I can do all the questions apart from part iv). After some work, I got $\displaystyle df_{z}(1,0)=\frac{|z|^2-2z^2}{|z|^4}$. Unfortunately, the image of this under the metric is not 1/im(z)^2 as it ought to be.
I can do all the questions apart from part iv). After some work, I got $\displaystyle df_{z}(1,0)=\frac{|z|^2-2z^2}{|z|^4}$. Unfortunately, the image of this under the metric is not 1/im(z)^2 as it ought to be.
let $\displaystyle v=df(1,0)=\frac{(y^2-x^2, -2xy)}{(x^2+y^2)^2}$, $\displaystyle w=df(0,1)=\frac{(-2xy,x^2-y^2)}{(x^2+y^2)^2}$
$\displaystyle g(v,v)=\frac{v \cdot v}{y^2/(x^2+y^2)^2}=\frac{1}{(x^2+y^2)^2}\cdot\frac{(x^2+ y^2)^2}{y^2}=1/y^2$
If you take the real part of what I got above, you get $\displaystyle Re(\frac{|z|^2-2z^2}{|z|^4})=\frac{x^2+y^2-2(x^2-y^2)}{(x^2+y^2)^2}=\frac{3y^2-x^2}{(x^2+y^2)^2}$ which disagrees with your answer due to the 3y^2 instead of y^2.
Ok I've had another go. Can you tell me where I go wrong. Think of R^2 as C and let z=x+iy. Then, a(t)=z+t is a generating curve for e=(1,0).
$\displaystyle f(z)=\frac{z}{|z|^2}$ so $\displaystyle f(a(t))=\frac{z+t}{|z+t|^2}$ whose derivative is $\displaystyle \frac{|z+t|^2-(z+t)(2+x)t}{|z+t|^4}$so that evaluating at t=0 gives $\displaystyle \frac{1}{|z|^2}$ which appears even further from the truth!. I think it hinges on what the derivative of $\displaystyle |z+t|^2$ is. My reasoning was $\displaystyle |z+t|^2=|z|^2+t^2+2Re(tz)=|z|^2+t^2+2tx$ which differentaing w.r.t t gives $\displaystyle 2t+2x$. Thanks