1. Given z1 = 2(cos pi/6 + i sin pi/6) and z2 = 3 (cos pi/4 + i sin pi/4), find z1z2 where 0 ≤ theta ≤ 2pi 2. Find the product of z1 = 2/3 (cos60° + i sin60°) and z2 = 9 (cos20° + i sin20°) where 0 ≤ theta ≤ 360° 3. Given z1 = 12 (cos pi/3 + i sin pi/3) and z2 = 3 (cos 5pi/6 + i sin 5pi/6), find z1/z2 where 0 ≤ theta ≤ 2pi 4. Find the quotient of z1 = cos 2pi/3 + i sin 2pi/3 and z2 = 2 (cos pi/12 + i sin pi/12) where 0 ≤ theta ≤ 2pi

a)z1.z2=cos (5pi/2) + i  sin(5pi/2)

b)z1.z2=6[cos 80 + isin 80]

c)z1/z2= 3[cos 3pi + i sin 3pi]

d)z1/z2 = (cos 7pi/12 + i sin 7pi/12 )

Step-by-step explanation:

a) z1=2(cos pi/6 + i sinpi/6) , z2=3(cospi/4 + i sin pi/4)

z1.z2=[2(cos pi/6 +i sinpi/6)] . [3(cospi/4 + i sin pi/4)]  

z1.z2=6[cos(pi/6 + pi/4) + i sin(pi/6 + pi/4)]

z1.z2=6[cos (5pi/12) + i sin(5pi/12)]

z1.z2=cos (5pi/2) + i  sin(5pi/2)

Hence the answer is this.

b)z1= 2/3 (cos60° + i sin60°) , z2=9 (cos20° + i sin20°)

z1.z2=2/3 *9[(cos 60 +i sin60)+(cos20 + i sin20)]

z1.z2=18/3[cos(60+20) + i sin(60+20)]

z1.z2=6[cos 80 + isin 80]

Hence the answer is this

c) z1 = 12 (cos pi/3 -+i sin pi/3) , z2 = 3 (cos 5pi/6 + i sin 5pi/6)

z1/z2= (12/3)[(cos pi/3 + i sin pi/3) - (cos 5pi/6 + i sin 5pi/6)]

z1/z2= 6[cos(pi/3 - 5pi/6) + i sin(pi/3 - 5pi/6)]

z1/z2= 6[cos(2pi- pi/2) + i sin(2pi-pi/2)]

z1/z2= 6[cos 3pi/2 + i sin 3pi/2]

z1/z2= 3[cos 3pi + i sin 3pi]

Hence the answer is this

d) z1 = cos 2pi/3 + i sin 2pi/3  and z2 = 2 (cos pi/12 + i sin pi/12)  

z1/z2 = (1/2)(cos(2pi/3-pi/12) + i sin (2pi/3 -pi/12))

z1/z2 = (cos 7pi/12 + i sin 7pi/12 )

Hence the answer is this

there are infinitely many solutions.