## IMC 2013 Problem 6

**Problem 6.** Let be a complex number with . Prove that .

*Solution:* Write in its trigonometric form:

Then the condition translates to

which is equivalent to

Since must be positive, we have as a consequence

The condition is equivalent to

which translates to

Thus it is enough to prove that

Note that the result we need to prove is not trivial if and only if and on that interval is positive.

Denote then and we want to prove that

If we denote then .

If we evaluate then we obtain

which is a sum of two convex functions (first two and last two terms) among which one is strictly convex, and therefore is strictly convex (on ).

This means that is the global minimum of on and therefore is positive on that interval which finishes the proof.

## IMC 2013 Problem 5

**Problem 5.** Does there exist a sequence of complex numbers such that for every positive integer we have that converges if and only if is not a prime?

## Geometric mean for n complex numbers

Let be a closed disc in the complex plane. Prove that for all positive integers , and for all complex numbers there exists a such that .

*Romanian TST 2004*

## Every expansion has a 0

Suppose is entire such that for each at least one of the coefficients of the expansion is equal to 0. Prove that is a polynomial.

Read more…

## Power series & number theory

Let where denotes the number of divisors of . Calculate the radius of convergence of this series and prove that . Furthermore, prove that for .

## Amazing property of entire functions

Prove that if are non-constant, non-vanishing entire functions with then there exists an entire function such that are constant multiples of .

Read more…

## Blaschke factors

Take such that . Prove that if and and also if or .

Prove that for any fixed in the unit disk , the mapping has the following properties:

1) maps to and is holomorphic.

2) interchanges and .

3) .

4) is an involution on and thus is bijective.

5) Construct a bijective, holomorphic function, which maps the unit disk to itself, takes the unit circle into itself and interchanges two given complex numbers with .