Let B = {z:|z|<3, Re z + Im z <= 0}
prove that the function f(z) = cosh(1+iz^2)/cos z
is bounded on the set delta B
Could anyone give me a steer on how I should go about solving this?
Many thanks
Graham
Your question seems weird to me; a typo, maybe?
The value $\displaystyle -\frac{\pi}{2}$ is in $\displaystyle B$, right? It is such that $\displaystyle \cos\left(-\frac{\pi}{2}\right)=0$, while $\displaystyle \cosh(1+i\left(-\frac{\pi}{2}\right)^2)$ has no reason to be equal to 0, so that $\displaystyle f$ diverges at $\displaystyle -\frac{\pi}{2}$, hence $\displaystyle f$ is not even defined on $\displaystyle B$...
I'm not sure to be honest. The approach I have taken is as follows:
|cosh(1+iz^2)/cos z|
= |1/2 (e^(1+iz^2)+e^-(1+iz^2)) / 1/2(e^iz+e^-iz)|
= e^1+|z^2|+e^1+|z^2| / |e^z|+|e^z|
= e^1+|z^2| / e^|z|
When |z|< 3, this equates to e^7
It looks right but I can't find a way to check this.
It isn't right. Most notably (I assume you mean $\displaystyle \leq$ instead of $\displaystyle =$): you can't use the inequality $\displaystyle |e^{iz}+e^{-iz}|\leq |e^{iz}|+|e^{-iz}|$ at the denominator since it would give you an upper bound. And again, look at what happens when $\displaystyle z=-\frac{\pi}{2}$: the denominator is 0, so you can't expect to bound it away from 0.