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I want to take the absolute value of this fraction $$ \frac{2-ixT}{2+ixT} \tag 1 $$ I know for a complex number $z=a+ib$ we have $$ \lvert z\rvert =\sqrt{ a^2+b^2}= \sqrt{(\Re\{ a\})^2 + (\Im \{b\})^2}\tag 2 $$ So \begin{align} \lvert \frac{2-ixT}{2+ixT}\rvert = \sqrt{\frac{2^2+(xT)^2}{2^2+(xT)^2}} =\sqrt1 =1 \tag 3 \end{align} But using the complex conjugate I get a different answer, why is that?

My attempt: \begin{align} \frac{2-ixT}{2+ixT} &= \frac{2-ixT}{2+ixT}\cdot \frac{2-ixT}{2-ixT}\\ &=\frac{(2-ixT)(2-ixT)}{4+x^2T^2}\\ &= \frac{4+x^2T^2-i4Tx}{4+x^2T^2}\\ &= \frac{4+x^2T^2}{4+x^2T^2}-i\frac{4Tx}{4+x^2T^2} \tag 4 \\ &=1-i\frac{4Tx}{4+x^2T^2} \tag 5 \end{align} But using the absolute value $\lvert z\rvert =\sqrt{(\Re\{ a\})^2 + (\Im \{b\})^2}$ now gives \begin{align} \lvert 1-j\frac{4Tx}{4+x^2T^2} \rvert &= \sqrt{ 1^2 + \bigg(\frac{4xT}{4+x^2T^2}\bigg)^2} \tag 6\\ &=\sqrt{ \frac{(4+x^2T^2)^2+(4xT)^2}{(4+x^2T^2)^2} } \tag 7 \end{align} I'm stuck here.

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3 Answers 3

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$$\dfrac{a-ib}{a+ib}=\dfrac{(a-ib)^2}{a^2+b^2}=\dfrac{a^2-b^2+i(-2ab)}{a^2+b^2}$$

Now $$(a^2-b^2)^2+(-2ab)^2=(a^2-b^2)^2+4a^2b^2=(a^2+b^2)^2$$

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The third line in your answer is wrong. You will have $-x^{2}T^{2}$ instead of $x^{2}T^{2}$.

Once you make this correction you will get the answer as $1$ using the identity $(4-x^{2}T^{2})^{2}+16x^{2}T^{2}=(4+x^{2}T^{2})^{2}$

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Supposing $x, T \in \mathbb{R}$ and calling $z = \frac{2-i xT}{2+i x T}$ we have

$$ |z|=\sqrt{z\bar z} = \sqrt{ \frac{2-i xT}{2+i x T} \frac{2+i xT}{2-i x T}} = \sqrt{\frac{4+x^2T^2}{4+x^2T^2}}=1 $$

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