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Fp2 (complex number modulus question)

sulexk

If I have |(x+5)-iv| and |(x+5)+iv|

Are these both equal to (x+5)^2 + v^2

Thank you :smile:

Reply 1

Daniel Freedman

Yes (but no). They're both the same, but equal to the square root of what you've written.

Reply 2

sulexk

So what |(x+5)-iv| be the negative square root?

Reply 3

username325135

Yes. I am assume the "v" in your expressions is a constant.

You basically apply Pythagoras' theorem. If, on an argand diagram, you draw a line between the point representing the complex number and the origin...you'll see.

Reply 4

sulexk

Original post by dknt

No. I can kind of see what you're trying to do, but there are 2 main things wrong. The first is your squaring of it. Considering you're doing FP2 I assume you know how to expand ((x+5)-iv)^2 correctly. e.g. (x+y)^2 \not = x^ + y^2

Secondly, even if you had sqaured it correctly it wouldn't be "equal" to each other.

|-1| \not = (-1)^2

You've sqaured one side, but not the other.

In the Fp2 book(the old one) it had this equation:

|2u+2iv|=|2-i(u+iv)|

4u^2 + 4v^2 = |(2+v)-iu|

4u^2 + 4v^2 = (2+v)^2 + u^2

This is from a worked example.
Are they wrong?
This is for fp2 complex numbers loci

Reply 5

username325135

Original post by dknt

Secondly, even if you had sqaured it correctly it wouldn't be "equal" to each other.

You've sqaured one side, but not the other.

Surely...if you want to find the modulus of a complex number, you can square the real and imganiary parts separately, then add them together and then take the square root of the result.

Reply 6

dknt

Original post by KiraMayz

Surely...if you want to find the modulus of a complex number, you can square the real and imganiary parts separately, then add them together and then take the square root of the result.

Oh my yes, I completely ignored the complex numbers bit! :colondollar:

Reply 7

sulexk

Original post by KiraMayz

Surely...if you want to find the modulus of a complex number, you can square the real and imganiary parts separately, then add them together and then take the square root of the result.

Yes I believe so.

although how would you sketch |(u+5)-iv|

Reply 8

sulexk

Original post by KiraMayz

Surely...if you want to find the modulus of a complex number, you can square the real and imganiary parts separately, then add them together and then take the square root of the result.

I believe it is the same!

I sketched it, and found out that two lines have the same length(i.e. modulus) it would be valid to assume that

|(u+5)-iv|=|(u+5)+iv)|

Thank you