The Modulus Sign #4

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.

Modulus of a Complex Number

Modulus of a Complex Number

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.

    pingbacks / trackbacks

    Leave a Comment

    Contact Us

    CONTACT US We would love to hear from you. Contact us, or simply hit our personal page for more contact information

    Not readable? Change text. captcha txt

    Start typing and press Enter to search