4)
If z = r (cos θ+ i sin θ), then the n distinct nth roots of z are:
w(k) = r^(1/k) ( cos ((θ + 2pi k)/ n) , i sin (θ +2pi k) /n)), k=0,1,2,.....(n-1)
-1 = (cos(pi) + i sin(pi))
θ = pi
First root:
1^(1/2) ( cos(pi+2pi(0)) / 2) +i sin (pi+ 2pi(0))/2) )
= 1 (cos(pi/2) + i sin(pi/2)
= 1( 0 + i)
= i
Second root:
1^(1/2) ( cos ((pi+ 2pi(1))/2) + i sin ((pi+2pi(1))/2))
= 1 ( cos((pi+2pi)/2) + isin((pi+2pi)/2))
= 1 ( cos(3pi/2) + i sin(3pi/2))
= 1 ( 0 + i(-1))
= -i
The two roots are: i and -i
5.
z = 12 (cos(pi/3) + i sin(pi/3))
θ = pi/3
12^(1/2) ( cos ((pi/3+2pi(0))/2) + i sin ((pi/3+2pi(0))/2) )
= 2sqrt(3) ( cos(pi/6) + i sin(pi/6))
= 2 sqrt(3) ( sqrt(3)/2 + i/2)
= 3 + sqrt(3)i
12^(1/2) ( ( cos ((pi/3+2pi(1))/2) + i sin ((pi/3+2pi(1))/2) )
= 2sqrt(3) ( cos(7pi/6) + i sin(7pi/6))
= 2 sqrt(3) ( -sqrt(3)/2 -i /2)
= -3 - sqrt(3) i
The two roots are 3 + sqrt(3)i and -3 - sqrt(3)i
3)
z = -36+ 36 sqrt(3) i
a = -36
b= 36sqrt(3)
(a+bi)^2 = -36 +36 sqrt(3)i
a^2-b^2 +2abi = -36 +36sqrt(3)i
2ab = 36sqrt(3)
Compare real and imaginary parts:
a^2-b^2 = -36
|z|^2= |z^2|
a^2+b^2 = sqrt( (-36)^2 + (36sqrt(3))^2 ) = 72
a^2-b^2 =-36
a^2+b^2 = 72
add:
2a^2 = 36
a^2=18
a = +/- sqrt(18) = +/- 3sqrt(2)
a^2+b2=72
18+b^2 = 72
b^2 = 54
b= +/-sqrt(54) = +/- 3sqrt(6)
2ab= 36sqrt(3)
a abd b are of the same sign
a = 3sqrt(2) and b= 3sqrt(6)
Yes DeMoivre's Theorem would be appropriate to use to solve these.
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Answers & Comments
4)
If z = r (cos θ+ i sin θ), then the n distinct nth roots of z are:
w(k) = r^(1/k) ( cos ((θ + 2pi k)/ n) , i sin (θ +2pi k) /n)), k=0,1,2,.....(n-1)
-1 = (cos(pi) + i sin(pi))
θ = pi
First root:
1^(1/2) ( cos(pi+2pi(0)) / 2) +i sin (pi+ 2pi(0))/2) )
= 1 (cos(pi/2) + i sin(pi/2)
= 1( 0 + i)
= i
Second root:
1^(1/2) ( cos ((pi+ 2pi(1))/2) + i sin ((pi+2pi(1))/2))
= 1 ( cos((pi+2pi)/2) + isin((pi+2pi)/2))
= 1 ( cos(3pi/2) + i sin(3pi/2))
= 1 ( 0 + i(-1))
= -i
The two roots are: i and -i
5.
z = 12 (cos(pi/3) + i sin(pi/3))
θ = pi/3
First root:
12^(1/2) ( cos ((pi/3+2pi(0))/2) + i sin ((pi/3+2pi(0))/2) )
= 2sqrt(3) ( cos(pi/6) + i sin(pi/6))
= 2 sqrt(3) ( sqrt(3)/2 + i/2)
= 3 + sqrt(3)i
Second root:
12^(1/2) ( ( cos ((pi/3+2pi(1))/2) + i sin ((pi/3+2pi(1))/2) )
= 2sqrt(3) ( cos(7pi/6) + i sin(7pi/6))
= 2 sqrt(3) ( -sqrt(3)/2 -i /2)
= -3 - sqrt(3) i
The two roots are 3 + sqrt(3)i and -3 - sqrt(3)i
3)
z = -36+ 36 sqrt(3) i
a = -36
b= 36sqrt(3)
(a+bi)^2 = -36 +36 sqrt(3)i
a^2-b^2 +2abi = -36 +36sqrt(3)i
2ab = 36sqrt(3)
Compare real and imaginary parts:
a^2-b^2 = -36
2ab = 36sqrt(3)
|z|^2= |z^2|
a^2+b^2 = sqrt( (-36)^2 + (36sqrt(3))^2 ) = 72
a^2-b^2 =-36
a^2+b^2 = 72
add:
2a^2 = 36
a^2=18
a = +/- sqrt(18) = +/- 3sqrt(2)
a^2+b2=72
18+b^2 = 72
b^2 = 54
b= +/-sqrt(54) = +/- 3sqrt(6)
2ab= 36sqrt(3)
a abd b are of the same sign
a = 3sqrt(2) and b= 3sqrt(6)
Yes DeMoivre's Theorem would be appropriate to use to solve these.