Question

find the root of z = (-1)^(1/4)
given (-1)^(1/4)
wherein:
n = 4
r = -1
with the formulas:
R = (r)^(n)
phi = (deta + 2(pi)(k))/n

and w = R(cos phi + isin phi)

i know r and n but how do i find deta to solve for phi?

Answer 1

you are looking for the fourth roots of \(-1\) right?

Answer 2

yes

Answer 3

that is the unit circle in the complex plane
should be more or less clear that the angle made by \(-1\) is \(\pi\)

Answer 4

divide that by \(4\) and get \(\frac{\pi}{4}\)

Answer 5

then what if it has an i ?

i.e: (-1 + i)^(1/3)

x = -1 y = 1

tan deta = 1/-1 = -1

deta = -45 = pi/4

but when i used my calculator the answer is (3pi)/4

can you please tell me what I missed?

Answer 6

what if what was \(i\) ?

Answer 7

im looking for deta. (like in (-1)^(1/4) where its deta = pi)

Answer 8

ooh i see ok lets go slow because you made at least two mistakes above

Answer 9

the calculator says its deta is (3pi/4) but when i tried to solve it i got pi/4 instead (without the 3)

Answer 10

first of all it is not true that \(\theta=\tan^{-1}(\frac{b}{a})\)

Answer 11

\(\tan^{-1}(\theta)\)is always a number between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) but your angle may not be in that range

Answer 12

ahhh...so what should I use?

Answer 13

your example was \(-1+i\) so first you should see where that is

Answer 14

it should now be clear, that even though \(\tan^{-1}(-1)=-\frac{\pi}{4}\) your angle is not \(-\frac{\pi}{4}\) but rather \(\frac{3\pi}{4}\)

Answer 15

you have to know where you are is all
if you are in quadrant II or III then you cannot use \(\tan^{-1}(\frac{b}{a})\) except as a reference

Answer 16

uhm...i dont understand the graph...

Answer 17

\(-1+i\)
left \(1\) unit, up \(1\) unit

Answer 18

just like plotting the point \((-1,1)\)

Answer 19

ah, i understand how you graph it. what i dont understand is how (-1,1) = (3pi)/4

Answer 20

it is not equal to \(\frac{3\pi}{4}\)
the angle it makes with the positive real axis is \(\frac{3\pi}{4}\)

Answer 21

ah, i undestandhow you got (-1,1). what i dont undestand was how (-1,1) = 3pi/4

Answer 22

did you get your answer to this?

Answer 23

there is a straightforward way to answer this.
you want to solve z^4 = -1
then
z^4=1*[cos (pi + 2pi*k) + i sin(0+ 2pi*k)]
then we use demoivre's theorem
z = 1^(1/4)*[ cos (pi/4 +2pi/4*k) + i sin(0 + 2pi/4*k)] , k = 0,1,2,3