Complex analysis question including maximum of real part of $x+i$

Question

I'm asked to prove that:

$$\max_{x\in \mathbb{R} } \operatorname{Re}(x+i)^{10} = -\left(\sin \left(\frac{\pi}{18}\right)\right)^{-9}$$

any ideas? i don't even know where to start.

Answer 1

You can start by expanding $x+i$ :

$$(x+i)^{10}={10 \choose 0}x^{10}+ {10\choose 1}x^9i^1+{10\choose 2}x^8i^2...{10 \choose 10}i^{10}=\\\\x^{10}+ {10\choose 1}x^9i^1-{10\choose 2}x^8...-{10 \choose 10}\\\\$$

Then, any term left over that still contains $i$ could be changed to complex notation, $$i=e^{i\pi/2} => i^m = e^{i^m(\frac{\pi}{2})^m}=cos(\frac{\pi^m}{2^m})+sin(\frac{\pi^m}{2^m})i^m$$

I think that $x^{10}$ would win out in the end. You can show that any term without $i$ will be smaller than $x^{10}$ because it's power of x will be smaller, and any term multiplied by a $sin$ or $cos$ will also be smaller because $sin$ and $cos$ are always smaller than 1.