To convert z=x+iy in CC into polar /trigo. form, we have to find
r>0, and theta in (-pi,pi] such that, x=rcostheta, y=rsintheta.
Condsider, z_1=sqrt3+i
:. rcostheta=x=sqrt3, rsintheta=y=1 rArr x^2+y^2=r^2=3+1=4
:. r=2.
"Then, rcostheta=2costheta=sqrt3 rArr costheta=sqrt3/2 >0, and,
sintheta=1/2 >0". We conclude that, "theta in (0,pi/2), &, theta=pi/6.
Altogether, z_1=2(cos (pi/6)+isin(pi/6))....................(1).
Similarly, we can show that, z_2=2(cos(-5pi/6)+isin(-5pi/6)).........(2).
Now, it can easily proved that,
z_j=r_1costheta_j+isintheta_j, j=1,2,
rArr z_1*z_2=r_1*r_2{cos(theta_1+theta_2)+isin(theta_1+theta_2)}.
Accordingly, in our case, we have,
z_1*z_2=2*2{cos(pi/6-5pi/6)+isin(pi/6-5pi/6)}
=4(cos(-4pi/6)+isin(-4pi/6))
=4(cos(2pi/3)-isin(2pi/3))
=4(-cos(pi/3)-isinsin(pi/3)
=4(-1/2-isqrt3/2)
=-(2+i2sqrt3)
=-2(1+isqrt3).
We can verify that, z_*z_2=(sqrt3+i)(-sqrt3-i)=-(sqrt3+1)^2
=-(3+2sqrt3i+i^2)=-(2+2sqrt3i)=-2(1+sqrt3i)
Enjoy maths.!
