Limit, Continuity, and Differentiability

The definitions for continuity and differentiability of Complex Functions are very similar to the definition in \(\mathbb R\). However, note that if we consider complex numbers as on the complex plane, it's actually a ball instead of a segment, it's somewhat similar to \(\mathbb R^2\), but take special note that they are not identical to each other.

Limits

Example evaluate limits

\[\begin{align*} \lim_{z\rightarrow i} z+z^{-1} &= \lim_{z\rightarrow i} z+\frac{\bar z}{|z|} = i + \frac{-i}{1} = 0\\ \lim_{z\rightarrow i}\sinh z&= \sinh i = i\sin 1 \end{align*}\]

Continuity

Known that any \(f:\mathbb C\rightarrow \mathbb C\) can be decomposed into real and imaginary parts \(f = u+iv, u:\mathbb R\rightarrow \mathbb R, v:\mathbb R\rightarrow \mathbb R\) with \(z = x+iy\). Therefore, if \(u,v\) is continuous at \((x_0, y_0)\), then f is continuous at \(z_0=x_0+iy_0\) with limit

\[\lim_{z\rightarrow z_0}f(z) = \lim_{(x,y)\rightarrow (x_0, y_0)} u(x, y) + iv(x, y) = u(x_0, y_0) + iv(x_0, y_0)\]

Conversely, we can also prove \(f\) continuous implies \(u\) and \(v\) are both continuous.

Differentiation Laws

Differentiation laws on sum, product, quotient, composition, power, trigonometric all hold on complex numbers. The proof is similar to real number ones, hence some are omitted.

Example. LHospitals Rules for both 0

If \(f(z)\) and \(g(z)\) have formal power series about \(a\) and \(f(a)=f'(a) = \cdots = f^{(k)}=g(a)=g'(a) = \cdots = g^{(k)} = 0\) with \(f^{(k+1)}(a)\) and \(g^{(k+1)}(a)\) note simultaneously \(0\). Then, \(\lim_{z\rightarrow a}f(z)/g(z) = f^{(k+1)}(a)/g^{(k+1)}(a)\).

proof For every \(k\),

\[\begin{align*} \lim_{z\rightarrow a}\frac{f^{(k)}(z)}{g^{(k)}(z)} &= \lim_{z\rightarrow a}\frac{f^{(k)}(z)-f^{(k)}(a)}{g^{(k)}(z)-g^{(k)}(a)}\\&= \lim_{z\rightarrow a}\frac{\frac{f^{(k)}(z)-f^{(k)}(a)}{z-a}}{\frac{g^{(k)}(z)-g^{(k)}(a)}{z-a}} \end{align*}\]

Note that \(f^{(k)}(z), g^{(k)}(z)\) are differentiable, so that

\[\lim_{z\rightarrow a}\frac{\frac{f^{(k)}(z)-f^{(k)}(a)}{z-a}}{\frac{g^{(k)}(z)-g^{(k)}(a)}{z-a}} = \lim_{z\rightarrow a}\frac{f^{(k+1)}(z)}{g^{(k+1)}(z)}\]

If \(\lim_{z\rightarrow a}\frac{f^{(k+1)}(z)}{g^{(k+1)}(z)}\) exist to be \(\frac{f^{(k+1)}(a)}{g^{(k+1)}(a)}\), then all limits ends to be \(\frac{f^{(k+1)}(a)}{g^{(k+1)}(a)}\).

Cauchy Riemann Equations

Claim Consider some complex function \(f:\mathcal R\rightarrow \mathbb C, f(z)=u(z)+iv(z)\) defined on some regions \(\mathcal R\subseteq \mathbb C\). \(f\) is differentiable IFF

\[\partial_x u = \partial_yv \land \partial_x v = -\partial_yu\]

Note that this automatically implies that \(u,v\) need to have derivative, hence also are continuous.

Define \(\tilde f: \tilde{\mathcal R}\rightarrow \mathbb R^2: f(x, y) = (\tilde u(x, y), \tilde v(x, y))\). Since \(f\) is differentiable, the limit will exist from any approach, and the derivate is \(f'(z) = a(x)+ib(y)\) for \(a(x), b(y)\in\mathbb R\).
Consider approach only from the real line, let \(t\in\mathbb R\)

\[\begin{align*} &\lim_{t\rightarrow 0} \frac{f(z_0+t) - f(z_0)}{t}\\ = &\lim_{t\rightarrow 0} \frac{u(z_0+t) + iv(z_0+t)}t - \frac{u(z_0) + iv(z_0)}{t}\\ = &a(x_0) + ib(y_0) \end{align*}\]

On the projected \(\mathbb R^2\) function,

\[\lim_{t\rightarrow 0}\frac{\tilde u(x_0+t, y_0) - u(x_0,y_0)}t +\frac{\tilde v(x_0+t, y_0) - v(x_0,y_0)}{t} = \partial_x\tilde u + \partial_x\tilde v\]

So that

\[(\partial_x\tilde u ,\partial_x\tilde v) = (a, b)\]

Similarly, if we approach only from the imaginary line, we have

\[\begin{align*} \lim_{t\rightarrow 0}\frac{f(z_0+it) - f(z_0)}{it} &= \lim_{t\rightarrow 0}-i\frac{f(z_0+it) - f(z_0)}{t}\\ &= \lim_{t\rightarrow 0} \frac{v(z_0+it) - v(z_0)}t - i\frac{u(z_0+it) - u(z_0)}t \end{align*}\]

So that

\[(\partial_y\tilde v, - \partial_y\tilde u) = (a, b)\]

Therefore, we have the Jacobian as

\[D_{\tilde f}(x_0, y_0) = \begin{bmatrix}a(z_0)&-b(z_0)\\b(z_0)&a(z_0)\end{bmatrix}\]

Therefore, we may notice that \(\mathbb R^2\)-differentiable only approach from \(2\) lines, while \(\mathbb C\)-differentiable can approach from other curves. So \(C\)-differentiable implies \(\mathbb R^2\)-differentiable, but ont the converse. For example

\[f(z) = \bar z = x -iy, u = x, v = -y\]

However,

\[\lim_{\Delta z\rightarrow 0}\frac{\overline{(z_0 + \Delta z)} - \bar z_0}{\Delta z} = \lim_{\Delta z\rightarrow 0}\frac{\overline{\Delta z}}{\Delta z}\]

Taking \(\Delta z = re^{i\theta}, \overline{\Delta z} = re^{-i\theta}\), the limit becomes \(\lim_{\Delta z\rightarrow 0}e^{-2i\theta}\), note that \(\Delta z = r(\cos\theta + i\sin\theta)\) so that if we approach with fixed \(\theta\) and \(r\rightarrow 0\), it gives different limits. Hence the limit does not exist.

Derivative from Cauchy Riemann Equations

Note that from the proof above, we have shown that if Cauchy Riemann conditions hold, then the derivative of some complex function \(f\) is

\[f'(z) = u_x + iv_x = v_y - iu_y\]

Cauchy Riemann Equation in Polar Form

Theorem \(f = u+iv\) is analytic at \(z = re^{i\theta}, r>0, \theta\in[0, 2\pi)\) IFF

\[u_r = r^{-1}v_\theta, v_r = -r^{-1}u_\theta\]

and the partial derivatives are continuous.

proof. Note that \(re^{i\theta} = r\cos(\theta) + ri\sin(\theta)\) so that \((x, y) = (r\cos\theta, r\sin\theta)\). Follow chain rule

\[\begin{align*} u_r &= u_x x_r + u_yy_r= u_x\cos\theta + u_y\sin\theta\\ u_\theta &= u_x x_\theta + u_yy_\theta= -r(u_x\sin\theta + u_yr\cos\theta)\\ v_r &= v_x x_r + v_yy_r= v_x\sin\theta + u_y\cos\theta\\ v_\theta &= v_x x_\theta + v_yy_\theta= r(v_x\cos\theta - u_y\sin\theta) \end{align*}\]

Then, using CR equations, we can establish the equalities

Derivative in Polar Form

Using the relationship above, we can then apply chain rule to compute \(f'\) as

\[f'(z) = (\cos\theta - i\sin\theta)(u_r + iv_r) = e^{-i\theta}(u_r + iv_r)\]

Corollary The level set of \(u\), \(C = \{(x, y) : u(x, y) = c_1, c_1\in\mathbb R\}\), are orthogonal to the level set of \(v\), at point \(z_0\).

proof. \(\nabla u\cdot \nabla v = u_xv_x + u_yv_y = -u_xu_y + u_yu_x = 0\), so that their gradient is orthogonal at \(z_0\), and then each of the level-set is orthogonal to their gradient, hence also orthogonal to each other.

Example Show that \(Re(z)\) and \(Im(z)\) are nowhere differentiable.
Let \(\tilde f(x, y) = x\) so that its Jacobian is \(\begin{bmatrix}1&0\\0&0\end{bmatrix}\) hence Cauchy Riemann equations does not hold, hence not differentiable. (\(Im(z)\) is similar).

Analytic Function

\(f:\mathcal R\rightarrow \mathbb C\) is analytic at \(z_0\) IFF it is differentiable within the neighborhood of \(z_0\). \(f\) is an analytic (or sometimes holomorphic) is it is analytic on each point on the domain.

Singular point is some point \(z_0\) that \(f\) fails to be analytic.

In fact, a complex function is analytic, then it is also infinitely differentiable (as of a power series function on the reals).

Also, take a note that the Laplacians of u and v, given CR conditions, satisfy that

\[\nabla^2 u = \frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = \frac{\partial}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial}{\partial y}\frac{\partial v}{\partial x} = 0, \nabla^2v = 0\]

So that \(u,v\) are harmonic conjugate of each other. We can obtain \(v\) given \(u\), then as well as \(f\).

Example Take \(u(x,y) = y^2 - x^2\). Then \(u_x = -2x = v_y, u_y = 2y = -v_x\), implies that \(v = -2xy + c\) for some constant \(c\). So that

\[f(z) = y^2-x^2 + i(-2xy+c) = -z^2 + ic\]

Differentiability at Infinity

The limit at infinity \(\lim_{z\rightarrow\infty}f(z) = L\) is defined as \(\forall \epsilon > 0. \exists N > 0. \forall z\in\mathbb C. |z|> N\implies |f(z)-L| < \epsilon\).

Define \(g:R\rightarrow\mathbb C, g(z):=\begin{cases}f(z^{-1})&z\neq 0\\L &z=0\end{cases}\).
Then, \(f\) is differentiable/analytic at \(\infty\) is \(g(z)\) is differentiable/analytic at \(0\).

Example Consider \(f(z) = z^{-1}\), then \(g(z) = z\) is differentiable at 0, so that \(z^{-1}\) is differentiable at \(\infty\). \(z\) is not differentiable at \(\infty\), since \(z^{-1}\) is not differentiable at \(0\).