HSC Tips - Complex Numbers (1 Viewer)

[spice girl]

humph....funny, i thought it's usually the other way round... i know how to solve complex numbers questions using circular geometry...

and ok, keep the personal stuff out of the tips pages... the long list of sticky threads is already pissin me off

btw KeypadSDM, u've reason to be happy and show things off, but plz... ur noobness shows learn from the likes of turtle, phenol, etc,.... experienced players show off too, but they do it more subtly

 

[freaking_out]

Originally posted by spice girl
...btw KeypadSDM, u've reason to be happy and show things off, but plz... ur noobness shows learn from the likes of turtle, phenol, etc,.... experienced players show off too, but they do it more subtly

lol, yeah, i hardly notice when turtle, phenol, spice girl p), etc. show off.

 

[KeypadSDM]

Originally posted by freaking_out
lol, yeah, i hardly notice when turtle, phenol, spice girl p), etc. show off.

At least give me a week to calm down. I'm still on an excruciatingly high peak.

For Spice Girl: And I really shouldn't have posted that here. Delete if you will (I can't actually do that to my own posts, go figure)

And to justify my post's placing here; How the hell do you do complex numbers in circle geometry?

You can do the converse by proving (2003 HSC, Q2) e)) this:

|z| = 1

0 <= Arg(z) <= pi/2

Prove 2Arg(z+1) = arg(z)

Use the circle proof, angle at center = 2 * angle at circumference.

Can you prove that with algebra? If you can, you can prove the above identity using complex numbers. Just a thought.

 

[Grey Council]

but...but.., thats like the ONLY way to do it graphically, isnt it? If thats all that was meant by doing complex numbers by circular geometry, ah well. Thanks anyway.

 

[KeypadSDM]

I'm pretty sure you should be able to solve it using algebra, but I've either forgotten or never learnt how to do it.

I was trying something like this:

2Arg(z+1) - Arg(z)
= Arg((z+1)^2) - Arg(z)
= Arg(z^2 + 2z + 1) - Arg(z)
= Arg(z + 2 + 1/z)

But if you draw it out, and remember that |z| = 1, then
Arg(z + 1/z) = 0 (Using the e^(i@) notation)
And, similarly
Arg(z + 1/z + 2) = 0

:. 2Arg(z+1) - Arg(z) = 0
2Arg(z+1) = Arg(z)

Oh, well maybe that's how you do it?

 

[MAICHI]

Complex numbers can be used to find cos(ax) and sin(bx) in terms of their powers or vice versa.

Like if you want to convert cos6x into an expression involving just cos^n(x), you can use de Moivre's theorem on (cosx + isinx) and also binomial expansion, then equating the real part.

If you want to go the other way like converting cos^6(x) to an expression involving only cos(ax), you would use [(cisx + cis(-x))/2] = cosx. Just bring it up to the power of 6 and expand the thing, then convert it back to cos(ax). It's kinda hard to explain, but yeah, ask your teacher if you don't understand what I am saying. Just learn it.

 

[joey_prince42]

correction for response

I'm pretty sure you should be able to solve it using algebra, but I've either forgotten or never learnt how to do it.

I was trying something like this:

2Arg(z+1) - Arg(z)
= Arg((z+1)^2) - Arg(z)
= Arg(z^2 + 2z + 1) - Arg(z)
= Arg(z + 2 + 1/z)

But if you draw it out, and remember that |z| = 1, then
Arg(z + 1/z) = 0 (Using the e^(i@) notation)
And, similarly
Arg(z + 1/z + 2) = 0

:. 2Arg(z+1) - Arg(z) = 0
2Arg(z+1) = Arg(z)

Oh, well maybe that's how you do it?

If you're gonna prove it using algebra try this.

Firstly,

2*Arg(z+1) = [Arg{(z+1)^2}

= Arg (z^2 +2z + 1)

Taking out z
= Arg {z (z +2 +1/z) i)

Now with 1/z multiply both the numerator and the denominator by the conjugate of z. This gives you {conjugate of z}/|z|^2. And since |z|=1, then the expression equals to the conjuagte of z.

Now from i)

= Arg (z) + Arg (z + 2 + conjugate of z)

and since z + 2 + {conjuagte of z} is a postive real number and that the arg of any postive real number is 0.

Therefore,

2Arg(z+1) = Arg (z)

 

[Stefano]

When it comes to complex numbers just remember:

75% of them are actually complex and 75% of them are imaginary

 

[Riviet]

When it comes to complex numbers just remember:

75% of them are actually complex and 75% of them are imaginary

That doesn't add up LOL.

 

[pLuvia]

That doesn't add up LOL.

lmao .

 

[buchanan]

Here's a question on circle geometry using complex numbers:

Given any 3 non-collinear points, there is 1 and only 1 circle through those points. You can prove it by proving that for any 3 non-collinear complex numbers z₁, z₂, z₃, the locus of points in the complex plane defined by

is a circle passing through the points, and conversely, any circle through the three points must satisfy that equation.

 

[~shinigami~]

Are there anymore examples of solving circle geometry via complex numbers?

 

[Affinity]

You can prove simpson's thereom which states that if you have a circle and a triangle inscribed in it, pick any point on the circle, then the feet of the perpendiculars from that point to the 3 sides of the triangle are colinear. you can find thisin the harder 3 unit chapter of the cambridge book.

It's probably alot longer coz your tools are not developed enough. give it a try though

 

[~shinigami~]

You can prove simpson's thereom which states that if you have a circle and a triangle inscribed in it, pick any point on the circle, then the feet of the perpendiculars from that point to the 3 sides of the triangle are colinear. you can find thisin the harder 3 unit chapter of the cambridge book.

It's probably alot longer coz your tools are not developed enough. give it a try though

Thanks Affinity. Luckily I have a copy of cambridge, I'll go look it up right now. Although, I should probably study for my assessment on monday instead.

 

[airie]

Just a quick note, it's called Simson's line, not Simpson's

 

[h3ll h0und]

applying trig stuff like double angle formulas in complex numbers? e.g find the modulus/argument of 1 + costheta + i sin theta or something similar to that.

 

[h3ll h0und]

Are there anymore examples of solving circle geometry via complex numbers?

find equation of circle for z-1/z+i = +- pi/4 ... stuff like that... i remember i had to use circle geometry and some easy trig.

 

[ninetypercent]

anyone have some good complex number questions?

 

[adomad]

anyone have some good complex number questions?

all questions are good, its just some are better than others. look at trial HSC papers and do them. if they are good enough, they are in the HSC

 

[Mature Lamb]

Not sure if this has been mentioned yet (cbf reading the other posts in this thread) but there's a Pol button on most calculators. If you need to check the modulus and argument of something, e.g. 6+7i, enter in your calculator: Pol(6,7) and it will show the modulus and argument.

Very useful if you're not confident with complex numbers yet