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| -rw-r--r-- | ClassJune26.page | 4 |
1 files changed, 2 insertions, 2 deletions
diff --git a/ClassJune26.page b/ClassJune26.page index b5454a4..b5dff79 100644 --- a/ClassJune26.page +++ b/ClassJune26.page @@ -216,7 +216,7 @@ angle, and $df$ is linear, its image must bisect too, that is, the little square must become a little *square*. Numerically, this means that $\left(\begin{array}{c} \frac{\partial u}{\partial y}\\ -\frac{\partial u}{\partial y}\end{array}\right)$ is $\left(\begin{array}{c} +\frac{\partial v}{\partial y}\end{array}\right)$ is $\left(\begin{array}{c} \frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial x}\end{array}\right)$ rotated by $\pi/2$. If we write $\left(\begin{array}{c} @@ -226,7 +226,7 @@ b\end{array}\right)=\left(\begin{array}{c} \frac{\partial v}{\partial x}\end{array}\right)$ , then $\left(\begin{array}{c} \frac{\partial u}{\partial y}\\ -\frac{\partial u}{\partial y}\end{array}\right)=\left(\begin{array}{c} +\frac{\partial v}{\partial y}\end{array}\right)=\left(\begin{array}{c} -b\\ a\end{array}\right).$ Put another way, the linear transformation $df(z)$ has the form |