How to Find an Analytic Function Given Specific Conditions?

[sparkster]

So let f be analytic in the open unit disk and continuous on the closed unit disk. Also, |f(z)|=1 for |z|=1, all zeros are simple zeros at 0, and f'(0)=-1/2.

I need to find f.

I've tried using the cauchy integral formula for f' but that's not getting me anywhere. Can anyone point me in the right direction?

 

[ObsessiveMathsFreak]

Perhaps you could use the fact that the real an imaginary parts of a complex differentiable function are harmonic, i.e.
$$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 v}{\partial x^2}+\frac{\partial^2 v}{\partial y^2}= 0$$
in the disc.

Also on the boundary
$$|f|=1=|f|^2=u^2+v^2=1$$

And for the derivative you have
$$f'(0)=\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}=\frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y}=-\frac{1}{2}$$

This gives
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=-\frac{1}{2}$$
$$\frac{\partial v}{\partial x}=\frac{\partial u}{\partial y}=0$$
At the origin.

But is this enough to solve the partial differential equation?

 

[tacman]

Based on the given information, we can use the Maximum Modulus Principle to show that f(z) must be a constant function. Since |f(z)|=1 for |z|=1, this means that f(z) cannot attain any values outside of the unit circle. Therefore, by the Maximum Modulus Principle, f(z) must be a constant function.

To determine the specific value of this constant, we can use the fact that f'(0)=-1/2. Using the Cauchy-Riemann equations, we can write f'(0)=u_x(0)+iv_x(0)=-1/2, where u(x,y) and v(x,y) are the real and imaginary parts of f(z) respectively. Since u and v are harmonic functions, we can use the Laplace equation to solve for u and v.

Solving for u first, we have u_x(0)=u_y(0)=0. So u(x,y)=c1 (a constant). Similarly, solving for v, we have v_x(0)=-v_y(0)=1/2. So v(x,y)=c2-y/2.

Therefore, f(z)=u(x,y)+iv(x,y)=c1+(c2-y/2)i.

Finally, using the condition that all zeros of f(z) are simple zeros at 0, we can set f(0)=0. This gives us the equation c1+(c2-0/2)i=0, which means c1=0 and c2=0. Therefore, f(z)=-yi/2.

In conclusion, f(z)=-yi/2 is the unique function satisfying the given conditions.