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UPSC 2026 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Laurent's series in an annulus · Complex Analysis · asked 8× in 14 yrs · Read the full method →

Question

Prove that

where

Technique

The trigger is the shape of the claim: a two-sided series in $z$ for a function that is analytic everywhere except at $z=0$ — that is a Laurent expansion, and the printed real integral is nothing but the Laurent coefficient formula evaluated on the unit circle. Two things must be proved, not asserted: (i) that $c_{-n}=c_n$, which is what makes the two-sided series collapse into the printed $a_n(z^n+z^{-n})$ form — this comes from the functional symmetry $f(1/z)=f(z)$ together with uniqueness of Laurent coefficients; and (ii) that the complex coefficient integral is in fact real, i.e. that its imaginary part $\int_0^{2\pi}\cosh(2\cos\theta)\sin n\theta\,d\theta$ vanishes.

Solution

Step 1 — The function, its singularity, and the annulus of validity.

Put

The map $z\mapsto z+\dfrac1z$ is analytic on the punctured plane $\mathbb{C}\setminus\{0\}$, and $\cosh$ is entire; a composition of analytic maps is analytic, so

This punctured plane is an annulus, namely $A=\{z:\;r_1<|z|<r_2\}$ with $r_1=0$, $r_2=\infty$. This is the widest annulus available, and it is the region on which everything below is valid. (We record for later that $z=0$ is an isolated essential singularity of $f$: $\cosh w\to\infty$ along $w$ real and $\cosh w$ oscillates boundedly along $w$ imaginary, so $f$ has no limit, finite or infinite, at $0$.)

Step 2 — Laurent’s theorem.

Laurent’s theorem. If $f$ is analytic in the annulus $A=\{r_1<|z|<r_2\}$ then $f$ has a unique representation

the series converging absolutely on $A$ and uniformly on every compact subset of $A$, with

where $C_r:\;|w|=r$ is any positively oriented circle with $r_1<r<r_2$ (the value of $c_n$ is independent of $r$, by Cauchy’s theorem for the annulus).

Applying this to our $f$ on $0<|z|<\infty$:

Step 3 — The symmetry $c_{-n}=c_n$ (this is what collapses the series).

The function satisfies the functional equation

because $w\mapsto w$ is unchanged under $z\leftrightarrow 1/z$ in the argument $z+\frac1z$.

Crucially, the annulus $0<|z|<\infty$ is invariant under $z\mapsto 1/z$, so we may substitute $1/z$ into (1) and stay inside the region of validity:

the re-indexing $m=-n$ being legitimate because the Laurent series converges absolutely on $A$, so its terms may be rearranged freely.

By (2) the left side equals $f(z)=\sum_m c_m z^m$. Thus the same function has two Laurent expansions in the same annulus $0<|z|<\infty$; by the uniqueness clause of Laurent’s theorem the coefficients must agree:

Now split (1), which is permissible term-by-term by absolute convergence:

So the expansion already has the printed shape; it remains only to identify $c_n$ with the printed $a_n$.

Step 4 — Evaluate $c_n$ on the unit circle.

The coefficient formula holds for every radius $r\in(0,\infty)$. Choose $r=1$ — this is the choice that manufactures the printed integral, because on $|w|=1$ the combination $w+\frac1w$ becomes real:

Hence

Step 5 — The imaginary part vanishes.

Let

Substitute $\theta=2\pi-\phi$ (so $d\theta=-d\phi$, and $\theta:0\to2\pi$ corresponds to $\phi:2\pi\to0$). Using $\cos(2\pi-\phi)=\cos\phi$ and $\sin n(2\pi-\phi)=-\sin n\phi$,

Therefore $2I_n=0$, i.e.

(The substitution is the reflection of $[0,2\pi]$ about $\theta=\pi$; the weight $\cosh(2\cos\theta)$ is symmetric about $\theta=\pi$ while $\sin n\theta$ is antisymmetric about it.)

Substituting (6) into (5):

which is exactly the printed coefficient. In particular each $c_n$ is real.

Step 6 — Conclusion.

Combining (4) and (7):

valid for every $z$ with $0<|z|<\infty$, the convergence being absolute there and uniform on every closed sub-annulus $0<\rho_1\le|z|\le\rho_2<\infty$. $\qquad\blacksquare$

Step 7 — A second, independent proof of the symmetry (worth one line in the hall).

Formula (7) holds for all integers $n$, and $\cos(-n\theta)=\cos n\theta$; hence directly

This re-derives (3) without invoking uniqueness. Either route earns the mark; quoting both costs two lines and shows the symmetry is structural, not accidental.

Remark 1 — Half the coefficients are zero.

Replace $\theta$ by $\theta+\pi$ in the defining integral (a shift by a period of the $2\pi$-periodic integrand, so the value is unchanged). Since $\cos(\theta+\pi)=-\cos\theta$ and $\cosh$ is even, $\cosh(2\cos(\theta+\pi))=\cosh(2\cos\theta)$; and $\cos n(\theta+\pi)=(-1)^n\cos n\theta$. Hence

This is also visible from $f$ itself: $f(-z)=\cosh\!\left(-z-\frac1z\right)=f(z)$, so $f$ is an even function of $z$ and only even powers can occur.

Remark 2 — The coefficients in closed form.

Expanding directly, $\cosh w=\sum_{m\ge0}\dfrac{w^{2m}}{(2m)!}$ with $w=z+\frac1z$, and $(z+\frac1z)^{2m}=\sum_{j=0}^{2m}\binom{2m}{j}z^{2m-2j}$, the coefficient of $z^{2s}$ is

where $I_\nu$ is the modified Bessel function of the first kind. Equivalently, from the generating function $e^{\frac{x}{2}(t+1/t)}=\sum_{n}I_n(x)t^n$ with $x=\pm2$ and $I_n(-x)=(-1)^nI_n(x)$, one gets $\cosh(z+\frac1z)=\sum_{n\ \mathrm{even}}I_n(2)z^n$ at once. Numerically $a_0=I_0(2)=\sum_{k\ge0}\frac{1}{(k!)^2}$. Since infinitely many $a_{2s}>0$, the principal part of (1) is infinite — confirming that $z=0$ is an essential singularity, as claimed in Step 1.

Answer

Moreover $a_n=0$ for odd $n$, and $a_{2s}=I_{2s}(2)=\sum_{k\ge0}\dfrac{1}{k!\,(k+2s)!}$.