Derivatives

Enter homework on derivatives here.

## Question 1

#### a) $f(z) = z^{3}$

(1)
$$z = x+iy$$
(2)
$$(x+iy)^{3}$$
(3)
$$(x+iy)(x+iy)(x+iy)$$
(4)
$$x^{3}+x^{2}iy+2x^{2}iy-2xy^{2}-xy^{2}-iy^{3}$$
(5)
$$u = x^{3}-3xy^{2}$$
(6)
$$v = 3x^{2}y - y^{3}$$
(7)
\begin{align} \frac{\partial u}{\partial x} = 3x^{2}-3y^{2}\;\;\; \frac{\partial v}{\partial y} = 3x^{2}-3y^{2} \end{align}
(8)
\begin{align} \frac{\partial v}{\partial x} = 6xy \;\;\; -\frac{\partial u}{\partial y} = 6xy \end{align}
(9)
\begin{align} \frac{\partial w}{\partial z}=\frac{\partial u}{\partial x} +\frac{i\partial v}{\partial x} \end{align}
(10)
\begin{align} \frac{\partial w}{\partial z}= 3x^{2}-3y^{2}+6xyi \end{align}
(11)
\begin{align} \frac{\partial w}{\partial z}= 3(x^{2}-y^{2}+2xyi) \end{align}
(12)
\begin{align} \frac{\partial w}{\partial z}= 3(x+iy)^{2} \end{align}
(13)
\begin{align} \frac{\partial w}{\partial z}= 3z^{2} \end{align}

Therefore the derivatives of $z^{3}$ is $3z^{3}$

### Question 2b

Show that $f(z) = \bar{z}^2$ is not analytic anywhere.

Differentiability is a stronger condition in complex analysis than it is in real analysis. In real analysis $h$ approaches zero in only two different directions. In complex analysis, $h$ can approach zero along an unlimited number of paths in the complex plane. The geometric interpretations of differentiation are also different. In real analysis, derivatives are interpreted as local linearity whereas in complex analysis the derivative has an "amplitwist" interpretation. In other words, in complex analysis differentiable functions map disks to disks (if a disk is amplified and twisted, it remains a disk).

Definition : The derivative of $w = f(z)$ is defined as $f'(z_0) =\lim_{h \rightarrow 0}\dfrac{f(z_0+h)-f(z_0)}{h}$ provided the limit exists. In this case, $f(z)$ is called differentiable at the point $z_0$.

Although differentiability is a strict requirement, there is another class of functions in complex analysis which meet even more severe requirements called analytic functions.

Definition : A function $w = f(z)$ is analytic at a point $z_0$ if it is differentiable at each point in a neighborhood of $z_0$.

Analyticity is a neighborhood property defined over an open set, a function cannot be analytic at a point. The following theorem provides necessary and sufficient conditions for analyticity.

Theorem [Necessary condition for analyticity]: Suppose $f(z)$ is differentiable at the point $z_0$. Then the first-order partial derivatives of $u$ and $v$ exist and satisfy the Cauchy-Riemann equations

(14)
\begin{align} \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and } \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \end{align}

Theorem [Sufficient condition for analyticity]: Suppose $u$ and $v$ are continuous and have continuous first-order partial derivatives in an open neighborhood $D$ of the point $z_0$. If the first-order partial derivatives satisfy the Cauchy-Riemann equations, then $f(z)$ is analytic in $D$.

Example: Show that $f(z) = \bar{z}^2$ is not analytic anywhere.

Set $w = u + i v$ and $z = x + i y$ to separate the function into its real and imaginary parts.

(15)
$$u + i v = (x + i y)^2 = x^2-y^2+i2xy$$

Hence,

(16)
\begin{eqnarray} u & = & x^2-y^2 \\ v & = & 2xy \end{eqnarray}

Check to see of the Cauchy-Riemann equations are satisfied.

(17)
\begin{array} {ll} \frac{\partial u}{\partial x} = 2x & \frac{\partial v}{\partial y} = -2x \\ \frac{\partial u}{\partial y} = -2y & -\frac{\partial v}{\partial x} = 2y \end{array}

The Cauchy-Riemann equations are satisfied at the point $z=0$. Since there is no open disk about the point $z=0$ for which the Cauchy-Riemann equations are satisfied, we conclude that $f(z) = \bar{z}^2$ is not analytic anywhere.

Comment: Differentiability is a separate issue and we can use the limit definition to show that $f(z) = \bar{z}^2$ is differentiable at $z=0$, but nowhere else. In the calculation below, we make use of the fact that $h$ can approach 0 from any direction. For example, we can let $h$ approach 0 along rays passing through the origin by setting $h = r e^{i \theta}$ and letting $r \rightarrow 0$.

(18)
\begin{eqnarray} f'(z) & = & \lim_{h \rightarrow 0}\dfrac{f(z+h)-f(z)}{h} \\ & = & \lim_{h \rightarrow 0}\dfrac{(\bar{z}-h)^2-\bar{z}^2}{h} \\ & = & \lim_{h \rightarrow 0}\dfrac{2 \bar{z} \bar{h} + \bar{h}^2}{h} \\ & = & \lim_{h \rightarrow 0}\dfrac{2 \bar{z} r e^{-i \theta} + r^2 e^{-i 2\theta}}{r e^{i \theta}} \\ & = & 2 \bar{z} e^{-i 2\theta} \end{eqnarray}

From this calculation, we can see that $f'(0) = 0$ and that for all other values of $z$, the value of the limit is path dependent and the derivative does not exist.

### Question 3b

Linear approximation to $f(z)=z^3$ near $z=1+i$

Function:

(19)
$$f(z)=z^3$$

Derivative:

(20)
$$f'(z)=3z^3$$

Use the linearization $L(z)=f(z_0)+f'(z_0)(z-z_0)$ and plug the given $z_0$ into it to get the linearization

(21)
$$z_0=1+i$$

Plug it in:

(22)
\begin{eqnarray} L(z) & = & (1+i)^3+3(1+i)^2 (z-1+i)\\ & = & -2+2i+3(2i)(z-(1+i))\\ & = & -2+2i+6iz-6i+6\\ & = & 4-4i+6iz \end{eqnarray}

## Question 3c

##### Use the linearization to explain the mapping properties of $f(z)=z^{3}$ near the point z=1+i. Be sure to explain the amount of translation, rotation and magnification.

Using the equation from 3B.
=-4i+4+6iz
=(4-4i)+6iz

this indicates:
-translation to the right 4 units
-translation down 4 units
-magnification of 6
-rotation of $\pi /2$