- #1
chwala
Gold Member
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- 388
- Homework Statement
- See attached
- Relevant Equations
- Complex numbers
π
My take; i multiplied by the conjugate of the denominator...
$$\dfrac{z_1}{z_2}=\dfrac{2(\cos\dfrac{π}{3}+i \sin \dfrac{π}{3})}{3(\cos\dfrac{π}{6}+i \sin \dfrac{π}{6})}⋅\dfrac{3(\cos\dfrac{π}{6}-i \sin \dfrac{π}{6})}{3(\cos\dfrac{π}{6}-i \sin \dfrac{π}{6})}=\dfrac{2(\cos\dfrac{π}{3}+i \sin \dfrac{π}{3})}{3}⋅\dfrac{3(\cos\dfrac{π}{6}-i \sin \dfrac{π}{6})}{3}$$
...This will also realise the required result;though with some work by making use of,
##\cos a⋅\cos b-i\cos a⋅\sin b + i\sin a⋅cos b + \sin a⋅\sin b##
##=\cos a⋅\cos b+\sin a⋅\sin b-i\cos a⋅\sin b+i\sin a⋅\cos b##
##=\cos(a-b)-i\sin (a-b)##
for our case, and considering the argument part of the working we shall have,
##=\cos\left[\dfrac{π}{3}- - \dfrac{π}{6}\right]-i(\sin \left[\dfrac{π}{3}- - \dfrac{π}{6}\right]= \cos\left[\dfrac{π}{3}+\dfrac{π}{6}\right]-i(\sin \left[\dfrac{π}{3}+\dfrac{π}{6}\right]##
##=\cos\left[\dfrac{π}{2}\right]-i\sin \left[\dfrac{π}{2}\right]##
My take; i multiplied by the conjugate of the denominator...
$$\dfrac{z_1}{z_2}=\dfrac{2(\cos\dfrac{π}{3}+i \sin \dfrac{π}{3})}{3(\cos\dfrac{π}{6}+i \sin \dfrac{π}{6})}⋅\dfrac{3(\cos\dfrac{π}{6}-i \sin \dfrac{π}{6})}{3(\cos\dfrac{π}{6}-i \sin \dfrac{π}{6})}=\dfrac{2(\cos\dfrac{π}{3}+i \sin \dfrac{π}{3})}{3}⋅\dfrac{3(\cos\dfrac{π}{6}-i \sin \dfrac{π}{6})}{3}$$
...This will also realise the required result;though with some work by making use of,
##\cos a⋅\cos b-i\cos a⋅\sin b + i\sin a⋅cos b + \sin a⋅\sin b##
##=\cos a⋅\cos b+\sin a⋅\sin b-i\cos a⋅\sin b+i\sin a⋅\cos b##
##=\cos(a-b)-i\sin (a-b)##
for our case, and considering the argument part of the working we shall have,
##=\cos\left[\dfrac{π}{3}- - \dfrac{π}{6}\right]-i(\sin \left[\dfrac{π}{3}- - \dfrac{π}{6}\right]= \cos\left[\dfrac{π}{3}+\dfrac{π}{6}\right]-i(\sin \left[\dfrac{π}{3}+\dfrac{π}{6}\right]##
##=\cos\left[\dfrac{π}{2}\right]-i\sin \left[\dfrac{π}{2}\right]##
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