I received this from a student of mine who was really confused with the transition from line 2 to line 3. Students should give it a thought before reading my explanation below.

Firstly, my student felt that $|(2x-1)+2yi| \le |(x+1)+yi|$ was being squared on both sides to remove the modulus and the next step should be $[(2x-1)+2yi]^{2} \le [(x+1)+yi]^{2}$. She is very mistaken, as $(2x-1)+2yi$ is a complex number, which is a vector, and not a mere scalar. So the transition from step 2 to step 3 was really an evaluation of the modulus. If we do not skip any steps, the following should clarify how we got from step 2 to step 3.

$|(2x-1)+2yi| \le |(x+1)+yi|$

$\sqrt{(2x-1)^{2}+(2yi)^{2}} \le \sqrt{(x+1)^{2}+(yi)^{2}}$

$\sqrt{(2x-1)^{2}+(2yi)^{2}} \le \sqrt{(x+1)^{2}+(yi)^{2}}$

$(2x-1)^{2}+(2yi)^{2} \le (x+1)^{2}+(yi)^{2}$

I do hope this raise awareness for students to treat modulus carefully and question themselves, when they are dealing with a scalar or vector.

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