Algebra question

Smythe

Could someone please fill in any missing lines to explain to me how the equation in the attachment went from the square of the modulus to the last part?

I've been working on this for hours and I can't figure it out.

Thank you.

MathsManiac

It's squaring out and getting relevant moduli, noting the following:
i^2 = -1, which can be ignore because of the modulus sign.
e^(-ika) has modulus 1, for any real k and real a.
modulus of (a+ib) is sqrt(a^2 + b^2).

Does that clear it all up for you?

Yakuza

It looks wrong to me; for a start the -i in the exponent of e should still be around after the squaring.
[latex]{(e^{-i})}^2 = e^{-2i}[/latex]

MM is much more informed than me in this field - ignore what I said
I haven't dealt with imaginary numbers much since, well, lets just say when mullets and PLO scarves were in fashion.

MathsManiac

Yakuza wrote: »

It looks wrong to me; for a start the -i in the exponent of e should still be around after the squaring.
[latex]{(e^{-i})}^2 = e^{-2i}[/latex]

That's what I thought when I looked at it first, but then I remembered from De Moivre's theorem / Euler's identity that, for any theta, exp(-i*theta) = cos(theta) + i*sin(theta), and so the modulus of this is 1.

(And thanks for the compliment, although i'm not sure it's deserved!)

Yakuza

Ah yes! Good old DeMoivre - now that brings back memories. I remember being able to use it to prove that [latex]e^{i \pi} = -1 [/latex]. That single fact alone blew me away, and was / is one of the things that keeps my interest in maths very much alive, some quarter-century after doing the Leaving. (Albeit some areas are a bit rustier than others )

TheBody

Complex analysis is my favourite area of mathematics. Blows my mind how beautiful it is.

Edit: Here is a great video on Mobius transformations for example:

https://www.youtube.com/watch?v=JX3VmDgiFnY

Smythe

MathsManiac wrote: »

It's squaring out and getting relevant moduli, noting the following:
i^2 = -1, which can be ignore because of the modulus sign.
e^(-ika) has modulus 1, for any real k and real a.
modulus of (a+ib) is sqrt(a^2 + b^2).

Does that clear it all up for you?

Thanks MM, that's helped a lot.

So when using the modulus of (a+ib) is sqrt(a^2 + b^2), does that mean my denominator of

|-(α - ik)^2|^2 = (α^2 + k^2)^2 as it should, and it does not retain the minus sign between the 'α' and 'k'?

ray giraffe

Smythe wrote: »

Thanks MM, that's helped a lot.

So when using the modulus of (a+ib) is sqrt(a^2 + b^2), does that mean my denominator of

|-(α - ik)^2|^2 = (α^2 + k^2)^2 as it should, and it does not retain the minus sign between the 'α' and 'k'?

See problem 4 and 5 on this page:
http://www.mathwarehouse.com/algebra/complex-number/absolute-value-complex-number.php

Smythe

Thanks ray giraffe!

Michael Collins

Smythe wrote: »

Thanks MM, that's helped a lot.

So when using the modulus of (a+ib) is sqrt(a^2 + b^2), does that mean my denominator of

|-(α - ik)^2|^2 = (α^2 + k^2)^2 as it should, and it does not retain the minus sign between the 'α' and 'k'?

The nice thing about the modulus is that it is a 'multiplicative function'. This means if I have two complex numbers, multiplied like so:

[latex] \displaystyle z = z_1 \cdot z_2, [/latex]

and I take the modulus

[latex] \displaystyle |z| = |z_1 \cdot z_2|, [/latex]

I can actually apply it individually to each factor, like this

[latex] \displaystyle |z| = |z_1| \cdot |z_2|. [/latex]

This makes things a lot easier, since, in your case above, it means

[latex] \displaystyle |-(a-ik)^2|^2 = (|-1| \cdot |(a-ik)^2|)^2 [/latex],

and since the modulus of -1 is just 1, we can rewrite this as

[latex] \displaystyle |-(a-ik)^2|^2 = |(a-ik)^2|^2 [/latex].

(Incidentally the modulus also obeys the following rule

[latex] \displaystyle z = z^2 \Rightarrow |z| = |z^2| = |z|^2 [/latex],

so the denominator above can be simplified further by taking the square outside the modulus, to get

[latex] \displaystyle |-(a-ik)^2|^2 = (|(a-ik)|^2)^2=(a^2+k^2)^2 [/latex],

but it will of course work by multiplying out what's inside the modulus - I don't want to confuse you with this so if it doesn't make sense don't worry, just do it the way you're familiar with - or if you're really interested you can ask me to explain it further!)