Example 4.31. Differentiate

1.

\(f(z) = 3z^4 - 5z^3 + 2z\)

Using the sum rule, we have \(\, f'(z) = 12z^3 - 15z^2 + 2\)

2.

\(f(z) = \frac {z^2}{4z + 1}\)

Quotient rule \(\, f'(z) = \frac {2z (4z + 1) - z^2 (4)}{(4z + 1)^2} = \frac {4z^2 + 2z}{(4z + 1)^2}\)

3.

\(f(z) = (iz^2 + 3z)^5\)

Chain rule \(\, f(z) = (iz^2 + 3z)^5\)

Let \(g(z) = iz^2 + 3z \implies g'(z) = 2iz + 3\)

So \(\, f'(z) = g'(z)\cdot (g(z))^4\cdot 5= 5(2iz + 3)(iz^2 + 3z)^4\)

Example 4.32. Show that the function \(f(z) = x + 4iy \) is not differentiable at any point \(z\).

Solution
Let \(z\) be any point in the complex plane with \(\Delta z = \Delta x + i\Delta y\). \begin {align*} \lim \limits _{\Delta z \rightarrow 0} \frac {f(z + \Delta z ) - f(z)}{\Delta z } & = \lim \limits _{\Delta z \rightarrow 0}\frac {(x + \Delta x) + 4i(y + \Delta y ) - x - 4yi}{\Delta z}\\\\ & = \lim \limits _{\Delta z \rightarrow 0}\frac {\Delta x + 4i \Delta y}{\Delta x + i \Delta y} \end {align*}

Now if we let \(\Delta z \longrightarrow 0\) along a line parallel to the \(x-\)axis then \(\Delta y = 0\) and \(\Delta z = \Delta x\) and \[\lim \limits _{\Delta z \rightarrow 0} \frac {f(z + \Delta z ) - f(z)}{\Delta z } = \lim \limits _{\Delta z \rightarrow 0} \frac {\Delta x}{\Delta x} = 1\] On the other hand, if we let \(\Delta z \longrightarrow 0\) along a line parallel to the \(y-\)axis, then \(\Delta x =0\) and \(\Delta y = i\Delta y\) so that \[\lim \limits _{\Delta z \rightarrow 0} \frac {f(z + \Delta z ) - f(z)}{\Delta z } = \lim \limits _{\Delta z \rightarrow 0}\frac {4i\Delta y }{i\Delta y } = 4\] We see that the limits are different so \(f(z) = x + 4iy\) is nowhere differentiable.

Example 4.33. Show that \(f: \mathbb {C}\longrightarrow \mathbb {C}\) defined by \(f(z) = z\cdot \operatorname {Re}(z)\) is differentiable only at 0.

Solution \begin {align*} \text {Let}\quad z \in \mathbb {C}\qquad \lim \limits _{h\rightarrow 0} \frac {f(z + h) - f(z)}{h} & = \lim \limits _{h\rightarrow 0}\frac {(z + h)\,\operatorname {Re}(z + h) - z\, \operatorname {Re}(z)}{h}\\ & = \lim \limits _{h\rightarrow 0}\frac {z\cdot \operatorname {Re}(z) + z\cdot \operatorname {Re}(h) + h\cdot \operatorname {Re}(z) + h\cdot \operatorname {Re}(h) - z\cdot \operatorname {Re}(z)}{h}\\\\ & = \lim \limits _{h\rightarrow 0}\Big (\frac {z\cdot \operatorname {Re}(h)}{h} + \operatorname {Re}(z) + \operatorname {Re}(h)\Big )\\\\ & = \lim \limits _{h\rightarrow 0}\Big (\frac {z\cdot \operatorname {Re}(h)}{h}\Big ) + \operatorname {Re}(z) \end {align*}

If \(z\neq 0\) then \(\lim \limits _{h\rightarrow 0} \frac {z \cdot \operatorname {Re}(h)}{h}\) does not exist.
Because if \(h\longrightarrow 0\) through purely imaginary numbers then \(\operatorname {Re}(h) = 0\) or \(\frac {Re(h)}{h} = 0\) for \(h\) approach zero through such numbers
Whereas if \(h\longrightarrow 0\) through real numbers then \(\operatorname {Re}(h) = 1\) so \(\frac {Re(h)}{h} = 1\).
So \(\lim \limits _{h\rightarrow 0} \frac {z \cdot \operatorname {Re}(h)}{h}\) does not exist.

If \(z = 0, \quad \frac {z\cdot \operatorname {Re}(h)}{h} = 0\) so \(\lim \limits _{h\rightarrow 0} \frac {z \cdot \operatorname {Re}(h)}{h}=0\) and so \(f(z)\) is differentiable only at \(z= 0\) and \(f'(0) = 0\).

Assume that \(f\) is differentiable at a point \(z = x + iy \in D\). Then \[f'(z) = \lim \limits _{h\rightarrow 0 } \frac {f(z + h) - f(z)}{h}\] Since \(h\) can approach zero through real numbers \begin {align*} f'(z) & = \lim _{\substack {h \rightarrow 0\\h\in \mathbb {R}}} \frac {f(x + iy + h) - f(x + iy)}{h}\\\\ & = \lim _{\substack {h \rightarrow 0\\h\in \mathbb {R}}}\frac {u(x + h,y) + iv(x + h, y) - (u(x,y) + iv(x,y))}{h}\\\\ & = \lim _{\substack {h \rightarrow 0\\h\in \mathbb {R}}} \frac {u(x+ h, y) - u(x,y)}{h} - i\,\lim _{\substack {h \rightarrow 0\\h\in \mathbb {R}}}\frac {v(x + h, y) - v(x,y)}{h}\\ \end {align*}

\[\therefore \quad f'(z) = \frac {\partial u}{\partial x}\Bigg |_{(x,y)} + i \frac {\partial v}{\partial x}\Bigg |_{(x,y)}\]

If \(h \longrightarrow 0\) through purely imaginary numbers, then write \(h = ik\) where \(k\in \mathbb {R}\) and then \begin {align*} f'(z) & = \lim _{h \rightarrow 0 } \frac {f(z + h) - f(z)}{h} = \lim _{\substack {k \rightarrow 0\\k\in \mathbb {R}}}\frac {f(z + ik) - f(z) }{ik}\\\\ & = \lim _{\substack {k \rightarrow 0\\k\in \mathbb {R}}} \frac {u(x, y + k) - u(x,y)}{ik} + i\,\lim _{\substack {k \rightarrow 0\\k\in \mathbb {R}}}\frac {v(x,y + k) - v(x,y)}{ik}\\\\ & = -i\,\lim _{\substack {k \rightarrow 0\\k\in \mathbb {R}}}\frac {u(x, y + k) - u(x,y)}{k} + \lim _{\substack {k \rightarrow 0\\k\in \mathbb {R}}}\frac {v(x, y + k) - v(x,y)}{k}\\\\ & = -i\,\frac {\partial u}{\partial y }\Bigg |_{(x,y)} + \frac {\partial v}{\partial y }\Bigg |_{(x,y)}\\ \end {align*}

\[\therefore \quad f'(z) = -i\,\frac {\partial u}{\partial y }\Bigg |_{(x,y)} + \frac {\partial v}{\partial y }\Bigg |_{(x,y)} = \frac {\partial u}{\partial x}\Bigg |_{(x,y)} + i \frac {\partial v}{\partial x}\Bigg |_{(x,y)}\]

\[\boxed {\text {So}\quad u_x(x,y) = v_y(x,y) \quad \text {and}\quad v_x(x,y) = -u_y(x,y)\qquad \text {Cauchy-Riemann equations}}\]

Example 4.34. Let \(f(z) = z^3\), then \(f'(z) = 3z^2\)

\(f(z) = (x^3 - 3xy^2) + i (3x^2 y - y^3)\)

\(\implies \quad u(x,y) = x^3 - 3xy^2\quad ;\quad v(x,y) = 3x^2y - y^3\)

\(\frac {\partial u}{\partial x} = 3x^2 - 3y^2\quad ;\quad \frac {\partial v}{\partial y} = 3x^2 - 3y^2\)

\(\frac {\partial u}{\partial y} = -6xy\quad ;\quad \frac {\partial v}{\partial x} = 6xy\)

1.

Complex differentiability implies complex continuity.

2.

Geometric interpretation of the complex derivatives. \[Df\Big |_{(a,b)} = \begin {pmatrix} \frac {\partial u}{\partial x} & \frac {\partial u}{\partial y}\\\\ -\frac {\partial u}{\partial y} & \frac {\partial u}{\partial x} \end {pmatrix}\,\Bigg |_{(a,b)}\]

\[J_{aC}(f) \Big |_{(a,b)}= \Bigg (\frac {\partial u}{\partial x}\Bigg )^2 + \Bigg (\frac {\partial u}{\partial y}\Bigg )^2\Bigg |_{(a,b)}\]

\[f'(a + ib) = \frac {\partial u}{\partial x} + i \frac {\partial v}{\partial x}\,\Bigg |_{(a,b)} = \frac {\partial v}{\partial y} - i\frac {\partial u}{\partial y}\,\Bigg |_{(a,b)}\]

\[\left |f'(a + ib)\right |^2 = \Bigg (\frac {\partial u}{\partial x}\Bigg )^2 + \Bigg (\frac {\partial u}{\partial y}\Bigg )^2 \,\Bigg |_{(a,b)}\]

Assume \(f'(a + ib) \neq 0 \implies \left |f'(a + ib)\right | \neq 0\).

\[\begin {bmatrix} \alpha _1\\ \alpha _2\\ \end {bmatrix} \iff \alpha _1 + i\alpha _2\]

\begin {align*} Df \Bigg (\begin {bmatrix} \alpha _1\\ \alpha _2\\ \end {bmatrix}\Bigg ) & = \begin {pmatrix} \frac {\partial u }{\partial x} & \frac {\partial u}{\partial y}\\\\ -\frac {\partial u }{\partial y} & \frac {\partial u}{\partial x}\\ \end {pmatrix}\,\Bigg |_{(a,b)}\,\begin {pmatrix} \alpha _1\\ \alpha _2\\ \end {pmatrix}\\\\ & = \alpha _1\,\frac {\partial u}{\partial x} + \alpha _2\,\frac {\partial u}{\partial y} + i \Big (-\alpha _1\frac {\partial u}{\partial y} + \alpha _2\frac {\partial u}{\partial x}\Big )\\ \end {align*}

\begin {align*} \left |Df\Bigg (\begin {bmatrix} \alpha _1\\ \alpha _2\\ \end {bmatrix}\Bigg )\right | & = \Big (\alpha ^2_1 + \alpha _2^2\Big )\Bigg (\Big (\frac {\partial u}{\partial x}\Big )^2 + \Big (\frac {\partial u}{\partial y}\Big )^2\Bigg )\Bigg |_{(a,b)}\qquad \alpha = \alpha _1 + i\alpha _2\\\\ \left |Df\Bigg (\begin {bmatrix} \alpha _1\\ \alpha _2\\ \end {bmatrix}\Bigg )\right | & = \left |\alpha \right |\left |f'(a + ib)\right | \end {align*}

\[arg\Bigg (Df\Bigg (\begin {bmatrix} \alpha _1\\ \alpha _2\\ \end {bmatrix}\Bigg )\Bigg ) = \operatorname {arg}(f'(a + ib)) + \operatorname {arg} (\alpha )\] If \(r_1\) and \(r_2\) are two curves which intersect at \(a + ib\) at angle \(\theta \). Then \(f(r_1)\) and \(f(r_2)\) intersect at \(f(a + ib)\) at angle \(\theta \). \(f\) is conformal (angle preserving) at \(a+ ib\) if \(f'(a + ib) \neq 0\).
If \(f'(z_0) \neq 0\) then we say \(f\) is conformal at \(z_0\).