# Real and Complex number systems

• Jan 6th 2009, 07:26 AM
Conorsmom
Real and Complex number systems
Hi there... I'm back in school after ten years, and I must say my math is a little rusty. I could use some help-with some explanation....so I can truly follow and understand.

I am given this:

x=r(cos u + i sin u) and y=t(cos v + i sin v)

and I'm being asked to (1) prove that the modulus of (xy) is the product of their moduli, and (2) prove that the amplitude of (xy) is the sum of their amplitudes, showing each step of both proofs.

I understand modulus with respect to the time of day using a 12 hour clock, but I'm finding it hard to apply, nevermind prove this situation. Just need some direction!!!! Thanks so much.
• Jan 6th 2009, 09:12 AM
Soroban
Hello, Conorsmom!

Quote:

Given: . $\begin{array}{ccc}
x &=&r(\cos u + i\sin u) \\y &=&t(\cos v + i\sin v) \end{array}$

(a) Prove that the modulus of $xy$ is the product of their moduli

(b) Prove that the amplitude of $xy$ is the sum of their amplitudes.

$x\!\cdot\!y \:=\:\bigg[r(cos u + i\sin u)\bigg]\,\bigg[t(\cos v + i\sin v)\bigg]$

. . . $= \;rt\bigg[\cos u\cos v + i\sin u\cos v + i\sin v\cos u + i^1\sin u\sin v\bigg]$

. . . $= \;rt\bigg[\underbrace{(\cos u\cos v - \sin u\sin v)}_{\text{This is }\cos(u+v)} + i\underbrace{(\sin u\cos v + \sin v\cos u)}_{\text{This is }\sin(u+v)}\bigg]$

. . . $= \;rt\bigg[\cos(u+v) + i\sin(u+v)\bigg]$

$\text{Then: }\;|x\!\cdot\!y| \:=\:\sqrt{r^2t^2\underbrace{\left(\cos^2(u+v) + \sin^2(u+v)\right)}_{\text{This is 1}}} \quad\Rightarrow\quad |x\!\cdot\!y| \:=\:rt
$
.[1]

$|x| \:=\:\sqrt{(r\cos u)^2 + (r\sin u)^2} \:=\:\sqrt{r^2\underbrace{(\cos^2\!u + \sin^2\!u)}_{\text{This is 1}}} \:=\: \sqrt{r^2} \:=\:r$

$|y| \:=\:\sqrt{(t\cos v)^2 + (t\sin v)^2} \:=\:\sqrt{t^2(\cos^2\!v+\sin^2\!v)} \:=\:\sqrt{t^2} \:=\:t$

$\text{Then: }\;|x|\!\cdot|y| \;=\;rt$ .[2]

(a) Therefore: . $|x\!\cdot\!y| \;=\;|x|\!\cdot\!|y|$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: . $x \:=\:r(\cos u + i\sin u) \qquad y \:=\:t(\cos v + i\sin v)$
. . . . . . . . . . . . . . $\uparrow\qquad\quad\; \uparrow \qquad\qquad\qquad\quad \uparrow\qquad\quad\,\uparrow
$

. . . . . . . . . . . . .
amplitude of x . . . . . . . . amplitude of y

And:. . $x\!\cdot\!y \;=\;rt\bigg[\cos(u+v) + i\sin(u+v)\bigg]$
. . . . . . . . . . . . . . . . $\uparrow \qquad\qquad\quad\; \uparrow$
. . . . . . . . . . . . . . . .
amplitude of xy

(b) Therefore, the amplitude of $xy$ is the sum of the amplitudes of $x\text{ and }y.$

• Jan 6th 2009, 09:36 AM
Mush
Quote:

Originally Posted by Conorsmom
Hi there... I'm back in school after ten years, and I must say my math is a little rusty. I could use some help-with some explanation....so I can truly follow and understand.

I am given this:

x=r(cos u + i sin u) and y=t(cos v + i sin v)

and I'm being asked to (1) prove that the modulus of (xy) is the product of their moduli, and (2) prove that the amplitude of (xy) is the sum of their amplitudes, showing each step of both proofs.

I understand modulus with respect to the time of day using a 12 hour clock, but I'm finding it hard to apply, nevermind prove this situation. Just need some direction!!!! Thanks so much.

By modulus, I don't think it means modular arithmetic. I think it means modulus in the sense that if $z = x+iy$, then the modulus of z is given by $|z| = \sqrt{x^2+y^2}$. I.e., by modulus they mean absolute value!
• Jan 7th 2009, 04:10 AM
HallsofIvy
Also "amplitude" is not the correct word here. What is meant is the "argument" of the complex number.
• Feb 11th 2009, 03:33 PM
BeachLover
as it turns out amplitude and argument are the same thing.
one is older text, and one is newer text.