Find the modulus and argument of ##\dfrac{z_1}{z_2}##

[chwala]

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]##

 

[pasmith]

Use [tex]\cos \theta + i\sin \theta = e^{i\theta}.[/tex]

 

[chwala]

Use [tex]\cos \theta + i\sin \theta = e^{i\theta}.[/tex]

Fine, let me check on this again...

 

[chwala]

##\dfrac{z_1}{z_2}=\dfrac{2}{3}e^{i\left[\dfrac{π}{3}+\dfrac{π}{6}\right]}=\dfrac{2}{3}e^{i\left[\dfrac{π}{2}\right]}##

Therefore

Modulus =##\dfrac{2}{3}##

and

Argument= ##\dfrac{π}{2}##