Note: While everyone should have a look at these problems, the assignment is specifically aimed at those who scored 20 points or few on the first quiz. For such folks successful completion is worth 5 points.
The assignment is simple - respond to this question with the statement and a solution to one of the following problems from chapter 2:
15, 18, 20, 21, 22, 23
2.18)
Define $f(z)=0$ if $Re(z)Im(z)=0$, and $f(z)=1$ if $Re(z)Im(z)\neq0$.
We can then break $f(z)$ into its real and imaginary parts, $f(z)=u(x,y)+iv(x,y)$.
Therefore, since $f(z)=0$ if either $Re(z)=0$ or $Im(z)=0$ $$u(x,0)=v(x,0)=0$$ and $$u(0,y)=v(0,y)=0$$The Cauchy-Riemann equations state that a function is differentiable if $$\frac{-\partial u}{\partial y}=\frac{\partial v}{\partial x}$$ and $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$.
So $$\frac{-\partial u}{\partial y}=0=\frac{\partial v}{\partial x}=0$$ and $$\frac{\partial u}{\partial x}=0=\frac{\partial v}{\partial y}=0$$.
Theorem 2.15 (b) is not contradicted because f is not continuous at $z=0$.
2.15
$\color{red}{\text{Please state the problem}}$.
2.15) Prove that if $f(z)$ is given by a polynomial in $z$ then $f$ is entire. What can you say if $f(z)$ is given by a polynomial in $x=Re(z)$ and $y=Im(z)$?
Let $f(z)$ be a polynomial, thus $f(z)=a_nz^{n} + a_{n-1}z^{n-1}+ \cdots $
$\color{red}{\text{I edited the definition of this polynomial. Please edit the portion below to agree}}$.
Then $f'(z) = na_nz^{n-1} + (n-1)a_{n-1}z^{n-2}+\cdots $ always exists, as per the rules of differentiation. Thus, the function is entire.
Let $f(z)$ be a given polynomial in $x = Re(z)$ and $y = Im(z)$ . Generally, $f(z) = u(x,y) + iv(x,y)$.
If, however, $f$ depends only on $Re(z)$, we have $f(x+iy) = u(x) + iv(x)$ so, by the Cauchy-Riemann equations,
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = 0.$$
Similarly, ($\color{red}{\text{edit this next part}}$) if $f$ depends only on $Im(z)$, we have $f(x+iy) = u(y) + iv(y)$.
Also, therefore, $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} = 0$$
@jgorman This is a good start but there are still some issues. Please:
I went ahead and made some partial edits commenting on how you can finish it. My comments are in $\color{red}{\text{red}}$. In addition, I added a little sectioning.
exercice 2.20
Let's have the function g on the region G such as $g(z)=f(z)-\bar{f(z)}$
both $f(z)$ and $\bar{f(z)}$ are holomorphic on G thus $g(z)$ is holomorphic on G
$f(z)=u(x,y)+iv(x,y)$ then $\bar{f(z)}=u(x,y)-iv(x,y)$, thus $g(z)=2iv(x,y)$
g is holomorphic on G then we can apply the equation of Cauchy Riemann, we obtain then
$\frac{\partial v}{\partial y}=\frac{\partial v}{\partial x}=0$
therefore g is constant on G ($v(x,y)=K_1 \in \mathbb{C}$)
we do the same with the function h such as $h(z)=f(z)+\bar{f(z)}$
h is therefore holomorphic on G, we apply again the equation of Cauchy Riemann, we obtain then
$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}=0$
therefore h is constant on G ($u(x,y)=K_2 \in \mathbb{C}$)
we have now $f(z)=\frac{k_1+k_2}{2}$ for each z in G $\to$ f constant on G.
Problem 2.22
Suppose \(f\) is entire, with real and imaginary parts \(u\) and \(v\) satisfying \(u(z)v(z)=3\) for all \(z\). Show that \(f\) is constant.
The definition of a “entire” complex function is a function that is holomorphic over the entire complex plane. Since the problem states that the function is holomorphic, we can use the Cauchy-Riemann equations \[\begin{aligned}
&\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\\
&\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \end{aligned}\] Since \(u(z)v(z)=3\) we can infer that \(v(z)=\frac{3}{u(z)}\) with the derivative of \(v(z)\) being\(\frac{-3}{u^2}\) We can directly substitute this into the above Cauchy-Riemann equations \[\begin{aligned}
&\frac{\partial u}{\partial x} = \frac{-3}{u^2}\frac{\partial u}{\partial y}\\
&\frac{-3}{u^2}\frac{\partial u}{\partial x} = -\frac{\partial u}{\partial y}\\
&\frac{\partial u}{\partial y}=\frac{3}{u^2}\frac{\partial u}{\partial x}\\
&=>\frac{\partial u}{\partial x} = (\frac{-3}{u^2})(\frac{3}{u^2})\frac{\partial u}{\partial x}\\
&=>\frac{\partial u}{\partial x}=(\frac{-9}{u^4})\frac{\partial u}{\partial x}\\
&=>(u^4)\frac{\partial u}{\partial x}=(-9)\frac{\partial u}{\partial x}\\
&=>\frac{\partial u}{\partial x}(u^4+9)=0\end{aligned}\] Since \(u\) is a member of \(\mathbb{R}\) and \((u^4+9) > 0\) we know that \(\frac{\partial u}{\partial x}\) must be equal to zero. The same logic can be applied to find that \(\frac{\partial u}{\partial y}=0\). Since both of these partial derivatives equal 0, we can state that \(u\) is constant. Going back to the original Cauchy-Riemann equations, if u(z) is constant for this holomorphic function than v(z) must be constant as well. Thus, the function \(f=u+iv\) must also be constant.
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2.23 Prove that the Cauchy–Riemann equations take on the following form in polar coordinates:
$$\frac{du}{dr}=\frac{1}{r}\frac{dv}{d\theta}$$ and $$\frac{1}{r}\frac{du}{d\theta}=-\frac{dv}{dr}$$
Answer
Cauchy-Riemann equations state:
$$\frac{du}{dx}=\frac{dv}{dy}$$ $$\frac{du}{dy}=-\frac{dv}{dr}$$
We can use this to say:$$\frac{du}{dx}=\frac{dv}{dy}<=>\frac{du}{d\theta}\frac{d\theta}{dx}=\frac{dv}{dr}\frac{dr}{dy}$$
While:
$$x=r\cos(\theta)=>\frac{dx}{d\theta}=-r\sin(\theta)=>\frac{d\theta}{dx}=\frac{1}{-r\sin(\theta)}$$
$$y=r\sin(\theta)=>\frac{dy}{dr}=\sin(\theta)\frac{dr}{dy}=\frac{1}{\sin(\theta)}$$
We can say that:
$$\frac{du}{d\theta}\frac{1}{-r\sin(\theta)}=\frac{dv}{dr}\frac{1}{\sin(\theta)}=>\frac{1}{r}\frac{du}{d\theta}=-\frac{dv}{dr}$$
Ta-da.
also:
$$\frac{du}{dr}=\frac{du}{dx}\cos(\theta)+\frac{dv}{dy}r\cos(\theta)$$
$$\frac{dv}{d\theta}=\frac{dv}{dx}(-r\sin(\theta))+\frac{dv}{dy}(r\cos(\theta))$$
Divide both sides by r and combine to get:
$$\frac{1}{r}\frac{dv}{d\theta}=\frac{du}{dr}$$
Suppose that $f$ is entire and can be written $f(z)=u(x)+iv(y)$.
$\textbf{Claim}$:
$f(z)=az+b$ for some $a\in\mathbb{R}$ , $b\in\mathbb{C}$.
$\textbf{Proof}$:
$f(z)=u(x)+iu(y)$ implies that:
$$\begin{array}{rcccccl} \frac{\partial u}{\partial x}&=&\frac{\partial v}{\partial y}&=&u'(x)&=&v'(y) \end{array}$$
$$\textbf{and}$$
$$\begin{array}{rcccl} \frac{\partial u}{\partial y}&=&0&=&\frac{\partial v}{\partial x} \end{array}$$
In order for $u'(x)=v'(y)$, to be true, both must be constant (independent of both $x$ and $y$). If we let $u'(x)=a\in\mathbb{R}=v'(y)$ and $c$,$d \in\mathbb{R}$ be two arbitray constants, then $u(x)=au'(x) + c$ and $v(y)=av'(y)+d$. We can now rewrite $f$:
$$f(z)=au'(x)+c+av'(y)+d$$
Which reduces to:
$$f(z)=a \left( u'(x)+i(v'(y)) \right)+(c+d)$$
we can define $b=c+d$ and we have our function of the form:
$$f(z)=a \left( u'(x)+i(v'(y)) \right)+b$$
Which is What We Wanted!