# Help With a Proof using Contour Integration

• I
• dm4b
In summary, Astrom is evaluating an integral along a contour which makes up the imaginary axis. He has the following: -i\int_{-iR}^{iR} log(S(iw))dw=-i\int_{-iR}^{iR}log\left | S(iw) \right |dw -i\int_{-iR}^{iR}\angle S(iw)dw=-2i\int_{0}^{iR}log\left | S(iw) \right |dw. Taking a look at the second, or middle expression above, he makes the claim the second term, or imaginary part of the logarithm,

#### dm4b

TL;DR Summary
utilizing even and odd functions as the punchline
I am reading a proof in Feedback Systems by Astrom, for the Bode Sensitivity Integral, pg 339. I am stuck on a specific part of the proof.

He is evaluating an integral along a contour which makes up the imaginary axis. He has the following:

$$-i\int_{-iR}^{iR} log(S(iw))dw=-i\int_{-iR}^{iR}log\left | S(iw) \right |dw -i\int_{-iR}^{iR}\angle S(iw)dw=-2i\int_{0}^{iR}log\left | S(iw) \right |dw$$

Taking a look at the second, or middle expression above, he makes the claim the second term, or imaginary part of the logarithm, is odd, while the first term is even.

Given that the angle on the upper part of the y-ordinate is pi/2 and the lower part is -pi/2, the fact that the second term is odd seems obvious, so this integral vanishes.

I would like to show that the first term is even for any complex function.

This is easy to show for a simple example like $$f(z) = Re^{i \theta}$$

I am struggling to show it for any function $$f(z)=f(Re^{i \theta})$$

I wrote a MATLAB script that allows me to put in any f(z) I can think of and its always true to the claim. I would just like to "prove" it.

Well, I came up with the following if anyone is interested. I still don't feel good about this. Not even sure it's correct. If anyone has improvements, I am all ears!

We want to show that the integrand term is even in the following integral, for any function S(jw)
$$j\int_{-jR}^{jR}ln\left|S\left(jw\right)\right|dw$$
This would imply the following about the integrand.
$$ln\left|S\left(jw\right)\right|=ln\left|S\left(-jw\right)\right|$$
Given that S is a complex function of z, the following should also be true.
$$S\left(jw\right)=S\left(-jw\right)\rightarrow S\left(Re^{j\theta}\right)=S\left(Re^{-j\theta}\right)=S\left(Re^{j\theta}\right)^\ast$$

We also have the following general properties when considering the magnitude and complex conjugate of complex numbers.
$$\left|z\right|=\sqrt{zz^*}=\sqrt{z^*z}=\left|z^*\right|$$
This generalizes to any complex function of z, as follows.
$$\left|f\left(z\right)\right|=\sqrt{f\left(z\right)f\left(z\right)^\ast}=\sqrt{f\left(z\right)^\ast f\left(z\right)}=\left|f\left(z\right)^\ast\right|$$

We took advantage of the fact that complex numbers are commutative in the last two equations. Circling back around, the second equation above is the condition that must be satisfied for the integrand to be an even function. The properties of S(jw) shown in the third equation are immediately implied. We also know from the general properties of complex numbers that the magnitudes of a complex function and its complex conjugate are indeed equal. Therefore, we come back around to the first equation showing that the integrand is indeed an even function. This allows us to write the integrand as the following.

$$2j\int_{0}^{jR}ln\left|S\left(jw\right)\right|dw$$