Question

-16i in trigonometric form

Answer 1

16(cos(π2)+isin(π2))

Answer 2

-16i = 16 [cos 3pi/2 + i sin 3pi/2]

So, by De Moivre's Theorem, the fourth roots are given by
16^(1/4) [cos((3pi/2 + 2 pi i k)/4) + i sin((3pi/2 + 2 pi i k)/4)], where k = 0,1,2,3.

More specifically, they are
k = 0 ==> 2 [cos(3pi/8) + i sin(3pi/8)]
k = 1 ==> 2 [cos(7pi/8) + i sin(7pi/8)]
k = 2 ==> 2 [cos(11pi/8) + i sin(11pi/8)]
k = 3 ==> 2 [cos(15pi/8) + i sin(15pi/8)]
--------------------------------------...

25i = 25 [cos pi/2 + i sin pi/2]

So, by De Moivre's Theorem, the fourth roots are given by
25^(1/4) [cos((pi/2 + 2 pi i k)/4) + i sin((pi/2 + 2 pi i k)/4)], where k = 0,1,2,3.

Note that 25^(1/4) = (5^2)^(1/4) = 5^(1/2).

More specifically, the roots are
k = 0 ==> sqrt(5) [cos(pi/8) + i sin(pi/8)]
k = 1 ==> sqrt(5) [cos(5pi/8) + i sin(5pi/8)]
k = 2 ==> sqrt(5) [cos(9pi/8) + i sin(9pi/8)]
k = 3 ==> sqrt(5) [cos(13pi/8) + i sin(13pi/8)].

Answer 3

Do u understand any of it?

Answer 4

Yes thank you!

Answer 5

No probs

Answer 6

Anymore?

Answer 7

so I need to find the Square root of -16i…. I am using the theorem but i am just not coming up with the right answer.

Answer 8

I dont know any other way to explain g2g be on later tho.

Answer 9

4 -i

Answer 10

So how can i put that in to trigonometric form?