very simple complex numbers thought (1 Viewer)

[Rahul]

ever wondered how the modulus of a comlex number is found, not the application of the formula, but the actual derivation of it?

this may be stupid for u, but i jus found it out today...lol

does anyone know it?

 

[McLake]

Do you mean the fact that it is the distance that it is from the origin on the Argand Plane?

 

[Rahul]

yes, but if u were given a diagram with axis, and a complex number z, how would u work out |z|.

i'm probably not explaining the question properly...its simple really. a diagram i have may show u what i mean, but it would give it away....get ppl thinking

 

[Affinity]

You're not alluding to one of our greek friend's work are you.

 

[Toodulu]

i didn't realise there was a formula for finding |z|, i just work it out?

 

[Affinity]

yes, but if u were given a diagram with axis, and a complex number z, how would u work out |z|.

Measure with a ruler or for accuracy, use a micrometer gauge.

 

[underthesun]

im not too sure, but |z| means distance right?

then, if you're not calculating that using modulus, you do this

let z = x+iy

|z| = |x+iy| = |x - iy|

then, |z|² = |x+iy||x-iy|
= |x² + y²| .....(1)
then the number could be added, and then square root that expression (1).

 

[deepulse]

let z = x + iy

plotting it, you get the point (x,y)

and we know |z| = distance from origine to end of the vector thingy

so drawing a triangle

<pre> /|
/ |
/ |</pre>

(its supposed to look like a triangle, but it doesnt~!)

where the height is y
and base is x
and the bottom left corner is O

so |z| (or the hypotenuse) = sqrt ( x^2 + y^2) from pythagorus

 

[underthesun]

since |z|² = |x² + y²|
|z| = |sqrt(x² + y²)|
|z| = sqrt(x² + y²)

there ya go

 

[Rahul]

yep thats it underthesun and deepulse.

i never used to think of it as a right angled triangle, i jus applied the sqroot(x^2 + y^2)

good to know where formulas come from rather than applying it each time.

Originally posted by underthesun

let z = x+iy

|z| = |x+iy| = |x - iy|

then, |z|² = |x+iy||x-iy|
= |x² + y²|

i never thought of it that way. i learnt something

 

[Rahul]

Originally posted by Affinity
You're not alluding to one of our greek friend's work are you.

who are you talking about?

here is a pic that shows it...

 

[Newbie]

i just use a ruler hmmm no wonder im getting consistently <70% in exams

 

[TastesGoodBut]

bahahaha

 

[underthesun]

Originally posted by Rahul
i never thought of it that way. i learnt something

pfft you should have given me a discount for that geha book then

say.. $5?

 

[VanCarBus]

| z | = distance = real squared + i's value squared

 

[Rahul]

Originally posted by VanCarBus
| z | = distance = real squared + i's value squared

its sqrt(real squared + i's value squared)

Originally posted by underthesun
pfft you should have given me a discount for that geha book then

say.. $5?

errr....i'll get back to you