Show that $|\text{Re}(2+3z^2 +iz)| \leq 6$, when $|z|\leq 1$.
Not really sure where to go this inequality. I know that you need to start with the triangle inequality and do it two times but I am confused on the way it is supposed to turn out.
2026-03-25 01:17:08.1774401428
Show that $|\text{Re}(2+3z^2 +iz)| \leq 6$, when $|z|\leq 1$
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Notice that for any complex number $w$, $$|w|^2=|\text{Re}(w)|^2+|\text{Im}(w)|^2\geq |\text{Re}(w)|^2\implies |w|\geq |\text{Re}(w)|.$$ From the geometric point of view, it means that the hypotenuse is the longest side of a right-angled triangle. Hence $$|\text{Re}(2+3z^2 +iz)|\leq |2+3z^2 +iz|.$$ Can you take it from here?