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| -rw-r--r-- | ClassJune26.page | 8 |
1 files changed, 4 insertions, 4 deletions
diff --git a/ClassJune26.page b/ClassJune26.page index 27b30b9..06d0305 100644 --- a/ClassJune26.page +++ b/ClassJune26.page @@ -188,10 +188,10 @@ Consider a smooth map $f$ from the plane to itself; it takes a smooth 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 +to vectors based at $f(z)$. If we use rectangular coordinates $z\mapsto f(z)$ -$x+iy\mapsto u(x,y)+iv(x,y)$ +$x+iy\mapsto u(x,y)+iv(x,y)$ $\left(\begin{array}{c} x\\ @@ -201,7 +201,7 @@ v(x,y)\end{array}\right)$ then the derivative is $df=\left(\begin{array}{cc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ -\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{array}\right).$ +\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) standard basis to orthogonal vectors; the little square becomes a little rectangle. Since the diagonal of the square bisects the right @@ -230,7 +230,7 @@ b & a\end{array}\right),$ ie, it looks just like multiplication by the complex number $a+bi$. The function $f$ is conformal if its derivative acts like a nonzero complex number. Analytically, this condition is given by the following -differential equations, called the **Cauchy-Riemann equations**: +differential equations, called the **Cauchy-Riemann equations**: $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\mbox{ and }\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$ A complex function $f=u+iv$ is said to be **holomorphic** if $f$ satisfies the CR. We've shown that conformal $\Longrightarrow$ |