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At zero, $[f(h)-f(0)]/h\to 1$, so the complex derivative exists. At a nonzero point, write $r=|z|$. The Cauchy-Riemann equations for $x/(1+r)$ and $y/(1+r)$ force $x^2=y^2$ and $xy=0$, impossible when $r>0$. Hence the function is complex differentiable only at zero and is holomorphic on no nonempty open subset. A complex derivative at one point gives no neighborhood of complex differentiability.

Write $f=u+iv$ and $z=x+iy$. The complex derivative computed along a real increment is $f^{\prime }=u_x+i{v}_x$, while along an imaginary increment it is $f^{\prime }=v_y-i{u}_y$. Equality givesThese are the Cauchy–Riemann equations at every point where $f$ is holomorphic. The directional calculations also establish the existence of those partial derivatives. More explicitly, $f(z+h)-f(z)=f^{\prime }(z)h+o(|h|)$ gives both real-direction formulas directly. If a converse is desired, continuous first partial derivatives together with these equations make the real differential multiplication by $u_x+i{v}_x$, hence give complex differentiability; the necessity above is what is requested.

For planar incompressible flow and irrotational flow, combine a velocity potential and a streamfunction into the complex potential of a two-dimensional potential flow:The chosen conventions giveThese are the Cauchy-Riemann equations, so $w$ is a holomorphic function wherever the flow is regular and a consistent local potential is chosen. Its complex derivative is the complex velocityA spatially uniform flow has $w=(U-iV)z+$ a constant. The following contour integral identities also apply to nonuniform potential flow, including the specified line vortex. A complex potential of a two-dimensional potential flow may be multivalued around a source or vortex even though its derivative is single-valued.

Compute the complex derivative along horizontal and vertical increments. Along the real direction, $f^{\prime }=u_x+i{v}_x$; along the imaginary direction, $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equationsThe real and imaginary parts of a holomorphic function are smooth. Equality of their mixed partial derivatives then gives $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$, proving that both are harmonic functions.

For the first real part, $u_x=2x$ and $u_y=-2y$. The Cauchy-Riemann equations require $v_y=2x$ and $v_x=2y$, so its harmonic conjugate is $v=2xy+C$. ThereforeFor the second real part, $u=x/(x^2+y^2)$ is the real part of $1/z$. Its harmonic conjugate is $-y/(x^2+y^2)+C$, givingThese are the most general answers on a connected open domain: the difference of two candidates has zero real part, so the Cauchy-Riemann equations make its imaginary part constant. The second function cannot extend holomorphically through zero.

At a point of complex differentiability, the complex derivative is independent of the direction of approach. Taking increments along the real axis gives $f^{\prime }=u_x+i{v}_x$; taking imaginary increments gives $f^{\prime }=v_y-i{u}_y$. Equality of their real parts and imaginary parts proves the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$.

For the specified real part, observe that $z^2{e}^z$ has exactly that real part. Equivalently, integrating $v_y=u_x$ and checking $v_x=-u_y$ givesThusThe freedom is only a constant imaginary part: the difference of two such holomorphic functions has zero real part, so its Cauchy-Riemann equations make it constant on each connected domain.

At $z=x+iy$, compute the complex derivative first along real increments and then along imaginary increments. Holomorphy givesEquating real and imaginary parts proves the Cauchy-Riemann equations:The partial derivatives exist because these directional difference quotients are limits of the holomorphic derivative. This argument proves the necessary equations; it does not claim that their pointwise existence alone is sufficient for holomorphy.

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.

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.

At every $z=x+iy\in U$, the Cauchy-Riemann equations areTo derive them from complex differentiability, approach $z$ along real increments $h$. The complex derivative givesApproaching along imaginary increments $ih$ instead givesThese limits exist because $f$ is holomorphic. Equating their real and imaginary parts proves the equations, together with $f^{\prime }=u_x+i{v}_x=v_y-i{u}_y$.

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.

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.

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.