Complex No. Question (1 Viewer)

[elseany]

Show that (1 + cosθ + isinθ)ⁿ = 2ⁿ.cosⁿθ.[cos(ⁿθ/2) + isin(ⁿθ/2)] where n is a positive integer.

 

[haque]

there should be a cos@/2 outside not @ i think-
let z=1+cos@+isin@
=1+2cos^2@/2-1+2isin@/2cos@/2
=2cos@/2(cos@/2+isin@/2) and z^n=(1+cos@+isin@)^n

this=2^ncos^n@/2(cosn@/2 +isinn@/2)

 

[elseany]

yep your right, thanks for that but umm i don't understand your working .. :<

is it possible to make it simpler?

 

[jyu]

Show that (1 + cosθ + isinθ)ⁿ = 2ⁿ.cosⁿθ.[cos(ⁿθ/2) + isin(ⁿθ/2)] where n is a positive integer.

haque's way is simpler

 

[haque]

Jyu's method is correct-so whichever one u feel comfortable with-but what i did was expressed 1+cos@+isin@ in modulus argument for-in which i used the double angle formula for cos and sin.

 

[elseany]

i dont understand either :<

is there some polynomial in this or is it just simply complex numbers?


edit: just read your above post haque, i get it now, thanks lol

quite a blonde moment there >_<

(i still dont understand jyu's method though)

 

[sonic1988]

haque is simply using double angle result to express @ in term of @/2. Then he use Demorive theorem to expand z^n