Question bank — Complex analysis — analytic functions, Cauchy-Riemann equations

False — real differentiability only asks that a linear approximation exist; complex differentiability additionally forces that linear map to be a rotation-plus-scaling, i.e. the Jacobian must have the special form $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a=u_x=v_y$ and $b=v_x=-u_y$ (the real and imaginary parts of $f^{\prime }(z)=a+ib$), which is exactly CR.

True — the product rule for complex derivatives holds identically to the real case, so the complex limit $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$ exists at every point of the region; analyticity is closed under products.

True — the quotient is complex-differentiable wherever the denominator is nonzero, and we assumed $f\ne 0$, so no singularity appears; $(1/f{)}^{\prime }=-f^{\prime }/f^2$.

False — $e^z$, $\sin z$, $\cos z$ are entire but not polynomials; "entire" only means the complex derivative exists everywhere, which power series with infinite radius of convergence also satisfy (see Power series and radius of convergence).

False — both being harmonic is necessary but not sufficient; they must be harmonic conjugates (linked by CR: $u_x=v_y$, $u_y=-v_x$). Two unrelated harmonic functions glued together generally break CR.

True — CR gives $v_x=-u_y$ and $v_y=u_x$, so $v$ is recovered by integrating these known partials; the only freedom is an additive real constant of integration.

False — conformality requires the complex derivative $f^{\prime }(z_0)\ne 0$; where it vanishes the local rotation-scaling collapses and angles can be multiplied (e.g. $z^2$ at $0$ doubles angles). See Conformal mappings.

True — $u$ constant gives $u_x=u_y=0$; CR then forces $v_y=0$ and $v_x=0$, so $v$ is constant too, hence $f$ is constant.

True — constant modulus means $u^2+v^2$ is constant; differentiating and applying CR forces all partials to vanish (unless $|f|=0$, giving $f\equiv 0$), so $f$ is constant.

False — CR at a point is necessary but not sufficient without continuity of partials; you can construct functions satisfying CR at $0$ yet failing to be differentiable there because the directional limits still disagree along diagonal approaches.

There is no "half analytic" — analyticity needs both CR equations simultaneously; the first fails ($u_x=1\ne -1=v_y$), so $f=\bar{z}$ is nowhere analytic, full stop.

The minus sign is missing: the correct pair is $u_y=-v_x$. Recall $\Delta z=\Delta x+i\Delta y$; approaching along the imaginary axis makes $\Delta z=i\Delta y$, and dividing by it uses $1/i=-i$ — that factor is exactly where the sign is born. Symmetric-looking equations describe a reflection, not a rotation.

Analyticity demands differentiability on an open set, and $\{0\}$ is a single point with empty interior — no disk fits inside it. $f$ is merely complex-differentiable at $0$; there is no neighborhood where CR holds, so it is analytic nowhere.

Smoothness of $u,v$ as real functions is irrelevant on its own; you must feed them through CR. Here $v=0$ forces $u_x=u_y=0$, true only at the origin, so smoothness never rescues analyticity.

They pin down $v$'s partials, hence $v$ up to an additive real constant — integration always leaves a constant. And on a non-simply-connected region even existence can fail.

They are two formulas for the same single complex derivative — equating them is precisely how CR arises. When CR holds, both expressions are numerically identical.

A complex equation packs a real and an imaginary component; setting the real-axis derivative $u_x+i{v}_x$ equal to the imaginary-axis derivative $v_y-i{u}_y$ forces real parts to match ($u_x=v_y$) and imaginary parts to match ($u_y=-v_x$) — two constraints from one complex demand.

Because the local action is "multiply by the single complex number $f^{\prime }(z)=a+ib$" (with $a=u_x=v_y$, $b=v_x$), and multiplication by $a+ib$ is, in $2\times 2$ real matrix form, exactly that rotation-scaling matrix — CR is just this matrix condition written out. See Jacobian and the multivariable chain rule.

Analytic functions turn out to be infinitely differentiable, so $u,v$ are $C^2$ (their second partials exist and are continuous); that continuity is exactly what licenses equality of mixed partials $v_{yx}=v_{xy}$ (Clairaut's theorem). Cross-differentiating CR gives $u_{xx}=v_{yx}$ and $u_{yy}=-v_{xy}$, and adding these cancels the $v$ terms to leave $u_{xx}+u_{yy}=0$. See Harmonic functions and Laplace's equation.

Reflections reverse orientation — their Jacobian has determinant $-(a^2+b^2)<0$, which can never equal the $+(a^2+b^2)$ of a rotation-scaling; CR encodes orientation-preservation, so mirrors break it everywhere.

Existence of partials only controls the two axis directions; continuity is what guarantees the linear approximation is valid along every direction, closing the gap between "CR holds" and "the full directional limit of $\frac{f(z+\Delta z)-f(z)}{\Delta z}$ exists and agrees."

For a generic real map different directions of $\Delta z$ give different rates (the directional derivative varies); only when the Jacobian is a pure rotation-scaling does every direction get scaled by the same factor and rotated by the same angle — that single common factor is $f^{\prime }(z)$.

Multiplication by $f^{\prime }(z)=r{e}^{i\theta }$ rotates every local vector by the same $\theta$ and scales by the same $r$; equal rotation of all directions leaves the angle between any two curves unchanged. If $f^{\prime }=0$ the local picture degenerates and this fails.

No — analyticity requires differentiability throughout an open set; a lone point (or even a whole curve) has empty interior, so there is no region on which $f$ qualifies.

The function is still analytic there (complex-differentiable in a neighborhood), but it is not conformal at that point — angles get multiplied (for $z^2$, doubled), because the rotation-scaling degenerates when the scale factor is $0$.

No — $v\equiv 0$ gives $v_x=v_y=0$, so CR forces $u_x=u_y=0$, making $u$ constant; a real-valued analytic function on a region is necessarily constant.

Yes in spirit — complex differentiability is a genuinely 2D limit (the step $\Delta z$ must be free to point every way), so you cannot even state it without room to approach from all directions; on a mere curve there is no notion of analyticity, only of a restricted function.

No — ${\bar{z}}^2=x^2-y^2-2ixy$ gives $u=x^2-y^2$, $v=-2xy$, so $u_x=2x$, $v_y=-2x$; CR fails except where $x=0$, i.e. only on a line (no open set), so nowhere analytic.

Complex differentiability can fail at that discontinuity even though CR holds — the classic counterexample has directional limits that disagree along diagonals, so continuity of partials is genuinely doing work in the sufficiency theorem.

$z^2$ is entire (holomorphic in $z$ alone), while $z\bar{z}$ depends on $\bar{z}$, and any explicit $\bar{z}$-dependence violates CR off a measure-zero set — analytic functions are exactly those that "do not see $\bar{z}$."

Only if they differ by a purely imaginary constant $ic$ — CR fixes $v$'s partials from $u$, leaving just an additive real constant in $v$; so the functions are $f$ and $f+ic$.