Question

l. Assume that j : R-→ R-s C and satisfies what are known as the Cauchy-Riemann equations: (c) Let r-(r1, 2) and (s1, s2) be vectors in IR2 and suppose that (ri, 2)f(s1, 82) and Df(81,82)メ0. Show that f-1 satisfies the Cauchy-Riemann equations when evaluated at r. (Hint: Might I make a notational suggestion: Leta(s) = sim) = % (n, s) and b(s) 쓺(81, 82) =-警( )) 81,82 (d) For this last bit, drop the assumption that f satisfies the Cauchy-Riemann equations. Give a function whose derivative is Do for all z E R2. Explain why this f is not invertible (although Df(x)メ0).

Answer 1

a) We are using the suggested notation.

Since $Df(s_1,s_2)\neq 0$ we know that either $a(s)\neq 0$ or $b(s)\neq 0$ , and thus, in any case, $Df(s_1,s_2)\neq 0$ implies

$\det Df(s_1,s_2) =a(s)^2+b(s)^2> 0$

Since $(f^{-1}\circ f )(x,y)=(x,y)$ for all $(x,y)\in\mathbb R^2$ , we have

$D(f^{-1}\circ f)(x,y)= \begin{pmatrix}1&0\\0&1\end{pmatrix}$

for all $(x,y)\in\mathbb R^2$ . Using chain rule, we get

$\begin{align*}Df^{-1}(r)\cdot Df(s)&=D(f^{-1}\circ f)(s)\\ &= \begin{pmatrix}1&0\\0&1\end{pmatrix}\\ \Rightarrow\hspace{2cm}Df^{-1}(r)&=(Df(s))^{-1} \\ &={\frac 1{a(s)^2+b(s)^2}}\begin{pmatrix}a(s)&b(s)\\ -b(s)&a(s)\end{pmatrix} \end{align*}$

Since

$\begin{align*}Df^{-1}(r)=\begin{pmatrix}{\frac{\partial f^{-1}_1}{\partial x}}(r)&{\frac{\partial f^{-1}_1}{\partial y}}(r)\\ {\frac{\partial f^{-1}_2}{\partial x}}(r)&{\frac{\partial f^{-1}_2}{\partial y}}(r)\end{pmatrix} \end{align*}$

we conclude that

$\begin{align*}{\frac{\partial f^{-1}_1}{\partial x}}(r)&={\frac{a(s)}{a(s)^2+b(s)^2}}={\frac{\partial f^{-1}_2}{\partial y}}(r)\\ {\frac{\partial f^{-1}_1}{\partial y}}(r)&={\frac{a(s)}{a(s)^2+b(s)^2}}= -{\frac{\partial f^{-1}_2}{\partial x}}(r)\end{align*}$

Therefore, $\begin{align*}f^{-1}\end{align*}$ satisfies Cauchy-Riemann.

b) The function $\begin{align*}f(x,y)=(x,0)\end{align*}$ has derivative

$\begin{align*}Df(r_1,r_2)&=\begin{pmatrix}{\frac{\partial f_1}{\partial x}}(r)&{\frac{\partial f_1}{\partial y}}(r)\\ {\frac{\partial f_2}{\partial x}}(r)&{\frac{\partial f_2}{\partial y}}(r)\end{pmatrix}\\ &=\begin{pmatrix}1&0\\ 0&0\end{pmatrix}\end{align*}$

for all $x=(r_1,r_2)\in\mathbb R^2$ .

Since $f(x,0)=(x,0)=f(x,1)$ for all $x\in\mathbb R$ , the function $\begin{align*}f(x,y)=(x,0)\end{align*}$ ​​​​​​​ is not injective, hence, not invertible either.