4.7 Cauchy - Riemann Equations
Theorem 4.41. Suppose \(f(z) = u(x,y) + iv(x,y)\) is differentiable at a point \(z = x + iy\). Then at \(z\) the first order partial derivatives of \(u\) and \(v\) exist and satisfy the Cauchy-Riemann equations \[\frac {\partial u}{\partial x} = \frac {\partial v}{\partial y}\quad \text {and}\quad \frac {\partial u}{\partial y} = -\frac {\partial v}{\partial x}\]
Proof. By hypothesis the limit \[f'(z)=\lim _{\Delta z\rightarrow 0}\frac {f(z+\Delta z)-f(z)}{\Delta z}\] exists. A limit in the plane cannot depend on the direction of approach, and the proof is simply to compute it along two directions and equate.
Along the real axis
Take \(\Delta z=\Delta x\) real: \[f'(z)=\lim _{\Delta x\rightarrow 0} \frac {\big [u(x+\Delta x,y)-u(x,y)\big ]+i\big [v(x+\Delta x,y)-v(x,y)\big ]}{\Delta x} =\frac {\partial u}{\partial x}+i\frac {\partial v}{\partial x},\] so in particular both \(x\) partials exist.
Along the imaginary axis
Take \(\Delta z=i\Delta y\) and use \(\frac {1}{i}=-i\): \[f'(z)=\lim _{\Delta y\rightarrow 0} \frac {\big [u(x,y+\Delta y)-u(x,y)\big ]+i\big [v(x,y+\Delta y)-v(x,y)\big ]}{i\Delta y} =\frac {\partial v}{\partial y}-i\frac {\partial u}{\partial y}.\]
Equate
These are the same complex number, so real and imaginary parts agree: \[\frac {\partial u}{\partial x}=\frac {\partial v}{\partial y},\qquad \frac {\partial v}{\partial x}=-\frac {\partial u}{\partial y}.\] □
Remark 4.42. The equations are necessary but not sufficient: satisfying them at a point does not by itself make \(f\) differentiable there. Continuity of the partial derivatives closes the gap, which is the next theorem.
Example 4.43. The polynomial function \(f(z) = z^2 + z\) is analytic for all \(z\) and can be written as
\[f(z) = (x^2 - y^2 + x ) + i(2xy + y)\]
Thus \(u(x,y) = x^2 - y^2 + x\) and \(v(x,y) = 2xy + y\). For any point \((x,y)\) in the complex plane.
We see that the Cauchy-Riemann equations are satisfied
\[\frac {\partial u}{\partial x} = 2x + 1 = \frac {\partial v}{\partial y}\quad \text {and}\quad \frac {\partial u}{\partial y} = -2y = - \frac {\partial v}{\partial x}\]
Criterion for Non-Analyticity
If the Cauchy-Riemann equations are not satisfied at every point \(z\) in a domain \(D\), then the function \(f(z) = u(x,y) + iv(x,y)\)
cannot be analytic in \(D\).
Example 4.44. Show that the complex function \(f(z) = 2x^2 + y+ i(y^2 x)\) is not analytic at any point.
Solution
Here, we identify \(u(x,y) = 2x^2 + y\) and \(v(x,y) = y^2 - x\) from
\[\frac {\partial u}{\partial x} = 4x\quad \text {and} \quad \frac {\partial v}{\partial y } = 2y\]
\[\frac {\partial u}{\partial y}= 1 \quad \text {and} \quad \frac {\partial v}{\partial x} = 1\]
The equality \(\frac {\partial u}{\partial x} = \frac {\partial v}{\partial y}\) is satisfied only on the line \(y = 2x\). However, for any point \(z\) on the line about there is
no neighbourhood or open disc about \(z\) in which \(f\) is differentiable at every neighbourhood. We
conclude that \(f\) is nowhere analytic.
Example 4.45. \(f(x + iy) = x - iy = \overline {z}, \quad \forall x + iy \in \mathbb {C}\) \[u = \operatorname {Re}(f) = x\quad ,\quad v = \operatorname {Im}(f) = -y\]
\[\frac {\partial u}{\partial x} = 1\,,\quad \frac {\partial v}{\partial x} = 0\,,\quad \frac {\partial u}{\partial y }= 0\,,\quad \frac {\partial v}{\partial y} = -1\]
\(\frac {\partial u}{\partial x}\neq \frac {\partial v}{\partial y}\) at any point \((x,y)\). So \(f(z) = \overline {z}\) is not differentiable at any point in the complex plane.
Remark 4.46. Cauchy-Riemann equations do not ensure analyticity of a function \(f(z) = u(x,y) + iv(x,y)\) at a point \(z = x + iy\) . It is possible for Cauchy-Riemann equations to be satisfied at \(z\) yet \(f(z)\) may not be differentiable at \(z\).
Theorem 4.47 (Criterion for Analyticity). Suppose the real function \(u(x,y)\) and \(v(x,y)\) are continuous and have continuous first order partial derivatives in the domain \(D\). If \(u\) and \(v\) satisfy the Cauchy-Riemann equations at all points in \(D\), then the complex function \(f(z) = u(x,y) + iv(x,y)\) is analytic in \(D\).
Proof. Fix \(z_0=x_0+iy_0\) in \(D\) and put \(\Delta z=\Delta x+i\Delta y\). Because \(u\) and \(v\) have continuous first order partials near \(z_0\), each is differentiable there in the real sense: \[\Delta u=u_x\Delta x+u_y\Delta y+\varepsilon _1\left |\Delta z\right |,\qquad \Delta v=v_x\Delta x+v_y\Delta y+\varepsilon _2\left |\Delta z\right |,\] with \(\varepsilon _1,\varepsilon _2\rightarrow 0\) as \(\Delta z\rightarrow 0\). This is precisely where continuity of the partials is used; without it the error terms need not vanish.
Now eliminate the \(y\) derivatives using \(u_y=-v_x\) and \(v_y=u_x\): \begin {align*} \Delta f&=\Delta u+i\Delta v\\ &=\big (u_x\Delta x-v_x\Delta y\big )+i\big (v_x\Delta x+u_x\Delta y\big ) +\big (\varepsilon _1+i\varepsilon _2\big )\left |\Delta z\right |\\ &=\big (u_x+iv_x\big )\big (\Delta x+i\Delta y\big ) +\big (\varepsilon _1+i\varepsilon _2\big )\left |\Delta z\right | , \end {align*}
the middle step being the observation that the bracket factors as a complex product — which is exactly what the Cauchy–Riemann equations buy. Dividing by \(\Delta z\) and letting \(\Delta z\rightarrow 0\), the error term dies because \(\left |\Delta z\right |/\Delta z\) has modulus \(1\), leaving \[f'(z_0)=u_x+iv_x .\] So \(f\) is differentiable at \(z_0\), and as \(z_0\) was arbitrary, analytic in \(D\). □
Example 4.48. For the function \(f(z) = \frac {x}{x^2 + y^2} - \frac {iy}{x^2 + y^2}\) the real functions \(u(x,y) = \frac {x}{x^2 + y^2}\) and
\(v(x,y) = \frac {-y}{x^2 + y^2}\) are continuous except at the point where \(x^2 + y^2 = 0\), that is \(z = 0\).
Moreover, the first order partial derivatives
\[\frac {\partial u}{\partial x}= \frac {x^2 - y^2}{(x^2 + y^2)^2}\quad \text {and}\quad \frac {\partial u}{\partial y} = \frac {-2xy}{(x^2 + y^2)^2}\]
\[\frac {\partial v}{\partial x} = \frac {2xy}{(x^2 + y^2)^2}\quad \text {and}\quad \frac {\partial v}{\partial y} = \frac {y^2 - x^2}{(x^2 + y^2)^2}\]
are continuous except at \(z = 0\). Finally, we see that
\[\frac {\partial u}{\partial x} = \frac {y^2 - x^2}{(x^2 + y^2)^2} = \frac {\partial v}{\partial y}\quad \text {and}\quad \frac {\partial u}{\partial y} = \frac {-2xy}{(x^2 + y^2)^2} = \frac {-\partial v}{\partial x}\]
This means that the Cauchy-Riemann equations are satisfied except at \(z = 0\).
We conclude that \(f\) is analytic in any domain \(D\) not containing \(z = 0\).
Remark 4.49. If the real functions \(u(x,y)\) and \(v(x,y)\) are continuous and have first order partial derivatives in some neighbourhood of a point \(z\), and if \(u\) and \(v\) satisfy the Cauchy-Riemann equations at \(z\), then the complex function \(f(z) = u(x,y) + iv(x,y)\) is differentiable at \(z\) and \(f'(z)\) is \[\boxed {f'(z) = \frac {\partial u}{\partial x} + i\frac {\partial v}{\partial x} = \frac {\partial v}{\partial y} -i \frac {\partial u}{\partial y}\quad \cdots \quad (*)}\]
Example 4.50. We saw that the complex function \(f(z) = 2x^2 + y + i(y^2 - x)\) was nowhere analytic but the Cauchy-Riemann equations were satisfied on the line \(y = 2x\). But since the functions \[u(x,y) = 2x^2 + y\quad \frac {\partial u}{\partial x} = 4x \,, \quad \frac {\partial u}{\partial x} = 1\] \[v(x,y) = y^2 - x\quad \frac {\partial v}{\partial x} = -1\,,\quad \frac {\partial v}{\partial y} = 2y\] are continuous at every point, it follows that \(f\) is differentiable on the line \(y = 2x\). Moreover, from \((*)\) we see that the derivative of \(f\) at points on the line \(y = 2x\) is given by \[ f'(z) = 4x - i = 2y - i\]
Theorem 4.51. Suppose the function \(f(z) = u(x,y) + iv(x,y)\) is analytic in a domain \(D\)
- 1.
- if \(\left |f(z)\right |\) is constant in \(D\), then so is \(f(z)\).
- 2.
- if \(f'(z) = 0\), then \(f(z) = C\) in \(D\) where \(C\) is a constant.
Proof.
- 1.
- Suppose \(\left |f\right |=c\) on \(D\), so \(u^2+v^2=c^2\). If \(c=0\) then \(f\equiv 0\), so assume \(c\neq 0\). Differentiating with respect to \(x\) and then \(y\), \[uu_x+vv_x=0,\qquad uu_y+vv_y=0 .\] Substituting \(u_y=-v_x\) and \(v_y=u_x\) into the second gives the pair \[uu_x+vv_x=0,\qquad -uv_x+vu_x=0 .\] Read as a linear system in \(u_x\) and \(v_x\), its determinant is \(u^2+v^2=c^2\neq 0\), so \(u_x=v_x=0\). Then \(u_y=v_y=0\) by Cauchy–Riemann, and \(f'=u_x+iv_x=0\) on \(D\). Part (2) now completes the argument.
- 2.
- If \(f'=0\) then \(u_x+iv_x=0\), so \(u_x=v_x=0\), and \(u_y=v_y=0\) by Cauchy–Riemann. A real function on a domain — an open, connected set — with all partials zero is constant, since any two of its points can be joined by a path along which the directional derivative vanishes. Hence \(u\) and \(v\) are constant and so is \(f\).
Connectedness matters in (2): on a disconnected open set \(f'=0\) forces \(f\) to be constant on each component separately, and the constants may differ. □
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