Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 11 / a

Multiply numerator and denominator by the complex conjugate $2+i$:Thus the real part and imaginary part are

Powers of the imaginary unit repeat in blocks $i,-1,-i,1$, each with sum zero. The first $44$ terms comprise eleven complete blocks; the last is $i^{45}=i$. Applying Euler's formula to the resulting complex exponential gives $e^i=\cos 1+i\sin 1$. HenceThe angles here are in radians.

The exponential definitions of hyperbolic sine and hyperbolic cosine, together with Euler's formula, give $\sinh (iu)=i\sin u$ and $\cosh (iu)=\cos u$. The argument of the complex logarithm therefore reduces to $i\sin (\pi /2)+\cos (9\pi /2)=i$.

Using the principal logarithm, $\log i=i\pi /2$, soIf no branch of the complex logarithm is selected, its complete set of values is $i(\pi /2+2\pi k)$, $k\in \mathbb{Z}$. The real part remains zero, while the imaginary part is $\pi /2+2\pi k$. The original paper does not explicitly state a branch convention.

The literal expression in the original PDF is undefined. Indeed, hyperbolic tangent satisfies $\tanh w=\sinh w/\cosh w$, andAlready the $n=2$ summand contains $\tanh (i\pi /2)$, a pole. The shifted term has the same issue when $n\equiv 3(\mathrm{mod}4)$. Thus this is not a finite sum of defined complex numbers, and it has no literal real or imaginary part. Cancelling two undefined terms is not ordinary arithmetic.

There is nevertheless a natural interpretation by meromorphic telescoping cancellation. Insert the same small positive real shift $\epsilon$ into every hyperbolic tangent argument. Each term is then finite, and the telescoping sum isSince hyperbolic tangent has period $i\pi$, its first argument can be reduced to $\epsilon +i\pi /4$. Taking the limit givesThis is a common-shift continuation of the sum, plausibly the intended telescoping answer. It requires that additional convention; it is not the value of the expression as printed.