Exercise 2.2.17

Note: While everyone should think about this problem, it is specifically for those who received fewer than 39 points on our first quiz.

Have a look at exercise 2.2.17 in the text. There are 12 complex functions listed and you are asked assess their domain of differentiability and to compute the derivative where it's defined. Everyone who qualifies should post a response below with a detailed solution to one of those problems. If you do so by Wednesday, you earn five points added on to your first quiz.

2.2 17 f) If $f(z)=Im(z)$, then $f(z)=Im(x+iy)=y.$ By Cauchy-Riemann equations, $\frac{-\partial u}{\partial x}=\frac{\partial v}{\partial y}$. Since $f(z)=y$, $\frac{\partial u}{\partial x}=0$ and $\frac{\partial v}{\partial y}=1$, therefore not satisfying the Cauchy-Riemann equations, so $f(z)=Im(z)$ is no where differentiable and therefore is not holomorphic.

@opernie This is close - but not quite there. We need two different outputs to apply the partials to - a real part $u$ and an imaginary part $v$. Thus, we should have
$$-\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y},$$
rather than having an $f$ in both spots.

Chapter 2, Question 2.17, Part B

Claim. A complex function $f(z)=2x+ixy^2$ is nowhere differentiable. Therefore, $f(z)$ cannot be holomorphic on any region $R\subseteq\mathbb{C}$.

Proof. Let $f(z)=2x+ixy^2$. So $u(x,y)=2x$ and $v(x,y)=xy^2$. Note $$\begin{align}\frac{\partial u}{\partial x} &= 2, & \frac{\partial v}{\partial y}&=2xy \\ \frac{\partial v}{\partial x} &= y^2, & \frac{\partial u}{\partial y}&=0\end{align}$$ According to the Cauchy-Riemann equations, a complex function $f(x+iy)=u(x,y)+iv(x,y)$ is differentiable at a complex number $z_0=x_0+iy_0$, $x_0,y_0\in\mathbb{R}$, if and only if $$\begin{align}\Bigg(\frac{\partial u}{\partial x} &= \ \ \ \frac{\partial v}{\partial y}\Bigg)\Bigg\rvert_{x0,y0} & \text{and} \\ \Bigg(\frac{\partial v}{\partial x} &= -\frac{\partial u}{\partial y}\Bigg)\Bigg\rvert_{x0,y0}\end{align}$$ However $y_0^2=0\implies y_0=0$ so $2x_0y_0=(2x_0)\cdot0=0\not=2$. Therefore, the complex function $f(z)=2x+ixy^2$ is nowhere differentiable, so $f(z)$ cannot be holomorphic on any region $R\subseteq\mathbb{C}$. QED.

h- $f(z)=z \text{Im}(z) $

I clame that $f$ is differentiable at the point 0,indeed,
$$\lim_{z \to 0}\frac{f(z)-f(0)}{z}=\lim_{z\to 0} \text{Im}(z)=0$$
If $ z=x+iy$, then $f$ is written as $f(z)=(x+iy)y=xy+iy^2$
the derivatives $f_x$ and $f_y$ of f are :$f_x=y$ and $f_y=x+2iy$
we notice that $f_x(z)\neq -if_y(z)$ which means according to Cauchy Riemann that f is nowhere holomorphic on $\mathbb{C}$

Chapter 2.2.17.L $f(z)=z^2-\bar{z}^2$.
Let
\begin{align*}
z&=x+iy\\
\Rightarrow f(z)=f(x,y)&=(x+iy)^2-\overline{(x+iy)}^2\\
&=(x+iy)^2-(x-iy)^2\\
&=\big((x^2-y^2)+(2xy)i\big)-\big((x^2-y^2)+(-2xy)i\big)\\
&=(x^2-y^2-x^2+y^2)+(2xy+2xy)i\\
&=4xy\\
\Rightarrow U(x,y)=0, V(x,y)=4xy.
\end{align*}
This means the partial derivatives are as follows
\begin{align*}
&\frac{\partial U}{\partial x}=0 &\frac{\partial V}{\partial x}=4y\\
&\frac{\partial U}{\partial y}=0 &\frac{\partial V}{\partial y}=4x.
\end{align*}
By the Cauchy Reimann equations, for a function to be differential,
$$\frac{\partial U}{\partial x}=\frac{\partial V}{\partial y}\Rightarrow 0=4x\Rightarrow x=0$$
and
$$\frac{\partial U}{\partial y}=-\frac{\partial V}{\partial x}\Rightarrow 0=-4y\Rightarrow y=0.$$
Therefore, the only place this will be differential will be at the origin. Again by the Cauchy Reimann equations,
$$f'(0,0)=-\frac{\partial U}{\partial x}+\frac{\partial V}{\partial x}=0+4(0)=0.$$
This function is not holomorphic because it is only differentiable at one point.

2.17-(c)

Let $f(z) = x^2 + iy^2$. $f$ is only differentiable at $x = y$.

Checking the Cauchy-Riemann equations:

$$\frac{\partial U}{\partial y} = 0 = -\frac{\partial V}{\partial x}$$

However,

$$\frac{\partial U}{\partial x} = 2x, \frac{\partial V}{\partial y} = 2y$$

This equality is only satisfied when $x=y$, therefore the function is not holomorphic, as only a single line out of the disk surrounding any point $z_0$ on the line is differentiable.

Problem 2.17
Part a

$\textit{Claim}$: Given the function $f(z)=e^{-x}e^{-iy}$, where $z=x+iy$, the function is differentiable at all points and therefore holomorphic on $\mathbb{C}$.

$\textit{Proof}$: Rewriting $e^{-iy}$ as $\cos(y)-i\sin(y)$, $f(z)=e^{-x}(\cos(y)-i\sin(y))$.
$$u(x,y)=e^{-x}\cos(y)
\\ v(x,y)=-e^{-x}\sin(y)
\\ \frac{\partial u}{\partial x} = -e^{-x}\cos(y)
\\ \frac{\partial u}{\partial y} = -e^{-x}\sin(y)
\\ \frac{\partial v}{\partial x} = e^{-x}\sin(y)
\\ \frac{\partial v}{\partial y} = -e^{-x}\cos(y)$$

This allows us to clearly see that all functions are continuous and that $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and that $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ which means $f(z)$ is holomorphic. From our notes, we know that $f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}$ which means $f'(x) = e^{-x}\cos(y) + i e^{-x}\sin(y)$.

problem 2.17 part G

$f(z)=|z|^2=x^2+y^2$

$U(x,y)=x^2+y^2, V(x,y)=0$

finding the partial derivatives we get:

$\dfrac{\partial U}{\partial x}=2x, \textrm{ }\dfrac{\partial U}{\partial y}=2y$
$\dfrac{\partial V}{\partial x}=0, \textrm{ }\dfrac{\partial V}{\partial y}=0$

from here one can see that this function is only differentiable when $x=y=0$ and since it is only differentiable at one point the function is not holomorphic.

@cdunn I don't think you want the $i$ included with the $v$. Kinda like the imaginary part of $1-2i$ is $-2$, not $-2i$.

Problem 2.17 Part J

$f(z)$ = 4(Im$(z)$)(Re$(z)$) - $i(\overline{z})^2$
$$f(z) = 4xy - i(x-iy)^2$$ $$f(z) = 4xy -2xy -i(x^2 - y^2)$$ $$f(z) = 2xy +i(y^2 - x^2)$$

$\frac{du}{dx} = 2y, \frac{dv}{dy} = 2y$
$\frac{du}{dy} = 2x, -\frac{dv}{dx} = 2x$

Since the Cauchy-Riemann equations are satisfied at every point, the function is holomorphic at every point, making it entire.

BAM!

2.2.17, part d
$\textbf{claim}$:
Given the function $f(z)=e^{x}e^{-iy}$ where $z=x+iy$ and $x,y\in\mathbb{R}$, The function $f$ is only differentiable at the Imaginary axis.
$\textbf{Proof}$:
Start by rewriting $f(z)=e^xe^{-iy}$ in terms of it's real and imaginary parts:
$$\begin{array}{rcl}f(z)&=&e^xe^{-iy}\\&=&e^x( \cos(y)-i \sin(y))\\&=&e^x \cos(y)-(e^x \sin(y))i \end{array}$$

Now, let:
$$\begin{array}{rcl} u(x,y)&=&e^x \cos(y)\\&and&\\v(x,y)&=&-e^x \sin(y) \end{array}$$

According to the Cauchy–Riemann equations, $f$ is only differentiable where:
$$\begin{array}{rcl}\frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y}\\&and&\\ \frac{\partial u}{\partial y}&=&-\frac{\partial v}{\partial x} \end{array}$$

Given our $u(x,y)$ , $v(x,y)$:

$$\begin{array}{rcl}\frac{\partial u}{\partial x}=e^x \cos(y)&,&\frac{\partial v}{\partial y}=-e^x \cos(y)\\&&\\ \frac{\partial u}{\partial y}=-e^x \sin(y)&,&-\frac{\partial v}{\partial x}=e^x \sin(y) \end{array}$$

Because $e^x \neq 0$ for any $x$, these equations is true for all $x$ and only for $y=0$.

Problem 2.17 Part K

Assess the domain of differentiability and compute the derivative where it’s defined for \(f(z)=2xy-i(x+y)^2\).

First we need to expand the equation: \[\begin{aligned}
&f(z)=2xy-i(x+y)^2\\
&=2xy+i(-x^2-2xy-y^2)\end{aligned}\]

We can then apply the Cauchy-Riemann Equations with \(u=2xy\) and \(v=-x^2-2xy-y^2\). \[\begin{aligned}
&\frac{\partial u}{\partial x} = 2y\\
&\frac{\partial v}{\partial y} = -2x-2y\\
&\frac{\partial v}{\partial x} = -2x-2y\\
&\frac{\partial u}{\partial y} = 2x\end{aligned}\]



Since \(\frac{\partial u}{\partial x}\neq \frac{\partial v}{\partial y}\) and \(\frac{\partial u}{\partial y}\neq -\frac{\partial v}{\partial x}\) the function is nowhere holomorphic. It is, however, differentiable when \(2y=-2x-2y\) and \(2x=2x+2y\). It is apparent that when \(x\) and \(y = 0\) the function can be differentiated. Thus, \(f\) is differentiable at \((0,0)\).