Thread: Urgent advice please on a complex no question:

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

Originally Posted by grahamlee

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 is in , right? It is such that , while has no reason to be equal to 0, so that diverges at , hence is not even defined on ...

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.

Originally Posted by grahamlee

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 instead of ): you can't use the inequality at the denominator since it would give you an upper bound. And again, look at what happens when : the denominator is 0, so you can't expect to bound it away from 0.