Complex numbers - supremum

Question

For any complex numbers $z$ and $w$, find $\sup\limits_{\theta\in[0,2\pi]}\left|z+e^{i\theta}w\right|$.

I have no idea how to solve it. Thank you for help!

Answer 1

Suggestion: try to interpret geometrically, with vectors, what this formula means:

• the sum of two complex numbers: a sum of vectors (Chasles relation);
• the multiplication of a complex number by $e^{i \theta}$: a rotation of the vector;
• taking the modulus: evaluating a distance to the origin

Answer 2

Write in polar coordinates $z=r_z e^{i\arg(z)}$, $w=r_w e^{i \arg(w)}$. Then $z+e^{i\theta}w=r_z e^{i\arg(z)}+r_we^{i(\arg(w)+\theta)}.$ It is clear that $$ |z+e^{i\theta}w|\leq r_z+r_w. $$ We are done if we can find a $\theta$ such that the above is equality. How can you choose $\theta$ so that this is the case? As has been pointed out, a picture involving vectors may be helpful.

Answer 3

What about fixing $z$ and $w$ and work with the function $ f(\theta) = |z + we^{i\theta}|^2 $ ? One can use $\forall c\in\mathbb{C}, c\overline{c} = |c|^2$ to expand the modulus squared and then calculate the derivative to find the zeros.