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