ThenThe residue $\mathrm{Res}(f,z_j)$ is the coefficient of $(z-z_j{)}^{-1}$ in the Laurent series at $z_j$. Each interior isolated singularity contributes once because the winding number of this positively oriented contour about each such point is one. Reversing the orientation of a contour reverses the sign. This form of Cauchy's residue theorem applies to the rectangle below.

The three contacts in the strip. Work with the external-contact arrangement drawn in the original diagram: both disks lie between the two parallel lines and touch opposite boundaries. A rigid motion puts those boundaries at $x=0$ and $x=d$, with $d>0$, and makes $P=(0,0)$. Write $R=(d,h)$, and let the positive radii be $r_1,r_2$. Because a tangent to a circle is perpendicular to the corresponding radius, the centres areAt the external tangency, the two radial segments form the segment $O_1{O}_2$, with $|O_{1}Q|=r_1$ and $|Q{O}_2|=r_2$. Hence the internal division formula givesThusThe coefficient is strictly between zero and one, so $Q$ lies on the actual line segment $PR$. This proves collinearity for tangent circles between parallel lines. We will also use the fact that $PQ$ and $QR$ are chords of the respective circles; the entire line segment $PR$ therefore lies in the union of their closed disks.

The four contacts. Label the successive contact points $T_{12},T_{23},T_{34},T_{41}$, with $T_{jk}$ belonging to $C_j$ and $C_k$. The original diagram has four positive-radius circles with disjoint interiors, successive external tangencies, and distinct contact points. Set $T=T_{41}$ and apply the reciprocal Möbius transformationBy part (b), the two circles $C_1,C_4$ through $T$ become straight lines ${\ell }_1,{\ell }_4$. These straight lines cannot meet at a finite point, since their inverse images would give a second intersection of $C_1,C_4$. They are therefore distinct parallel lines. The other two circles avoid $T$, so their images $D_2,D_3$ are ordinary circles missing zero.

The three remaining contacts retain their tangencies. Indeed, away from $T$ the map and its inverse are holomorphic functions with nonzero derivatives, hence are local conformal maps. In particular they preserve a shared tangent line. Equivalently, bijectivity preserves the unique intersection of two generalized circles, which characterizes their tangency here.

The finite closed disks bounded by $D_2,D_3$ lie between ${\ell }_1,{\ell }_4$, have external tangency to each other, and touch opposite boundaries at the images of $T_{12}$ and $T_{34}$. To track the sides as well as the boundaries, a closed disk avoiding $T$ maps to the bounded closed disk of its image circle, while each disk with $T$ on its boundary maps to a closed half-plane. The two latter half-planes have disjoint interiors, so their complement is the strip containing zero. Disjointness of the original interiors puts $D_2,D_3$ inside that strip. Neither image closed disk contains zero, since the reciprocal Möbius transformation sends only infinity to zero.

The first part now makes the three finite transformed contact points collinear, on a straight line $\ell$. It is essential that $\ell$ misses zero. If it passed through zero, its segment between ${\ell }_1$ and ${\ell }_4$ would contain zero, since zero lies strictly within the strip. But the first part shows that this whole line segment is covered by the two image closed disks, neither of which contains zero. This contradiction proves $0\notin \ell$.

The inverse image of a straight line missing zero is a genuine circle through $T$, by the same coefficient transformation in part (b), or by applying the involution $u\mapsto 1/u$ in reverse and translating back. This circle contains $T_{12},T_{23},T_{34}$, while $T$ is the inverse image of the point at infinity on $\ell$. ThereforeThis is the four contact points of a circle chain result; the four contacts form a cyclic quadrilateral.

The pictured arrangement matters. Arbitrary mixtures of internal tangency and external tangency can instead give four distinct collinear contacts: take four circles whose diameter endpoints on the real axis are respectively $[0,3]$, $[0,1]$, $[1,2]$, and $[2,3]$. Their successive contacts are $0,1,2,3$, so no ordinary circle contains all four. The proof above establishes the requested genuine-circle conclusion for the external-contact arrangement actually drawn, rather than silently treating every generalized circle as an ordinary circle.

Let $F(z)=\overline{\sinh z}$. More generally, for a holomorphic function $h$, the increment expansion at $z_0$ givesAlong real increments $w$ the first factor $\overline{w}/w$ is $1$, while along purely imaginary increments it is $-1$. Thus complex differentiability of $\overline{h}$ requires $h^{\prime }(z_0)=0$. Conversely, if $h^{\prime }(z_0)=0$, the displayed expression tends to zero for arbitrary complex increments, so the complex derivative exists and equals zero.

For $h=\sinh$, this condition is $\cosh z_0=0$, equivalent to $e^{2z_0}=-1$. ConsequentlyThese points form a discrete subset of the complex plane and contain no nonempty open set. Every neighbourhood of each one contains points where $F$ is not complex differentiable. Therefore $F$ is holomorphic on no nonempty open set, even though it has complex derivatives at the specified points.

The identity theorem states that if two holomorphic functions on a connected open set agree at a sequence of distinct points with an accumulation point inside that set, then they agree everywhere on it. Equivalently, the zeros of a holomorphic function which is not identically zero cannot accumulate inside its domain.

Because $f,g$ never vanish, their logarithmic derivatives are holomorphic. The holomorphic functionvanishes at every $a_n$. Since the distinct $a_n$ converge to $a\in U$, the identity theorem gives $H\equiv 0$. Applying the quotient rule now givesA holomorphic function with zero derivative on a domain is constant. One can see this locally from its Taylor series, then propagate the constant using connectedness. Hence $f/g=c$ on $U$, with $c\ne 0$, andThe conclusion can fail if the accumulation occurs outside $U$. Take the punctured complex plane $U=\mathbb{C}\setminus \{0\}$ andBoth functions are nowhere vanishing and holomorphic on $U$, and $a_n\to 0\notin U$. The chain rule givesso these logarithmic derivatives agree at every $a_n$. However, $f$ is not constant: for example $f(1/\pi )=e^{-1}$ and $f(1/(2\pi ))=e$. Therefore no constant $c$ gives $f=cg$.

On the slit domain $D$, choose the complex argument $\theta ={\mathrm{Arg}}_{(0,2\pi )}z$. The resulting holomorphic logarithm is $L(z)=\log |z|+i\theta$, with $L^{\prime }(z)=1/z$. DefineIt is holomorphic as a composition of holomorphic functions, and $f(z{)}^2=e^{L(z)}=z$. At $z=-1$ the chosen argument is $\pi$, so $f(-1)=i$.

Suppose a holomorphic square root $g$ existed on the whole punctured complex plane. It would be nonzero, and differentiating $g^2=z$ would give $2g{g}^{\prime }=1$. There are several ways to express the resulting obstruction; a direct one uses the unit circle. SetThe chain rule and $g^{\prime }(z)=g(z)/(2z)$ imply $q^{\prime }(t)=0$, so $q$ is constant. ButThese cannot agree because $g(1{)}^2=1$. Thus no such global holomorphic square root exists. Going once around zero changes the sign of the locally chosen root.

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.

Use the same principal square root $S(z)=\sqrt{(z-a)/(z-b)}$, which is holomorphic on $A$ because its argument avoids the nonpositive real axis. Thenis a holomorphic function satisfyingThus it is a holomorphic branch of the requested square root. This construction does not require a global holomorphic logarithm of $z-a$ or $z-b$ separately. Its sign is fixed by $S(z)\to 1$ at infinity, so $g(z)/z\to 1$. The negative of this function is the other holomorphic branch; any two branches have a continuous quotient taking values in $\{1,-1\}$, which is constant because $A$ is connected.

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.

The identity theorem states: if $V\subseteq \mathbb{C}$ is a nonempty connected open set and $p,q$ are holomorphic functions on $V$, and their equality set has a limit point in $V$, thenThe interior location of the limit point and connectedness are essential hypotheses.

For the proof put $h=p-q$. At any $z_0\in V$, its convergent Taylor series either has every coefficient zero, making $h$ zero on a neighbourhood, or has a first nonzero coefficient, of degree $m\ge 0$. In the latter case,where $k$ is holomorphic near $z_0$. By continuity, $k$ is nonzero on a smaller disc. Thus any zero at $z_0$ is isolated unless $h$ vanishes on a neighbourhood. An accumulation of distinct zeros at the assumed interior limit point therefore forces the neighbourhood alternative.

Let $E$ be the set of points of $V$ with a neighbourhood on which $h=0$. We have just proved $E\ne \varnothing$, and $E$ is an open set. It is also a closed set relative to $V$: a point of its closure outside $E$ would be approached by distinct points of $E$, hence by zeros, and the same Taylor series argument would place it in $E$. Since $V$ is connected, $E=V$. Thus $h=0$ on $V$, proving the theorem.

A holomorphic function has a zero of order $n$ at $a$ when it is holomorphic on a neighbourhood of $a$ and factors asEquivalently, $f(a)=f^{\prime }(a)=\cdots =f^{(n-1)}(a)=0$ and $f^{(n)}(a)\ne 0$. The order of a zero of a holomorphic function is the degree of the first nonzero term of its Taylor series.

An isolated singularity at $a$ is a removable singularity when there is a holomorphic function $F$ on a full disc $B(a,r)$ agreeing with $f$ away from $a$. Equivalently,The Riemann removable singularity theorem also characterizes this by boundedness of $f$ on a sufficiently small punctured disc. A preassigned incorrect value at $a$ is replaced by $L$.