analyse the saddle point of $(x+iy)^n$ at (0,0)

Question

I've got the following question.

$f(x,y)=\sum_{k=0}^{\lfloor{n/2}\rfloor}{n\choose k}x^{n-2k}(-y^2)^k$

$(f(x,y) \equiv Re((x+iy)^n)$

I need to explain the function near $(0,0)$.

I've calculated that $f , \nabla$ and the Hessian of $f$ is zero at $(0,0)$.

Now I want to show this point is a saddle point , meaning I want to find $x,y$ such that $f(x,y)>0$ and $x_1,y_1$ such that $f(x_1,y_1)<0$

I've changed the coordinates to polar coordinates $(x=rcos\theta , y=rsin\theta$ , $r>0$ , $0\leq\theta<2 \pi)$ and got the next equation:

$f(x,y)=\sum_{k=0}^{\lfloor{n/2}\rfloor}{n\choose k}({rcos\theta})^{n-2k}(-({rsin\theta})^2)^k$ $\Rightarrow$
$f(x,y)=r^n({cos\theta})^{n}\sum_{k=0}^{\lfloor{n/2}\rfloor}{n\choose k}(-1)^{k}({sin\theta})^{2k}({cos\theta})^{-2k} $

Now I am looking on $0<\theta<\pi/2$ and I cannot understand is this sum converges to a negative number?

Or maybe there is another method which will help me o determine the directions which the function is increasing/decreasing near this saddle point.

Thank you, Michael

Answer 1

It is a saddle point because:

• $\nabla f(0,0)=0$.
• If $x>0$, $f(x,0)=x^n>0$.
• If $x>0$ and $\theta=\frac\pi n$, then $f(x\cos\theta,x\sin\theta)=-x^n<0$.

Answer 2

Try

$$ \left(\sqrt{x^2+y^2}e^{i\phi}\right)^n = (x^2+y^2)^{n/2}(\cos(n\phi)+i\sin(n\phi)) $$

with $\phi(x,y) = \arctan(y/x)$

Now, depending on the value of $\phi(x,y)$...