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Let $$ z = \frac{a-jw}{a+jw}. $$ Then the angle of $z$ is $$ -\tan^{-1}z\left(\frac{w}{a}\right) -\tan^{-1}\left(\frac{w}{a}\right) = -2\tan^{-1}\left(\frac{w}{a}\right). $$ How is that so?

Ali Caglayan
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le4m
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    $\arg (a)$ works similar to $\log (a)$ in that $\arg (\frac{a}{b})$ = $\arg(a) - \arg(b)$ – Nicholas Oct 27 '15 at 05:54
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    See http://math.stackexchange.com/questions/523625/showing-arctan-frac23-frac12-arctan-frac125/523626#523626 and https://en.wikipedia.org/wiki/Atan2#Definition_and_computation – lab bhattacharjee Oct 27 '15 at 06:03

4 Answers4

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They are conjugate complex numbers

$$ \dfrac{r \cdot e ^{-i \theta }}{{r \cdot e ^{i \theta }}{} } =e ^{- 2 i \theta } $$

Narasimham
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Multiply numerator and denomination by the conjugate of a+jw i.e. a-jw So the numerator becomes a2 -w2 - 2awj and the denominator will be real and you'll be able to solve it and get the final argument.

OR

Taking numerator as z1 and denominator as z2
We know z1=|z1|eiarg(z1)
Similarly z2=|z2|eiarg(z1)

Dividing we have, z=(|z1|/|z2|)ei(arg(z1)-arg(z2))

=>arg(z)= arg(z1)-arg(z2)
Also, arg of a complex number is tan-1(complex part/real part) which in case of numerator will be tan-1(-w/a) = -tan-1(w/a)

and for denominator , argument will be tan-1(w/a)
and since we are dividing than be another complex number which is denominator, we subtract their arguments to get the final argument.

Quark
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I suppose that it could be easier to see writing $$z = \frac{a-jw}{a+jw}= \frac{a-jw}{a+jw} \times \frac{a-jw}{a-jw}=\frac{(a-jw)^2}{a^2+w^2}=\frac{a^2-w^2}{a^2+w^2}-\frac{2 a w}{a^2+w^2}j$$ which makes the angle to be such that $$\tan(\theta)=-\frac{2aw}{a^2-w^2}=-\frac{2\frac aw}{1-(\frac aw)^2}$$ and to remember the development of $\tan(2x)$.

Claude Leibovici
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  • I have a small question please, I read on a copybook about complex numbers, a sentence that said "if $z$ is a real number then $$arg z= 0, \pi$$ deprived of the denominator". And a similar one "if $z$ is pure immaginary then $$arg z= \pm \pi/2$$, deprived of the numerator and denominator". Any idea what could this possibly mean? Thanks in advance ... By the way I am just asking about the sentence "deprived of ..." – Fareed Abi Farraj May 22 '19 at 08:26
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Assuming $a,w\in\mathbb{R}$:

$$\arg\left(\frac{a-wi}{a+wi}\right)=$$ $$\arg\left(\frac{a-wi}{a+wi}\cdot\frac{a-wi}{a-wi}\right)=$$ $$\arg\left(\frac{(a-wi)^2}{a^2+w^2}\right)=$$ $$\arg\left(\frac{a^2-w^2-2awi}{a^2+w^2}\right)=$$ $$\arg\left(a^2-w^2-2awi\right)-\arg\left(a^2+w^2\right)=$$ $$\arg\left(a^2-w^2-2awi\right)-0=$$ $$\arg\left(a^2-w^2-2awi\right)=$$ $$\arg\left(a^2-w^2-2awi\right)=$$ $$\arg\left((a-wi)^2\right)=$$ $$\arg\left((a-wi)(a-wi)\right)=$$ $$\arg\left(a-wi\right)+\arg\left(a-wi\right)=$$ $$2\arg\left(a-wi\right)=$$ $$-2\tan^{-1}\left(\frac{w}{a}\right)$$

Jan Eerland
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