VCE Specialist 3/4 Question Thread!

[paulsterio]

(Also: How do you get that Microsoft Word Equation Editor?)

[pi]

Install Microsoft Maths too (its legally free!): http://www.microsoft.com/education/ww/products/Pages/mathematics-4.0.aspx

[TrueTears]

Use texnic centre + Miktex!

[HERculina]

@ is the theta sign: Consider the complex number z=cos@ + isin@, where -pi < @ < pi. A) expand z^3 algebraically using the binomial theorem or otherwise. B) use DeMoivre's theorem to expand z^3 c)Hence, show that sin3@ = 3cos^2@sin@ - sin^3@ and cos3@ = cos^3@ - 3cos@sin^2@.

[TrueTears]

z^3= (cos@+isin@)^3

use this 'formula': http://en.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem

using demoivre's theorem means just apply this 'formula' http://en.wikipedia.org/wiki/De_Moivre%27s_formula

then c) is just manipulation, equate the expression in a) and b), then solve for sin3@

[HERculina]

Thanks but i'm still a bit lost with part c) . can you show me how you get the answer

[TrueTears]

what was your answer to part a) and b)?

[HERculina]

Is it a) z^3= cos^3@ - isin^3@ and b) z = cis@ so z^3 = cos3@ + isin3@ :/

[HERculina]

O i think my part a is wrong   EDIT: is a) z^3 = cos^3@ + 3icos^2@sin@- 3cos@sin^2@ - isin^3@

[TrueTears]

i dono, cant be bothered doing the workings, however working from what you have, you should try equate real and imaginary coefficients, then just manipulate to have sin3@ on one side and simplify the mess on the other side of the equation

EDIT: so yeah, do the same thing, group the reals and imaginary components, then just equate.

[HERculina]

Thanks TT, equating coeeficients at the end got me to the right answer anywho, i have another q.  : Show that |z + w|^2 = |z|^2 + 2Re(z(conjugate of w)) + |w|^2

[TrueTears]

i assume z and w are both an element of the complex field?

well just let z = a+bi and w = c+di, sub them in and expand and do your algebra, then do the same to the RHS and show that you get the same expression

[paulsterio]

anywho, i have another q.  : Show that |z + w|^2 = |z|^2 + 2Re(z(conjugate of w)) + |w|^2

Here's an alternative to TrueTears' way, his way leads to some really tough algebra, so I thought about it and it seems this works out more nicely

[monkeywantsabanana]

Not sure if this is valid...



let
let  
conjugate w =












Proven.

[paulsterio]

It's valid, except you probably need to make it look a little better.
When proving, you usually don't work with both sides of an equation, you work with one side (e.g. RHS) and prove that it equals the LHS.

Btw, good work, that method is even faster than mine!