Question

Complex numbers - nth roots of r(cosα + isinα).
What is k here??

Answer 1

Find the fourth roots of the following complex numbers, expressing your answers in polar form
with argument ∈ (−pi,pi ]:
2 −√12i

Answer 2

can anybody explain me please?

Answer 3

@amistre64

Answer 4

a complex number can be written trigily as:

r (cos(t)+i sin(t))

the nth root is simply

r^n (cos(t)+i sin(t))^n

what do you mean by: "what is k"?

Answer 5

given a number (a+bi), r = sqrt(a^2+b^2)
a = tan^-1 (b/a)

2 - sqrt(12) i ; find r; sqrt(2^2 + (sqrt12)^2)
sqrt(16) = 4

find a; tan^-1(-sqrt(12)/2)
tan^-1(-2sqrt(3)/2)
tan^-1(-sqrt3)
a = -60^o = 300^o


4 (cos(300) + i sin(300))

Answer 6

the 4th roots are simply adding multiples of 360/4 to the given angle

Answer 7

and of course expressing it along the interval stated instead of from 0 to 2pi :)

Answer 8

here what sample solution provided by tutors

Answer 9

can you explain what is k there please?

Answer 10

@amistre64

Answer 11

k is reserved for a "multiple" of pi/2

Answer 12

we need to restrict k so that the solutions fall within -pi and pi

Answer 13

so, how we define that?

Answer 14

when it starts from -1 or -2

Answer 15

well, our original angle is: -pi/3, which is also known as 5pi/3 between 0 and 2pi

5pi/3 + pi/2
10pi/6 + kpi/2

k=0; 10pi/6
k=1; pi/6
k=2; 4pi/6
k=3; 7pi/6

is the setup on a normal looking interval :/ subtract pi from each one

Answer 16

k=0; 10pi/6 - 6pi/6 = 4pi/6
k=1; pi/6 ....... = -5pi/6
k=2; 4pi/6 ....... = -2pi/6
k=3; 7pi/6 ........ = pi/6

Answer 17

those ks are leftovers, forgot to get rid of them; but thats the idea i have

Answer 18

soo, k=-1,0,1,2