Homework Help: PRECALCULUS

Posted by YOSHI on Sunday, April 24, 2011 at 3:47pm.

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

PRECALCULUS - bobpursley, Sunday, April 24, 2011 at 4:09pm
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.
PRECALCULUS - Damon, Sunday, April 24, 2011 at 4:18pm
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
PRECALCULUS - Damon, Sunday, April 24, 2011 at 4:33pm
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
PRECALCULUS - YOSHI, Monday, April 25, 2011 at 12:36am
thank you guys so much! I had a really hard time with these. I appreciate it :)

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