Let $f(z)$ be an entire function satisfying $z\dfrac{f^{'}(z)}{f(z)} = z^{2}\dfrac{f^{'}(z^{2})}{f(z^{2})}$ show that $f(z) = cz^{m}$.

Question

a) Let $f(z)$ be an entire function satisfying the condition $$z\dfrac{f^{'}(z)}{f(z)} = z^{2}\dfrac{f^{'}(z^{2})}{f(z^{2})}$$ whenever $f(z) \neq 0$. Show that if $f(0) = 0$, then $$z\dfrac{f^{'}(z)}{f(z)}$$ is a function that is analytic at $z=0$.

b) Show that $$f(z) = cz^{m}$$ for some constant $c \in \mathbb{C}$ and positive integer $m$.

For part a), please see if i did correctly,since we are given that $f(0) = 0$, we know that $f$ has a zero of order $m$ at $0$. Hence we can write $$f(z) = z^{m}g(z)$$ where $g(z)$ is analytic at $0$ and $g(0) \neq 0$. It follows that $$f^{'}(z) = mz^{m-1}g(z)+z^{m}g^{'}(z)$$

Hence by subsituting we derive $$z\dfrac{f^{'}(z)}{f(z)} =m+\frac{zg^{'}(z)}{g(z)}$$

Hence this function is analytic at $z=0$ because $g(z)$ is analytic at $0$. (Can someone explain to me: $g(z)$ analytic at $0$ means $g(z)$ is analytic in a small neighborhood of $0$. But how can i ensure that $g(z)$ does not have zeroes in this small neighborhood? If $g(z)$ has zero inside, then the expression $m+\frac{zg^{'}(z)}{g(z)}$ is not analytic at $0$.

And for part $b$, anyone can write out their solution? I do have a model answer from my professor but find it hard to understand.

Answer 1

For part a), we can ensure that $g(z)$ does not have a zero in the neighbourhood of $0$ that we choose by choosing a small enough neighbourhood.If $g$ had zeros in every neighbourhood of $0$, then $0$ would be a limit point of zeros of $g$, and hence we would have that $g(0)=0$ by continuity. (In fact we would have that $g$ is identically $0$, since non-constant analytic functions have isolated zeros.)

For part b), we can modify my argument in this question: Prove that a function that satisfies $f(x^{2^{n}})=f(x)^{2^{n}}$ is constant given $f(0)=1$

Define the function $$ h(z) = z \, \frac{f^\prime (z)}{f(z)} $$

We note that $$ h\left( z^2 \right) = z^2 \, \frac{f^\prime \left(z^2 \right)}{f\left(z^2 \right)} = z \, \frac{f^\prime (z)}{f(z)} = h(z) $$ for all complex numbers $z$ where $h$ is defined.

We have shown that $h$ is analytic in a neighbourhood of $0$.

We also note that you derived that $$ h(z) = m + z \, \frac{g^\prime(z)}{g(z)} $$ where $m \geq 1$ is the order of $f$ at $0$. This gives us that $h(0) = m$, and so we can write $$ h(z) = m k(z) $$ where $k$ is some function analytic in a neighbourhood of $0$ such that $k(0) = 1$. We can check that $h \left( z^2 \right) = h(z)$ gives us that $k \left( z^2 \right) = k(z)$.

Since $k$ is analytic in a neighbourhood of $0$, there is a ball $B = B(0, \varepsilon)$ on which $k$ is analytic, and we can choose $\varepsilon$ so that $\varepsilon \leq 1$.

Then for any $z \in B$, we have that $|z| < 1$, and so by following the argument in the question I linked, this gives us that $|k(z)| = 1$.

Thus $|k(z)|$, and hence $|h(z)|$ is constant on $B$, and as noted in the linked question, it is a well-known fact that this implies that $h$ itself is constant on $B$. We see that $h(z) = m$ for all $z$ in $B$.

But we have that $$ h(z) = m + z \, \frac{g^\prime (z)}{g(z)} $$ for all $z$ in $B$, and so it follows that $g^\prime (z) = 0$ for all $z \neq 0$ in $B$. Thus $g$ is constant on $B$, and so $f(z) = c z^m$ for all $z$ in $B$, where $c = g(0)$ is some constant.

But then it is a standard argument that this then implies that $f(z) = cz^m$ for all complex $z$. Indeed, the zeros function $z \mapsto f(z) - cz^m$ has a limit point, and so $f(z) - cz^m$ must be identically $0$.

Of course the standard argument to show that if the zeros of a function have a limit point then the function is identically $0$ goes by considering the Taylor expansion of the function around that limit point to show that the function is identically zero in a neighbourhood of that point, so this answer is probably just hiding the computations with Taylor series from the other answers that have been posted.

Answer 2

Concerning part b, let $h(z) = z\frac{f'(z)}{f(z)}$. Then $h$ is analytic at $0$ and statisfies $h(z) = h(z^2)$. Since $h$ is analytic, there is a Maclaurin series $$h(z) = \sum_n a_nz^n$$ But then $$\sum_n a_nz^n = \sum_n a_nz^{2n}$$

By differentiating both sides repeatedly, we see that it is only possible for two power series to be equal is if the coefficients of the same powers are the same in each. But let $k$ be the smallest index $> 0$ such that $a_n \ne 0$. The lhs has a non-zero term $a_kz^k$, but the rhs has no non-constant term of lower degree than $2k$, a contradiction. So no such $k$ can exist, which means $a_n = 0$ for all $n > 0$, and $h$ is constant.

I'll leave demonstrating that this means $f(z) = cz^m$ to you.

Answer 3

Alternative solution: if $f$ has a zero of order $m$ at zero, then $f'$ has a zero of order $m-1$ at zero. Then $zf'(z)/f(z)$ has a removable singularity at $z = 0$ and can be extended to a holomorphic function $g$ in a nhood of $0$ with $g(0) = m$. This $g$ verifies $g(z) = g(z^2)$, i.e. its Taylor series verifies: $$ a_0 + a_1 z + a_2 z^2 + a_3 z^3 + \dots = a_0 + a_1 z^2 + + a_2 z^4 + \dots $$ and this implies $$a_1 = a_3 = a_5 = \cdots = 0,\qquad a_{2k} = a_k.$$ Using both properties, $$\forall k>0:\ a_k = 0$$ and $g(z) = a_0 = g(0) = m$. In some open set: $$\frac{f'(z)}{f(z)} = \frac{m}z,$$ $$\log f(z) = m\log z + b,$$ $$f(z) = cz^m.$$ By the identity theorem, this is true in all the plane.

EDIT: $\log$-less version. $$zf'(z) = mf(z),$$ $$\sum_{n=1}^\infty n c_n z^n = \sum_{n=0}^\infty m c_n z^n,$$ I.e., $$c_0 = 0,\qquad\forall n>0:\ n c_n = m c_n.$$ $$\cdots$$

Answer 4

Can someone explain to me: analytic at  $0$ means  $f(z)$ is analytic in a small neighborhood of $0$.

Since analyticity is a neighborhood property, it implies that $f(z)$ is analytic in $|z|<\epsilon$.

But how can i ensure that $g(z)$ does not have zeroes in this small neighborhood?

Suppose $g(z)$ has zeroes in a neighborhood of $z=0$ then by Identity theorem $g(z)=0$ identically, which contradicts your assumption $f(z)=z^m.g(z)$ s.t. $g(0)\ne 0$.