start with 3)
3) -8=8@180
roots: 2@60 and at each additional 120: 60, 180, 300
lets do another just in case you need practice: fifth root of 32@55 deg
on, 1/5 of 55 is 11. 1/5 of 360 is 72 so angles will be 11, then each 72 degrees for four additional roots. Notice the symettry when you sketch.
now, the magnitude fifth root of 55 is 2.228
2) 1-i= sqrt2@-45= sqrt2 @315
take that to the tenth power: (sqrt2)^10 = 2^5=32
angle? 315x10=3150
lets do the negative angle -45. that times 10 is -450 or -90 = 270 Nice.
then the answer is -32i
check that.
1) in polar, magnitude is sqrt(16*3*2)=9.80
angle is 135
check my work,I did most of this in my head with a migraine.
1.) determine two pairs of polar coordinates for the point (-4sqrt3, 4sqrt3)with 0degrees less than or equal to theta less than or equal to 360degrees
2.) Use DeMoivre's Theorem to find (1-i)^10. write your answer in the form a+bi.
3.) Find the cube roots of -8. Write your answer in form a+bi
4 answers
1. x^2 + y^2 = r^2 = 16*3+16*3
= 16*6
so r = +/- 4 sqrt 6
theta = 3 pi/4 or -pi/4
= 16*6
so r = +/- 4 sqrt 6
theta = 3 pi/4 or -pi/4
2. r = sqrt(1+1) = sqrt 2
theta = tan^-1 (1) = -pi/4
so
r e^i theta = sqrt 2 e^(-ipi/4 )
sqrt 2^10 = 2^5 = 32
-10 pi/4 = -5/2 pi = -pi /2 or -90
32 (cos -90 + i sin -90)
-32 i
theta = tan^-1 (1) = -pi/4
so
r e^i theta = sqrt 2 e^(-ipi/4 )
sqrt 2^10 = 2^5 = 32
-10 pi/4 = -5/2 pi = -pi /2 or -90
32 (cos -90 + i sin -90)
-32 i
thank you guys so much! I had a really hard time with these. I appreciate it :)