Modulus and Argument of Cosh(iπ)?

[struggles]

Homework Statement

So I'm trying to find the modulus and argument of
cosh(iπ)

Homework Equations

The Attempt at a Solution

so far coshπi = ½(e^iπ+e^-iπ) I am now a bit stuck as what to do as i have two terms in the form e^ix and I'm not sure homework to combine them to get the argument?

 

[Samy_A]

Homework Statement

So I'm trying to find the modulus and argument of
cosh(iπ)

Homework Equations

The Attempt at a Solution

so far coshπi = ½(e^iπ+e^-iπ) I am now a bit stuck as what to do as i have two terms in the form e^ix and I'm not sure homework to combine them to get the argument?

What is the value of e^iπ?

 

[struggles]

ah is it -1? in that case you'd have ½(-1-1) = -1. so it would just have radius of -1 modulus of π?

 

[Samy_A]

ah is it -1? in that case you'd have ½(-1-1) = -1. so it would just have radius of -1 modulus of π?

cosh(iπ) is indeed equal to -1.

But what are the modulus and the argument? Remember that the modulus is always 0 or positive, never negative.

 

[struggles]

so modulus of 1 and argument of pi?

 

[Samy_A]

so modulus of 1 and argument of pi?

Correct.