 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2021/2/solution

Using the hyperbolic functions identities

sin (i x) = i sinh x

and

cos (i x) = cosh x

, divide by

i

and put

u = e^{x} \neq = 0

. The equation becomes

- u /2 - 5/ (2 u) = 3

, hence

u^{2} + 6 u + 5 = (u + 1) (u + 5) = 0.

Taking all branches of the complex logarithm gives

x = (2 k + 1) πi or x = ln 5 + (2 k + 1) πi, k \in Z .

x

is required to be real, there are no real solutions, since

e^{x}

cannot be negative. The complex branches are essential to the unrestricted answer.

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