Problem sheet 9 March 9th, 2004

MT290 Complex variable

Ex. 1 Remark: Since not everybody appreciated this problem, let’s do it again; work through the solution I gave out. You can hand it in. If you successfully did this problem before, then work out ∫_{0}^{∞} e^{−x^{2}} cos(ax) dx instead. This problem shows that a) not only semicircle contours help to evaluate integrals along the real line; b) that you can reduce an integral like ∫ e^{−x^{2}} cos ax dx to ∫ e^{−x^{2}}, dx. The latter is not really easy, but at least well known so that you can use it here. Let f : ℂ → ℂ be the function with f(z) = e^{−z^{2}}. Let R(K) denote the rectangle defined by the four points P_{1} = −K + 0i, P_{2} = +K + 0i, P_{3} = +K + \frac{1}{2}i, P_{4} = −K + \frac{1}{2}i. Let γ_{1} denote the path along the edge connecting P_{1} and P_{2}, Let γ_{2} denote the path along the edge connecting P_{2} and P_{3}, Let γ_{3} denote the path along the edge connecting P_{3} and P_{4}, Let γ_{4} denote the path along the edge connecting P_{4} and P_{1}.

Note: it is advisable to use a parametrisation for the contour lines that keeps z simple but shifts any difficulty to the boundaries. For example, for γ_{1} use φ(t) = z = t, where −K ≤ t ≤ K. This keeps e^{−z^{2}} much simpler than z = −K + 2tK, with 0 ≤ t ≤ 1. So, which simple parametrisation do you get for γ_{2} etc?

i) Draw the integration contour in the Argand diagram.

ii) Show that ∫_{∂R(K)} f(z) dz = 0. Here ∂R(K) denotes the boundary of the rectangle R(K).

iii) Show that lim_{K→∞} ∫_{γ_{2}} f(z) dz = 0, and similarly lim_{K→∞} ∫_{γ_{4}} f(z) dz = 0. Use

the above results, and (without proof) the well known result ∫_{−∞}^{∞} e^{−x^{2}} dx =

\sqrt{π} to conclude that ∫_{0}^{∞} e^{−x^{2}} cos x dx = \frac{\sqrt{π}}{2e^{1/4}} .

Ex. 2 (Use the method of Monday 8th of March.) Find power series for the following functions about the points stated and give the radius of convergence for each of the series.

a. \frac{1}{2 − z} about z = 0; b. \frac{1}{2 − z} about z = 12; c. \frac{5}{(1 − z)(4 + z)} about z = 0.

d. e^{z} about z = i. e. \frac{1}{3 − z} about z = 4i.

Ex. 3 Assuming that it is alright to integrate a power series term by term within its radius of convergence (it is !) use the series for (1 + z)^{−1} to obtain the power series:

log(1 + z) = ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} z^{n}.  (∗)

What is the radius of convergence of this series? Let z = iy in (*). Take the imaginary part to obtain the series for arctan y.

Ex. 4 Find the first few coefficients of the Taylor series of tan z in two different ways.

a) using c_{n} = \frac{f^{(n)}(0)}{n!}. When you are tired of differentiating try

b)

tan z = \frac{sin z}{cos z} = \frac{z − \frac{z^{3}}{3!} + \frac{z^{5}}{5!} ∓}{1 − \frac{z^{2}}{2!} + \frac{z^{4}}{4!} ∓} = c_{0} + c_{1}z + c_{2}z^{2} + …

Multiply by cos z and find c_{0}, c_{2}, c_{4}, … Then find, c_{1}, from this c_{3} etc.

Ex. 5 Complete the following explanation of Taylor’s theorem: Perhaps you wondered

where this formula c_{n} = \frac{f^{(n)}(0)}{n!} comes from. Let’s try to filter out the coefficient c_{k} from f(z) = ∑_{n=0}^{∞} c_{n}z^{n}. A good filter

is our favourite integral ∫_{|z|=1} z^{n} dz = \begin{cases}2πi & if n = −1 \\ 0 & otherwise. \end{cases}

∫_{|z|=1} f(z) dz would not lead anywhere, for a differentiable function this is just 0. But let’s try

∫_{|z|=1} \frac{f(z)}{z^{k+1}} dz = ∫_{|z|=1} \frac{∑_{n=0}^{∞} c_{n}z^{n}}{z^{k+1}} dz = ∫_{|z|=1} ∑_{n=0}^{∞} c_{n} \frac{z^{n}}{z^{k+1}} dz

Now go on, exchange the integration and summation (you are allowed to do it!), use our filter, combine with Cauchy’s integral formulae and find the expression for c_{k}.