# Help with complex integration problem: $\left| \int_{[\sqrt2,\sqrt2i]}\frac{1}{z-(1+i)}dz\right| \le 2(\sqrt2+1)$

I found a math problem (of a 2002 exam) I can't seem to solve, that is simply stated as:

Show that $$\left| \int_{\left[\sqrt2,\sqrt2i\right]}\frac{1}{z-(1+i)}dz\right| \le 2(\sqrt2+1)$$

I looked around and found two properties that might help me in this case. The first one is

$$\left| \int_a^bf(t)dt\right| \le \int_a^b\left|f(t)\right|dt$$

Parameterezing $[ \sqrt2,\sqrt2i]$ as $\sqrt2(1-t)+\sqrt2it, 0\le t \le 1$, a and b become 0 and 1, respectively, and this can be solved by substitution($f(z)\rightarrow f(\sqrt2(1-t)+\sqrt2it)\cdot(\sqrt2(1-t)+\sqrt2it)'$, wielding, hopefully, $2(\sqrt2+1)$.

However, I've been unable to solve the integral on the right, which leads me to the second property, which, at first glance, seems a bit easier and more directly applied, but, again, I'm not getting the results I hoped. It follows:

Let $\gamma : [a,b] \rightarrow D$ a path, and f be a complex function, so that f is continuous in $tr(\gamma)$ (? - see note) (or $\varphi(t)=f(\gamma(t))$ is continuous in [a,b]). If there is $M \in R^+$, such that $\left|f(z)\right|\le M, \forall z \in tr(\gamma)$, then:

$$\left|\int_\gamma f(z)dz\right|\le Ml(\gamma)$$ where $l(\gamma)$ is the length of the path $\gamma$.

In this case, $\gamma$ is simply a line segment, so $l(\gamma)$ is quite easy to calculate, $l(\gamma)=|\sqrt2i-\sqrt2| = \sqrt{\sqrt2^2+\sqrt2^2}=\sqrt4=2$, so all that's left is to show that $\left|f(z)\right|\le (\sqrt2+1)$, which I, again, seem to be unable to do.

Through derivation, I was able to find the minimum of $|z-(1+i)|$ (and then the maximum of $|f(z)|$ ), and got the result $\left|f(z)\right|\le \frac{2}{\sqrt{3-2\sqrt2}}$, which, technically, shows what was asked (since $\frac{2}{\sqrt{3-2\sqrt2}}=0.41<\sqrt2+1$, but I really doubt that was the intended result.

(Edit: As Daniel Fischer noted in his answer below, this calculation (that wields 0.41) is actually incorrect. I leave the error unedited, as a cautinary tale for the use of Windows calculator)

Any help you can provide that at least points me in the right direction, or highlights something I've missed and/or misinterpreted, will be greatly appreciated.

Note: The exercise and properties mentioned were not in english, and I translated them myself, so there may be some errors, particularly in notation/nomenclature. If you spot such errors, please either let me know by comment or edit the answer. Thank you.

The second method,

$$\left\lvert \int_\gamma f(z)\,dz \right\rvert \leqslant \operatorname{length} \gamma \cdot \max \{\lvert f(z)\rvert : z \in \operatorname{Tr} \gamma\},$$

yields the desired result. As you determined, the length of the segment connecting $\sqrt{2}$ and $\sqrt{2}\cdot i$ is $2$, and the maximum of $\lvert f(z)\rvert$ is the reciprocal of the smallest distance of the segment to $1+i$. By elementary geometry, the closest point on the segment to $1+i$ is the intersection of the segment with the line $\operatorname{Re} z = \operatorname{Im} z$, in your parametrisation corresponding to $t = \frac{1}{2}$, the point $\frac{1+i}{\sqrt{2}}$. The distance is $$\lvert 1+i\rvert - \left\lvert \frac{1+i}{\sqrt{2}}\right\rvert = \sqrt{2} - 1,$$

so $\max \{\lvert f(z)\rvert : z \in \operatorname{Tr} \gamma\} = \frac{1}{\sqrt{2}-1} = \sqrt{2} + 1$.

Note that $3 - 2\sqrt{2} = 2 - 2\sqrt{2}+ 1 = (\sqrt{2}-1)^2$, so

$$\frac{2}{\sqrt{3-2\sqrt{2}}} = \frac{2}{\sqrt{2}-1} = 2(\sqrt{2}+1),$$

and it looks like you introduced some errors when writing your computations in the question, the figures hint at your actual computations being correct.

• Oh, wow, I did get the right result, the 0.41 was an errorr - that's what I get for using Windows calculator. $t=\frac{1}{2}$ is what I got from my derivation, and then got the point $\frac{1+i}{\sqrt2}$. However, your method is much better - mine was about one page long, and downright confusing. Thank you for your clear explanation. Jan 22 '14 at 22:40