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| -rw-r--r-- | ClassJune26.page | 12 |
1 files changed, 6 insertions, 6 deletions
diff --git a/ClassJune26.page b/ClassJune26.page index 06d0305..5267204 100644 --- a/ClassJune26.page +++ b/ClassJune26.page @@ -189,17 +189,17 @@ curve $\gamma$ through $z$ to a smooth curve $f\circ\gamma$ through $f(z)$. What happens to the tangent of $\gamma$ at $z$? Given by the derivative $df(z)$, a linear map taking vectors based at $z$ to vectors based at $f(z)$. If we use rectangular coordinates -$z\mapsto f(z)$ + $z\mapsto f(z)$ -$x+iy\mapsto u(x,y)+iv(x,y)$ + $x+iy\mapsto u(x,y)+iv(x,y)$ -$\left(\begin{array}{c} + $\left(\begin{array}{c} x\\ y\end{array}\right)\mapsto\left(\begin{array}{c} u(x,y)\\ -v(x,y)\end{array}\right)$ -then the derivative is -$df=\left(\begin{array}{cc} +v(x,y)\end{array}\right)$ +then the derivative is + $df=\left(\begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).$ If $f$ is conformal, then this matrix had better take the (orthogonal) |