1. ## DeMoivre's Theorem

My question is when we use this theorem

z=4(cos60+isin60) * 4(cos 60 +i sin 60)
z^2= 16 (cos 120 +isin 120) *4(cos 60 +i sin 60)
z^3= 64 (cos 180 + i sin 180)

Why do we just add the 60 degrees to each
why do we not distribute? to get 16+ 8 cos 60 +8 i sin 60 + cos 60^2 + (i sin 60)^2

if that is correct how does that become line 2? thanks

Do we add the values because of the sum and difference formulas?

2. Originally Posted by algebraisabeast
My question is when we use this theorem

z=4(cos60+isin60) * 4(cos 60 +i sin 60)
z^2= 16 (cos 120 +isin 120) *4(cos 60 +i sin 60)
z^3= 64 (cos 180 + i sin 180)

Why do we just add the 60 degrees to each
why do we not distribute? to get 16+ 8 cos 60 +8 i sin 60 + cos 60^2 + (i sin 60)^2

if that is correct how does that become line 2? thanks

Do we add the values because of the sum and difference formulas?
http://en.wikipedia.org/wiki/De_Moivre%27s_formula

Also,

$Z = 4 (\cos 60 + i \sin 60)$

$Z^2 = 4^2 (\cos 60 + i \sin 60)^2$

$Z^2 = 16 (\cos 60 + i \sin 60)(\cos 60 + i \sin 60)$

$Z^2 = 16 (\cos^2 60 + i^2 \sin^2 60+i\sin 60. \cos 60 + i \cos 60.\sin 60)$

$Z^2 = 16 [(\cos^2 60 - \sin^2 60)+i(\sin 60. \cos 60 + \cos 60.\sin 60)]$

$Z^2 = 16 [\cos (60+60)+i\sin (60+ 60)]$

$Z^2 = 16 (\cos 120+i\sin 120)$

Also,

$Z^3 = 4^3 (\cos 60 + i \sin 60)^3$

$Z^3 = 4^2 (\cos 60 + i \sin 60)^2 \times 4(\cos 60 + i \sin 60)$

$Z^3 = 64 (\cos 120 + i \sin 120)(\cos 60 + i \sin 60)$ [substitute the value of $Z^2$, proved above.]

$Z^3 = 64 (\cos 120\cos 60 + i^2 \sin 120. \sin 60+i\sin 120. \cos 60 + i \cos 120.\sin 60)$

$Z^3 = 64 [(\cos 120\cos 60 - \sin 120. \sin 60)+i(\sin 120. \cos 60 + \cos 120.\sin 60)]$

$Z^3 = 64 [\cos (120+60)+i\sin (120+ 60)]$

$Z^3 = 64 (\cos 180+i\sin 180)$

Formula used:

$\cos A. \cos B -\sin A . \sin B = \cos (A+B)\;\;\;and\;\;\;
\sin A. \cos B +\cos A . \sin B = \sin (A+B)$