The discussion centers on proving inequalities involving complex numbers, specifically the expression |e^iz| under certain conditions. Participants clarify that for |z| >= 3, the parameter b must be less than or equal to 0, leading to the conclusion that |e^iz| is bounded by 1 when the imaginary part of z is non-negative. The conversation highlights the need for precise definitions of variables a and b, as well as the correct interpretation of complex magnitudes.
PREREQUISITESStudents studying complex analysis, mathematicians working with inequalities in complex numbers, and educators seeking to clarify concepts related to Euler's formula and complex magnitudes.
oddiseas said:Homework Statement
Show that |e^iz|<=1 for Im(z)>=0
b)show that if |z|>=3 then the following inequality holds
Homework Equations
The Attempt at a Solution
|e^iz|=e^-b(cosa+isina)
and since the inmaginery part is always one, the magnitude is always less or equal to one' provided b is <=0. Now the second part i am having trouble with.
What following inequality? It doesn't appear in either your original post or the edited post.oddiseas said:b)show that if |z|>=3 then the following inequality holds