This document provides the steps to find all solutions to the equation cosh(z) = 1. It first takes the imaginary part of the equation, then sets it equal to the definition of cosh(z). This leads to solving a quadratic equation for e^z, giving the solutions z = i2πk, where k is any integer.

1. Find all solutions to cosh(z) = 1.
(Hint: look at the imaginary part of the equation first.
Solution
Solving it from definition:
cosh z = (1/2)(e^z + e^(-z)) = 1
==> e^z + 1/e^z = 2
==> e^(2z) - 2e^z + 1 = 0
==> (e^z -1)^2 = 0
==> e^z - 1 = 0
==> e^z = 1
==> z = log(1) = ln |1| + i arg(1)
==> z = i(0 + 2?k) for any integer k