Separate real and imaginary part of a transfer function

Question

So I have this fourth order transfer function

$$ G(s) = \frac{2}{(1+i\tau\omega)^4} $$

I have figured out that multiplying with the complex conjugate let's you separate the real and imaginary part. However, I don't seem to find the right answer according to my book.

I end up with

$$ Re = \frac{2 - 12\tau^2\omega^2 + 2\tau^4\omega^4}{(1+\tau^2\omega^2)^4} $$

$$ Im = \frac{-8\tau\omega + 8\tau^3\omega^3}{(1+\tau^2\omega^2)^4} $$

Calculating the phase angle leaves me with

$$ \arctan(\frac{-8\tau\omega + 8\tau^3\omega^3}{2 - 12\tau^2\omega^2+2\tau^4\omega^4}) $$

While the book says the solution is $$ -4\arctan(\tau\omega) $$

Comparing the solution of the modulus suggests that my approach is correct, since the denominator is the same.. Am I missing something? I don't see how I can ever reduce the fraction to the one from the solutions.

Answer 1

\begin{eqnarray*} \tan^{-1}(A)+\tan^{-1}(B) &=& \tan^{-1}(\frac{A+B}{1-AB}) \end{eqnarray*} Is that helpful ? \begin{eqnarray*} -\tan^{-1}(A) &=& \tan^{-1}(-A). \end{eqnarray*} Maybe they are the same !

Answer 2

$$arg(1+irw)=\arctan\frac{rw}{1}=\arctan rw$$ $$arg\frac{\sqrt[4]{2}}{1+irw}=-\arctan rw$$ because numbers doesn't effect on arg, $$arg\frac{2}{(1+irw)^4}=arg\Big(\frac{\sqrt[4]{2}}{1+irw}\Big)^4=4arg\frac{\sqrt[4]{2}}{1+irw}=-4\arctan rw$$

Answer 3

$$\arg\frac {4}{(1+ir\omega)^4}=\arg 4-4\arg(1+ir\omega)=0-4\arctan r\omega$$

Answer 4

Observe that

$$\tan 4\theta=\frac{2\tan2\theta}{1-\tan^22\theta}=\frac{2\dfrac{2\tan\theta}{1-\tan^2\theta}}{1-\left(\dfrac{2\tan\theta}{1-\tan^2\theta}\right)^2}=\frac{4\tan\theta-4\tan^3\theta}{1-6\tan^2\theta+\tan^4\theta}$$

and compare the two solutions.