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I want to take the argument of the following complex fraction. Using the second method I get a different answer, why is that? $$ G(\omega)= \frac{1}{(1+2\omega i)^2} \tag 1 $$ Method 1: \begin{align} \arg\frac{1}{(1+2\omega i)^2} &=\arg1-\arg\Big((1+2\omega i)^2\Big) \tag 2\\ &=\arg1-\arg\Big((1+2\omega i)(1+2\omega i)\Big) \tag 3\\ &=\arg1-\arg(1+2\omega i)-\arg(1+2\omega i) \tag 4\\ &=\arctan\frac{0}{1}-\arctan\Big(\frac{2\omega}{1}\Big)-\arctan\Big(\frac{2\omega}{1}\Big) \tag 5\\ &=-2\arctan(2\omega) \tag 6 \\ \end{align}

Method 2: Expand the denominator: $$ (1+2\omega i)^2=1-4\omega^2+4\omega i $$ So I have \begin{align} \arg \frac{1}{(1+2\omega i)^2} &=\arg\frac{1}{1-4\omega^2+4\omega i} \tag 7\\ &=\arg1-\arg(1-4\omega^2+4\omega i)\tag 8\\ &=-\arctan\bigg (\frac{4\omega}{1-4\omega^2}\bigg) \tag 9 \end{align} So $(6)$ is not equal to $(9)$, What is wrong with method 2?

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There is nothing wrong with either method (besides not being cautious about range restrictions). The tangent double angle identity:

$$\tan(2A) = \frac{2\tan A}{1-\tan^2 A}$$

should answer your questions if you let $A = \arctan(2\omega)$

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Note that$$\tan\bigl(2\arctan(2\omega)\bigr)=\frac{4\omega}{1-4\omega^2}.$$

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  • $\begingroup$ Hi! Can you explain further? I know $\tan(\arctan x)=x$, but if we have a constant $K$ before $\arctan x$ we have $\tan(K\arctan x)\neq x$, or I'm wrong? $\endgroup$ – JDoeDoe Oct 22 at 10:40
  • $\begingroup$ I don't have an arbitrary constant $K$ in my answer! I am just using the fact that$$\tan(2x)=\frac{2\tan x}{1-\tan^2x}$$with $x=\arctan(2\omega)$. $\endgroup$ – José Carlos Santos Oct 22 at 10:42
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The problem is that given $x = x + i y$ you are using $\arctan(y/x)$ with one argument. You should use instead $\arctan(x,y)$. You can find those options in matlab or MATHEMATICA. Note that $\arctan(y/x)=\arctan((-y)/(-x))$ and $\arctan((-y)/x)=\arctan(y/(-x))$

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