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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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    $\begingroup$ Recall that $$\frac{1}{z}=\frac{\bar{z}}{|z|^2}$$ which should let you write $f$ in terms of its real and imaginary parts. $\endgroup$ – Alex Becker Feb 24 '12 at 21:36
  • $\begingroup$ @AlexBecker: Thank you, but $a,b$ are complex numbers and I am not sure how to split them... $\endgroup$ – bleu Feb 24 '12 at 21:40
  • $\begingroup$ If $f(z)=g(z)+c$ where $c$ is complex, then $f_1(z)=g_1(z)+\Re(c)$ and $f_2(z)=g_2(z)+\Im(c)i$. $\endgroup$ – Alex Becker Feb 24 '12 at 21:44
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    $\begingroup$ 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 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 $f'(z)=c+id$. $\endgroup$ – Michael Hardy Feb 24 '12 at 22:02
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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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  • $\begingroup$ Thank you very much, Michael. $\endgroup$ – bleu Feb 25 '12 at 9:14

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