3. find the four angles that define the fourth root of z1=1+ sqrt3*i

Question

z = 2 * (1/2 + i * sqrt(3)/2)
z = 2 * (cos(pi/3 + 2pi * k) + i * sin(pi/3 + 2pi * k))
z = 2 * (cos((pi/3) * (1 + 6k)) + i * sin((pi/3) * (1 + 6k)))
z^(1/4) = 2^(1/4) * (cos((pi/12) * (1 + 6k)) + i * sin((pi/12) * (1 + 6k)))
pi/12, 7pi/12, 13pi/12, 19pi/2

4. what are the fourth roots of z1=sqrt3 +i

2^(1/4) * (cos(pi/12) + i * sin(pi/12))
2^(1/4) * (cos(7pi/12) + i * sin(7pi/12))
2^(1/4) * (cos(13pi/12) + i * sin(13pi/12))
2^(1/4) * (cos(19pi/12) + i * sin(19pi/12))

Answer 1

Hmmm. YOu did this the hard way, very hard way.
Conver to polar..
2 * (1/2 + i * sqrt(3)/2) =2*sqrt(1/4+3/4)= 2*sqrt1=2
angle: arctan sqrt3/(2*1/2)=arctansqrt3= 1.04 radians= PI/3

so the vector is 2@PI/3
the fourth root is 2^1/4 @PI/12
and 2^1/4 at (Pi/12 +Pi/2) or angle (PI/12+PI) or (PI/12 + 3PI/4)

Answer 2

that's how they taught us. Is it right?