# Argument of a complex fraction, why different answers?

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?

$$\tan(2A) = \frac{2\tan A}{1-\tan^2 A}$$
should answer your questions if you let $$A = \arctan(2\omega)$$
Note that$$\tan\bigl(2\arctan(2\omega)\bigr)=\frac{4\omega}{1-4\omega^2}.$$
• 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? – JDoeDoe Oct 22 at 10:40
• 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 $x=\arctan(2\omega)$. – José Carlos Santos Oct 22 at 10:42
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 $$\arctan((-y)/x)=\arctan(y/(-x))$$