Modulus and Argument of Cosh(iπ)?

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The modulus and argument of cosh(iπ) are determined to be 1 and π, respectively. The calculation begins with cosh(iπ) expressed as ½(eiπ + e-iπ), which simplifies to -1 since eiπ equals -1. It is emphasized that the modulus must be non-negative, confirming that the modulus is 1. The final conclusion is that cosh(iπ) has a modulus of 1 and an argument of π.
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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(eiπ+e-iπ) I am now a bit stuck as what to do as i have two terms in the form eix and I'm not sure homework to combine them to get the argument?
 
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struggles said:

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(eiπ+e-iπ) I am now a bit stuck as what to do as i have two terms in the form eix and I'm not sure homework to combine them to get the argument?
What is the value of eiπ?
 
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?
 
struggles said:
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.
 
so modulus of 1 and argument of pi?
 
struggles said:
so modulus of 1 and argument of pi?
Correct.
 

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