2
$\begingroup$

According to the question, we must let f be a function, which is holomorphic on the plane and satisfies $f'(z) = f(z)$ and $f(0)=1$.

We are also told to assume that a function which is holomorphic on the plane with zero derivative is constant.

I know that a holomorphic function is a "complex-valued function of one or more complex variables that is complex differntiable", but I am still unsure how to answer the question above.

Could someone please offer some assistance?

Thank you :)

  • 3
    What is $g(0)$? What is $g'(z)$? – 2017-02-05
  • 0
    We are not told. All the information that was given, I've put above. – 2017-02-05
  • 5
    Sorry, that was not a request for clarification but a hint how to prove the assertion. – 2017-02-05

1 Answers 1

2

First of all, you don't need to know anything about holomorphic functions other than that they are differentiable here, which means that $f'(z)$ exists (which is trivially given when it is stated that $f(x) $ is equal to its derivative)

If you have to use $g(z) $, then, as Daniel Fischer said, I would note that $g(0)=f(0)^2 = 1$ and that $g'(z) = f'(z)f(-z) - f(z)f'(-z) = 0$. Since the derivative is zero, what does this say about $g(z)$?

If you didn't have to use $g(z)$, I would note that the differential equation $f'(z) = f(z) $ has only the non-trivial solution $f(z) = c_1 e^z$. Knowing that $f(0)=0$ gives an initial condition and allows us to conclude.