Question

Let z1 = √2(cos 7pi/4 + i sin 7pi/4) and z2 = 2(cos 5pi/3 + i sin 5pi/3), how do you find z1/z2?

Answer 1

`z_1/z_2=1/sqrt2[cos(pi/12)+isin(pi/12)]`

Explanation:

A complex number of form `rcostheta+irsintheta` can also be written as `r*e^(itheta)`.

Hence, `z_1=sqrt2(cos(7pi/4)+isin(7pi/4))=sqrt2e^(i7pi/4)`

and `z_2=2(cos(5pi/3)+isin(5pi/2))=2e^(i5pi/3)`

Hence `z_1/z_2=sqrt2/2xe^(i(7pi/4-5pi/3))=1/sqrt2*e^(i((21pi-20pi)/12))=1/sqrt2*e^(i(pi)/12)`

= `1/sqrt2[cos(pi/12)+isin(pi/12)]`