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What is the Jacobian of the function $f(u+iv)={u+iv-a\over u+iv-b}$?

I think the Jacobian should be something of the form $\left(\begin{matrix} {\partial f_1\over\partial u} & {\partial f_1\over\partial v} \\ {\partial f_2\over\partial u} & {\partial f_2\over\partial v} \end{matrix}\right)$

but I don't know what $f_1,f_2$ are in this case. Thank you.

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    You could just write $(u+iv-(a_1+a_2i))/(u+iv-(b_1+b_2i))$ where $u,v,a_1,a_2,b_1,b_2$ are real. And later, exploit the [Cauchy--Riemann equations](http://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations) to conclude that the matrix must have the form \begin{bmatrix} c & -d \\ d & c \end{bmatrix} where $c$ and $d$ are some real numbers and $f'(z)=c+id$.2012-02-24

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You could just write $(u+iv−(a_1+a_2 i))/(u+iv−(b_1+b_2 i))$ where $u,v,a_1,a_2,b_1,b_2$ are real. Then multiply the numerator and denominator by the complex conjugate of the denominator to find the real and imaginary parts.

Then later, exploit the Cauchy--Riemann equations to conclude that the matrix must have the form $\begin{bmatrix} {}\ \ \ c & d \\ -d& c\end{bmatrix}$ where $c$ and $d$ are some real numbers and f\;'(z)=c+id.

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    Thank you very much, Michael.2012-02-25