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$$z = \dfrac{2+2i}{4-2i}$$

$$|z| = ? $$

My attempt:

$$\dfrac{(2+2i)(4+2i)}{(4-2i)(4+2i)} = \dfrac{4+12i}{20} = \dfrac{4}{20}+\dfrac{12}{20}i = \dfrac{1}{5} + \dfrac{3}{5}i$$

Now taking its magnitude and we have that

$$|z| = \sqrt{\biggr (\dfrac 1 5 \biggr ) ^2 +\biggr (\dfrac 3 5 \biggr )^2} = \sqrt {\dfrac 2 5 }$$

Am I right?

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Yes, you are. You can do it also like this: $$\Big|{2+2i\over 4-2i}\Big|=\Big|{1+i\over 2-i}\Big|={|1+i|\over |2-i|}= {\sqrt{2}\over \sqrt{5}}$$

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It is better to use $$\left|\dfrac ab\right|=\dfrac{|a|}{|b|}$$

$|2+2i|=\sqrt{2^2+2^2}=2\sqrt2$

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  • $\begingroup$ I first made the denominator real and then did my calculations to avoid any errors . $\endgroup$ – Hamilton Dec 3 '18 at 17:55
  • $\begingroup$ @Hamilton, At least for modulus, the formula I believe is more suitable $\endgroup$ – lab bhattacharjee Dec 3 '18 at 17:56
  • $\begingroup$ However, is it a bad way? $\endgroup$ – Hamilton Dec 3 '18 at 18:11
  • $\begingroup$ Rationalization may be costly in some cases $\endgroup$ – lab bhattacharjee Dec 3 '18 at 18:17
  • $\begingroup$ @Hamilton I'd say it depends whether you're interested in the actual value of $z$ or not. This way gets you straight to the magnitude and saves you the slightly more error-prone algebra of making a real denominator to get $z$. $\endgroup$ – timtfj Dec 3 '18 at 19:53

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