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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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2 Answers

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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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