Relaxation Problem I

Calculus Level 5

Evaluate

$\large\left| \int_{|z| = 1} \frac{e^{z}}{z} \ dz \right|,$

where $z \in \mathbb{C}$ .

The answer is 6.28318530.

1 solution

Jake Lai
Jul 6, 2015

This is just a simple application of Cauchy's integral formula, which states that

$f(a) = \frac{1}{2i\pi} \oint_{\gamma} \frac{f(z)}{z-a} \ dz$

where $z$ lies on the boundary $\gamma$ of a disk and $a$ lies within that disk. Here, we obviously have $f(z) = e^{z}$ and $a = 0$ . Thus, we get that

$\oint_{\gamma} \frac{e^{z}}{z} \ dz = 2i\pi e^{0} = 2i\pi$

(Take note that the radius of $\gamma$ is irrelevant.) Hence, we have

$\left| \int_{|z|=1} \frac{e^{z}}{z} \ dz \right| = |2i\pi| = \boxed{2\pi}$

Moderator note:

Without resorting to Cauchy's formula, how can one evaluate $\int \frac{1}{z} \, dz$ about the unit circle?

@Calvin Lin Do a substitution $z = e^{i\theta}$ (the integral is over the unit circle so the parametrisation has $r = 1$ ) and the integral becomes

$\int_{0}^{2\pi} e^{-i\theta} \ de^{i\theta} = \int_{0}^{2\pi} e^{-i\theta} \cdot ie^{i\theta} \ d\theta = 2i\pi$

Jake Lai - 5 years, 11 months ago

Great! So another way of approaching this is to "secretly do the cauchy without explicitly mentioning it", by power-series converting $e^z = 1 + z + \frac{z^2}{2!} + \ldots$ .

That of course, is also an explanation for why Cauchy Integral Formula works out.

Calvin Lin Staff - 5 years, 11 months ago

Indeed. I recently bought Lars Ahlfors's Complex Analysis . Despite it being (apparently) a new edition, a little bit of the notation is foreign to me (but at least comprehensible). It's a great introduction to complex analysis. I'll probably be posting more problems as I read on.

Jake Lai - 5 years, 11 months ago

I think the question is suitable for level-1 or 2 but definitely not higher than level-3

Bhaskar Pandey - 2 weeks, 1 day ago

Relaxation Problem I

The answer is 6.28318530.

1 solution

Moderator note:

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