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

• chwala
In summary, the conversation discusses the process of multiplying complex numbers using the conjugate of the denominator. The final result can be expressed using the exponential form of a complex number, with a modulus of 2/3 and an argument of π/2.
chwala
Gold Member
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]##

Last edited:
Use $$\cos \theta + i\sin \theta = e^{i\theta}.$$

chwala
pasmith said:
Use $$\cos \theta + i\sin \theta = e^{i\theta}.$$
Fine, let me check on this again...

##\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}##

Last edited:

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