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.!
